Exercices

Exercices sélectionnés

TP1 — Revision on probability

Exercise 1 — Discrete joint distribution

Let us consider two discrete random variables (Y) and (Z) with joint pmf’s (probability mass function) presented in the following table:

	(Z=-1)	(Z=5)
(Y=-4)	(0.1)	(0.5)
(Y=6)	(0.2)	(0.2)

a) Calculate the expectation and variance of (Z).

b) Calculate the expectation and variance of (Y).

c) Calculate the covariance and correlation between (Z) and (Y).

Voir la solution

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Exercise 2 — Continuous distribution, quantiles and affine transformation

Let us consider a continuous random variable (X) with a pdf (probability distribution function) given by

$$f_X(x)= \begin{cases} k(1+x^2), & -1\le x\le 1,\\ 0, & \text{otherwise}. \end{cases}$$

a) Find the value of (k).

b) Calculate the expectation and variance of (X).

c) Find the equation satisfied by the quantiles (q) of (X) for a probability (alpha).

d) Let us define a random variable (Y:=3X+2). What is its pdf (f_Y(y)) and its support?

e) Find the covariance and correlation coefficient between (X) and (Y).

Voir la solution

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Exercise 7 — Quantiles from the statistical tables

Use the statistical tables to retrieve the following information:

a) The $2.5\%$ and $97.5\%$ quantiles of a Student $T$ random variable with $20$ degrees of freedom.

b) The $2.5\%$ and $97.5\%$ quantiles of a chi-square $\chi^2$ random variable with $20$ degrees of freedom.

c) The $2.5\%$ and $97.5\%$ quantiles of a Fisher $F_{n_1,n_2}$ random variable with $n_1=4$ and $n_2=10$.

d) The $2.5\%$ and $97.5\%$ quantiles of a Normal random variable $N(\mu,\sigma^2)$ with $\mu=10$ and $\sigma=3$ (hint: use standardization).

Voir la solution

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Exercise 12 — Normal distribution and inverse probability questions

Let (X) have a Normal distribution with mean (mu=3) and variance (sigma^2=1.25). Calculate using the statistical table:

1. The probability that (X) is at least (2.5).

3. The probability that (X) is at most (4).

6. Now find (a,b,c,d) such that

$$P(X\ge a)=0.674,\qquad P(X\le b)=0.037,\qquad P(X\le c)=0.81,\qquad P(X\ge d)=0.326,$$

respectively. Do you obtain results coherent with the first four questions?

Voir la solution

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Exercise 13 — Marginals, covariance and conditional distribution

Consider the following pdf of $X_1$, $X_2$ and $X_3$:

$$f(x_1,x_2,x_3) = 2x_2(x_1+x_3), \qquad 0<x_1,x_2,x_3<1.$$

1. Find the joint pdf $f_{X_1,X_3}(x_1,x_3)$.

2. Calculate the expectation of $X_1$ and its variance.

3. Calculate the covariance between $X_1$ and $X_2$. Are they independent?

4. Calculate the conditional pdf of $(X_1,X_2)$ conditionally on $X_3=x_3$, and calculate

$$E\!\left( \begin{pmatrix} X_1\\ X_2 \end{pmatrix} \middle|X_3=x_3 \right).$$

Voir la solution

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Exercise 16 — Bivariate Normal distribution

Let

$$X= \begin{pmatrix} X_1\\ X_2 \end{pmatrix}$$

be a bivariate Normal random variable with mean vector and covariance matrix

$$\mu= \begin{pmatrix} 2\\ 3 \end{pmatrix}, \qquad \Sigma= \begin{pmatrix} 4&2\\ 2&3 \end{pmatrix}.$$

1. Find the distribution of

$$Y=3X_1+X_2-3.$$

2. Retrieve the conditional distribution of $X_2$ given $X_1=x_1$.

Voir la solution

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TP2 — Parameters estimation

Selected exercises: 2, 3, 4, 6, 9 and 10. The statements are kept close to the official TP, with clearer notation where the PDF layout is ambiguous.

Exercise 2 — Method of moments, bias and piecewise likelihood

The time spent by a customer at the counter of a bank office is a random variable $X$ with pdf

$$f(x;\theta)= \begin{cases} \dfrac{1}{3\theta}, & x\in[0,\theta],\\[6pt] \dfrac{2}{3(21-\theta)}, & x\in(\theta,21],\\[6pt] 0, & \text{otherwise}, \end{cases} \qquad 0<\theta<21.$$

a) Determine $E(X)$. From a sample of observations $(x_1,\ldots,x_n)$, estimate $\theta$ by the method of moments.

b) Find the bias of the method-of-moments estimator.

c) Write the log-likelihood of the sample. Does the maximum-likelihood estimator admit an analytical closed form?

Voir la solution

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Exercise 3 — Method of moments, MLE and bias

Let $X$ be a random variable with pdf

$$f(x;\alpha)= \begin{cases} (\alpha+1)x^\alpha, & 0\le x\le1,\\ 0, & \text{otherwise}, \end{cases} \qquad \alpha>-1.$$

a) Determine $E(X)$ and $V(X)$, then estimate $\alpha$ by the method of moments.

b) From a sample $(x_1,\ldots,x_n)$, estimate $\alpha$ by maximum likelihood. Does the estimator admit an analytical form?

c) Find the bias of the maximum-likelihood estimator obtained in (b).
Hint: first find the distribution of

$$Y=-\ln X.$$

Voir la solution

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Exercise 4 — MLE for a Normal population

The height of Belgian male citizens is modeled by a Normal random variable

$$X\sim N(\mu,\sigma^2),$$

with unknown mean $\mu$ and unknown standard deviation $\sigma$. Its pdf is

$$f(x;\mu,\sigma) = \frac{1}{\sigma\sqrt{2\pi}} \exp\!\left[ -\frac12\left(\frac{x-\mu}{\sigma}\right)^2 \right].$$

From a random sample $(x_1,\ldots,x_n)$, we wish to estimate $\mu$ and $\sigma^2$ by maximum likelihood.

a) Write the log-likelihood of the sample.

b) Find the estimators $\hat\mu$ and $\hat\sigma^2_{\mathrm{MLE}}$, and calculate their biases.

c) Study the estimator of the mean obtained in (b): determine the exact probability distribution of

$$\hat\mu=\bar X,$$

give its variance, and compute

$$\operatorname{MSE}(\hat\mu) = E\!\left[(\hat\mu-\mu)^2\right].$$

Voir la solution

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Exercise 6 — Gamma estimation and sampling distribution

We consider a continuous random variable $X$ with pdf

$$f(x;\theta) = \frac{1}{\Gamma(2)} \frac{x}{\theta^2} e^{-x/\theta}, \qquad x\ge0,\quad \theta>0,$$

and observe a sample $(x_1,\ldots,x_n)$.

a) Estimate $\theta$ by the method of moments.

b) Write the log-likelihood and estimate $\theta$ by maximum likelihood.

c) Find the probability distribution of the estimator $\hat\theta$, its expectation and variance. Find the mean squared error (MSE) of $\hat\theta$.

d) Use a Normal approximation for $\hat\theta$ and determine an interval in which $\hat\theta$ lies with probability $0.95$. Can this interval be computed in practice?

Voir la solution

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Exercise 9 — Two estimators for the endpoint of a Uniform distribution

Let

$$X_1,\ldots,X_n\overset{iid}{\sim}U(0,\theta).$$

a) Check that two estimators of $\theta$ are

$$\hat\theta = \max(X_1,\ldots,X_n)$$

by maximum-likelihood estimation, and

$$\tilde\theta = 2\bar X$$

by the method of moments.

b) Calculate

$$E(\hat\theta),\qquad E(\tilde\theta),\qquad V(\hat\theta),\qquad V(\tilde\theta).$$

c) Which estimator do you prefer? Justify your answer using the mean squared error (MSE) rather than bias alone.

Voir la solution

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Exercise 10 — Exponential MLE, bias and confidence interval

Let $X_1,\ldots,X_n$ be iid random variables with pdf

$f_X(x;\lambda)=\frac{1}{\lambda}\exp\left(-\frac{x}{\lambda}\right), \qquad x>0,\quad \lambda>0.$

a) Find the maximum-likelihood estimator $\hat\lambda$ of $\lambda$.

b) What are its bias, variance and mean squared error (MSE)?

c) Assume $n=100$. Use a Normal approximation for $\hat\lambda$ and determine a confidence interval for $\lambda$ at level $\alpha=5\%$.

Hint from the statement: use the sample variance instead of $V(X)$.

Voir la solution

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TP3 — Properties of $\bar X$ and $S^2$

Exercise 2 — Confidence intervals for a Normal population

We assume that the height of a Belgian female citizen, denoted by $X$, follows a Normal distribution. The population standard deviation is unknown.

We measure the heights of a sample of $n=100$ UCL students and observe

$\bar x=174\ \mathrm{cm}, \qquad s^2=(11.7\ \mathrm{cm})^2.$

a) Find a confidence interval for $\mu=E(X)$ at level $\alpha=10\%$.

b) Find a confidence interval for $\sigma^2=V(X)$ at level $\alpha=10\%$.

Voir la solution

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Exercise 3 — Two Normal populations with a common variance

The heights of Belgian female and male citizens, denoted respectively by $X_f$ and $X_m$, follow Normal distributions. Their population standard deviations are unknown but assumed equal.

Samples of UCL female and male students give the following observations (the official correction uses $n_f=n_m=100$):

$\begin{aligned} \bar x_f&=173\ \mathrm{cm}, & s_f^2&=(8.7\ \mathrm{cm})^2,\\ \bar x_m&=177\ \mathrm{cm}, & s_m^2&=(9.3\ \mathrm{cm})^2. \end{aligned}$

a) Find a confidence interval for $\mu_m-\mu_f=E(X_m)-E(X_f)$ at level $\alpha=5\%$.

b) Find a confidence interval for the common variance

$\sigma^2=V(X_m)=V(X_f)$

at level $\alpha=5\%$.

Voir la solution

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TP4 — Hypothesis testing

Exercise 1 — Voltmeter: mean and variance

We test a voltmeter suspected of giving distorted measurements. We measure the voltage of a standard battery with nominal voltage 60 V nine times, obtaining:

Measurement	1	2	3	4	5	6	7	8	9
Voltage (V)	61.0	61.3	58.2	61.9	60.3	60.6	59.7	60.8	61.8

The statement gives the observed sample mean and sample variance:

$\bar x=60.62\ \mathrm V,\qquad s^2=1.31\ \mathrm V^2,\qquad n=9.$

a) Select the appropriate test to check whether the voltmeter is distorted. Calculate the p-value.

b) The manufacturer claims that the variance of its measurements exceeds $0.81\ \mathrm V^2$. Test this claim and calculate the p-value.

c) Give a confidence interval for the population mean of the measured voltages with $\alpha=0.05$.

d) Calculate a confidence interval for the population variance of the measurements with $\alpha=0.05$.

Voir la solution

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Exercise 2 — Dust emissions from two boiler types (a–b)

The quantities of dust emitted by 8 boilers of type A and 6 boilers of type B, used in similar conditions, were measured over a certain period (in grams):

Type	Observed dust quantities (g)
A	64, 75, 39, 47, 49, 81, 72, 61
B	31, 57, 63, 40, 54, 49

Determine whether:

a) The variances of dust emissions of the two boiler types differ, using $\alpha=0.02$.

b) The mean dust emission for boiler A is at least 5 g greater than for boiler B, using $\alpha=0.05$. Calculate the p-value.

Only (a) and (b) are selected here; (c) and (d) of the official exercise are not included.

Voir la solution

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Exercise 3 — Two phosphate measurement methods (paired observations)

To determine the quantity of phosphate in an ore, two methods are considered: gravimetric (A) and spectroscopic (B). We suspect that they do not give the same results.

Ten identical ore samples (with the same features) are each measured using both methods. Observed quantities in grams:

Sample	1	2	3	4	5	6	7	8	9	10
A	55.6	62.4	48.9	45.5	75.4	89.6	38.4	96.8	92.5	98.7
B	52.4	66.3	51.2	46.1	74.3	92.5	40.2	97.3	94.8	99.0

The statement assumes that the phosphate concentrations in the samples have a Normal distribution.

a) Propose a test to determine whether the two methods produce different measurements, with $\alpha=0.05$.

b) Calculate the p-value of your test.

Hint from the statement: the two methods are applied to the same samples, so a test for two independent populations is not appropriate.

Voir la solution

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Solutions

Solution — Exercise 1

a) Expectation and variance of (Z)

First obtain the marginal pmf of (Z) by summing the probabilities in each column:

$$P(Z=-1)=0.1+0.2=0.3,$$ $$P(Z=5)=0.5+0.2=0.7.$$

Hence

$$E(Z)=(-1)(0.3)+5(0.7)=3.2.$$

For the variance, first compute the second moment:

$$E(Z^2)=(-1)^2(0.3)+5^2(0.7) =0.3+17.5 =17.8.$$

Therefore

$$V(Z)=E(Z^2)-E(Z)^2 =17.8-3.2^2 =\boxed{7.56}.$$

So

$$\boxed{E(Z)=3.2,\qquad V(Z)=7.56.}$$

b) Expectation and variance of (Y)

The marginal pmf of (Y) is obtained by summing the probabilities in each row:

$$P(Y=-4)=0.1+0.5=0.6,$$ $$P(Y=6)=0.2+0.2=0.4.$$

Thus

$$E(Y)=(-4)(0.6)+6(0.4)=0.$$

The second moment is

$$E(Y^2)=(-4)^2(0.6)+6^2(0.4) =16(0.6)+36(0.4) =24.$$

Since (E(Y)=0),

$$V(Y)=E(Y^2)-E(Y)^2=24.$$

Hence

$$\boxed{E(Y)=0,\qquad V(Y)=24.}$$

c) Covariance and correlation

The key quantity is the cross moment (E(YZ)). For a discrete pair, each possible product (yz) is weighted by its joint probability:

$$\begin{aligned} E(YZ) &=(-4)(-1)(0.1)+(-4)(5)(0.5)\\ &\quad +(6)(-1)(0.2)+(6)(5)(0.2)\\ &=0.4-10-1.2+6\\ &=-4.8. \end{aligned}$$

The covariance is

$$C(Y,Z)=E(YZ)-E(Y)E(Z).$$

Since (E(Y)=0),

$$C(Y,Z)=-4.8-(0)(3.2)=\boxed{-4.8}.$$

Finally,

$$\rho_{Y,Z} = \frac{C(Y,Z)} {\sqrt{V(Y)}\sqrt{V(Z)}} = \frac{-4.8} {\sqrt{24}\sqrt{7.56}} \approx \boxed{-0.3563}.$$

The negative sign means that the linear association is negative; its magnitude is well below (1), so the relationship is not perfectly linear.

↑ Retour à l’exercice 1

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Solution — Exercise 2

a) Determine (k)

A probability density must integrate to (1):

$$\int_{-1}^{1} k(1+x^2)\,dx=1.$$

Therefore

$$k\left[x+\frac{x^3}{3}\right]_{-1}^{1}=1.$$

The integral equals

$$\left(1+\frac13\right)-\left(-1-\frac13\right) =\frac83.$$

Thus

$$k\frac83=1 \qquad\Longrightarrow\qquad \boxed{k=\frac38}.$$

b) Expectation and variance of (X)

The expectation is

$$E(X) = k\int_{-1}^{1}x(1+x^2)\,dx = k\int_{-1}^{1}(x+x^3)\,dx.$$

The integrand is odd and the interval is symmetric around (0), hence

$$\boxed{E(X)=0}.$$

For the variance, use

$$V(X)=E(X^2)-E(X)^2.$$

Since (E(X)=0),

$$V(X)=E(X^2).$$

Now

$$\begin{aligned} E(X^2) &=k\int_{-1}^{1}x^2(1+x^2)\,dx\\ &=k\int_{-1}^{1}(x^2+x^4)\,dx\\ &=\frac38\left(\frac23+\frac25\right)\\ &=\frac38\frac{16}{15}\\ &=\boxed{\frac25}. \end{aligned}$$

Therefore

$$\boxed{E(X)=0,\qquad V(X)=\frac25.}$$

c) Equation for the quantile of order (alpha)

A quantile $q_\alpha$ is defined by

$$P(X\le q_\alpha)=\alpha.$$

Because the support begins at (-1),

$$\int_{-1}^{q_\alpha} k(1+x^2)\,dx=\alpha.$$

Using (k=3/8),

$$\begin{aligned} \int_{-1}^{q_\alpha} \frac38(1+x^2)\,dx &= \frac38 \left[x+\frac{x^3}{3}\right]_{-1}^{q_\alpha}\\ &= \frac18 \left(q_\alpha^3+3q_\alpha+4\right). \end{aligned}$$

Hence the quantile is the solution in ([-1,1]) of

$$\boxed{ \frac18\left(q_\alpha^3+3q_\alpha+4\right)=\alpha. }$$

This is the inverse-CDF idea: the probability $\alpha$ is given, and we solve for the cutoff $q_\alpha$.

d) Distribution of (Y=3X+2)

First transform the support. Since

$$-1\le X\le1,$$

we have

$$-1=3(-1)+2\le Y\le3(1)+2=5.$$

Thus the support of (Y) is ([-1,5]).

For an affine transformation (Y=3X+2),

$$x=\frac{y-2}{3} \qquad\text{and}\qquad \left|\frac{dx}{dy}\right|=\frac13.$$

Therefore

$$f_Y(y) = \frac13 f_X\left(\frac{y-2}{3}\right).$$

For $-1\le y\le5$,

$$\begin{aligned} f_Y(y) &= \frac13\frac38 \left[ 1+ \left(\frac{y-2}{3}\right)^2 \right]\\ &= \frac18 \left[ 1+\frac{(y-2)^2}{9} \right]\\ &= \frac{1}{72}(y^2-4y+13). \end{aligned}$$

Hence

$$\boxed{ f_Y(y)= \begin{cases} \dfrac{1}{72}(y^2-4y+13), & -1\le y\le5,\\ 0, & \text{otherwise}. \end{cases} }$$

e) Covariance and correlation between (X) and (Y)

Because

$$Y=3X+2,$$

(Y) is an exact increasing affine function of (X). Therefore the linear relationship is perfect and positive:

$$\boxed{\rho_{X,Y}=1}.$$

For the covariance,

$$C(X,Y) = C(X,3X+2) = 3C(X,X) = 3V(X).$$

Since (V(X)=2/5),

$$\boxed{ C(X,Y)=3\frac25=\frac65. }$$

The same result follows from the official correction’s route:

$$V(Y)=9V(X)=\frac{18}{5},$$

so

$$C(X,Y) = \rho_{X,Y}\sigma_X\sigma_Y = 1\times \sqrt{\frac25} \sqrt{\frac{18}{5}} = \frac65.$$

↑ Retour à l’exercice 2

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Solution — Exercise 7

A quantile (q_p) is a value satisfying

$$P(X\le q_p)=p.$$

The goal of this exercise is therefore to identify the correct distribution, degrees of freedom and probability level before reading the statistical tables.

a) Student (T) with 20 degrees of freedom

Let (t_{20,p}) denote the (p)-quantile of a Student distribution with (20) degrees of freedom.

From the table,

$$t_{20,0.975}=2.086.$$

The Student distribution is symmetric around (0), hence

$$t_{20,0.025}=-t_{20,0.975}=-2.086.$$

Therefore

$$\boxed{ t_{20,0.025}=-2.086, \qquad t_{20,0.975}=2.086. }$$

b) Chi-square with 20 degrees of freedom

The chi-square distribution is not symmetric, so the two quantiles are read separately from the table:

$$\boxed{ \chi^2_{20,0.025}=9.59083, \qquad \chi^2_{20,0.975}=34.1696. }$$

There is no sign-symmetry shortcut here.

c) Fisher distribution (F_{4,10})

For the upper quantile, the table directly gives

$$\boxed{F_{4,10;0.975}=4.47}.$$

For the lower $2.5\%$ quantile, let

$$F\sim F_{4,10}$$

and seek (q) such that

$$P(F\le q)=0.025.$$

Use the reciprocal property

$$\frac1F\sim F_{10,4}.$$

Then

$$P(F\le q)=0.025$$

is equivalent to

$$P\left(\frac1F\ge\frac1q\right)=0.025.$$

Since the distribution is continuous,

$$P\left(\frac1F\le\frac1q\right)=0.975.$$

Thus $1/q$ is the $97.5\%$ quantile of $F_{10,4}$. From the table,

$$F_{10,4;0.975}=8.84.$$

Therefore

$$\frac1q=8.84 \qquad\Longrightarrow\qquad q=\frac1{8.84}\approx0.1131.$$

Hence

$$\boxed{ F_{4,10;0.025}\approx0.1131, \qquad F_{4,10;0.975}=4.47. }$$

d) Normal distribution $X\sim N(10,3^2)$

Standardize:

$$Z=\frac{X-10}{3}\sim N(0,1).$$

Equivalently,

$$X=10+3Z.$$

For the standard Normal distribution,

$$z_{0.025}=-1.96, \qquad z_{0.975}=1.96.$$

Therefore

$$x_{0.025} = 10+3(-1.96) = \boxed{4.12},$$

and

$$x_{0.975} = 10+3(1.96) = \boxed{15.88}.$$

The important reflex is:

$$\boxed{x_p=\mu+\sigma z_p}.$$

↑ Retour à l’exercice 7

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Solution — Exercise 12

We have

$$X\sim N(3,1.25),$$

so

$$\mu=3, \qquad \sigma=\sqrt{1.25}.$$

The standardized variable is

$$Z=\frac{X-3}{\sqrt{1.25}}\sim N(0,1).$$

For the official 2026–2027 Normal table used here, read probabilities in the form

$$P(Z<z).$$

1. Probability that (X) is at least (2.5)

We want

$$P(X\ge2.5).$$

Standardize:

$$\begin{aligned} P(X\ge2.5) &= P\left( Z\ge \frac{2.5-3}{\sqrt{1.25}} \right)\\ &\approx P(Z\ge-0.45). \end{aligned}$$

By symmetry of the standard Normal distribution,

$$P(Z\ge-0.45)=P(Z\le0.45).$$

The table gives

$$P(Z\le0.45)=0.6736.$$

Hence

$$\boxed{P(X\ge2.5)\approx0.6736}.$$

A useful check: (2.5) is below the mean (3), so a probability larger than (0.5) is coherent.

3. Probability that (X) is at most (4)

We want

$$P(X\le4).$$

Standardization gives

$$\begin{aligned} P(X\le4) &= P\left( Z\le \frac{4-3}{\sqrt{1.25}} \right)\\ &\approx P(Z\le0.89). \end{aligned}$$

From the table,

$$P(Z\le0.89)=0.8133.$$

Therefore

$$\boxed{P(X\le4)\approx0.8133}.$$

This is equivalent to the official correction’s complementary form

$$1-P(Z\ge0.89)=1-0.1867=0.8133.$$

6. Find (a,b,c,d)

This time the probability is known and the cutoff value is unknown: these are inverse probability, or quantile, questions.

Find $a$ such that $P(X\ge a)=0.674$

Standardize:

$$P\left( Z\ge \frac{a-3}{\sqrt{1.25}} \right)=0.674.$$

Equivalently, the left-tail probability is

$$P\left( Z\le \frac{a-3}{\sqrt{1.25}} \right)=0.326.$$

From the table,

$$P(Z\le-0.45)\approx0.326.$$

Thus

$$\frac{a-3}{\sqrt{1.25}}=-0.45,$$

so

$$a = 3-0.45\sqrt{1.25} \approx \boxed{2.496}.$$

Find $b$ such that $P(X\le b)=0.037$

We need

$$P\left( Z\le \frac{b-3}{\sqrt{1.25}} \right)=0.037.$$

The corresponding standard Normal quantile is approximately

$$z=-1.79.$$

Hence

$$\frac{b-3}{\sqrt{1.25}}=-1.79,$$

and

$$b = 3-1.79\sqrt{1.25} \approx \boxed{1.00}.$$

Find $c$ such that $P(X\le c)=0.81$

We need

$$P\left( Z\le \frac{c-3}{\sqrt{1.25}} \right)=0.81.$$

From the table,

$$P(Z\le0.87)\approx0.81.$$

Therefore

$$\frac{c-3}{\sqrt{1.25}}=0.87,$$

so

$$c = 3+0.87\sqrt{1.25} \approx \boxed{3.97}.$$

Find $d$ such that $P(X\ge d)=0.326$

Convert to a left-tail probability:

$$P(X\le d)=1-0.326=0.674.$$

Hence

$$P\left( Z\le \frac{d-3}{\sqrt{1.25}} \right)=0.674.$$

From the table,

$$P(Z\le0.45)\approx0.674.$$

Thus

$$\frac{d-3}{\sqrt{1.25}}=0.45,$$

and

$$d = 3+0.45\sqrt{1.25} \approx \boxed{3.50}.$$

Finally,

$$\boxed{ a\approx2.50, \qquad b\approx1.00, \qquad c\approx3.97, \qquad d\approx3.50. }$$

These values are coherent with the first four questions: question 6 simply reverses the probability calculations performed there.

↑ Retour à l’exercice 12

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Solution — Exercise 13

The joint density is

$$f(x_1,x_2,x_3)=2x_2(x_1+x_3), \qquad 0<x_1,x_2,x_3<1.$$

1. Joint pdf of $(X_1,X_3)$

To keep only $X_1$ and $X_3$, integrate out $X_2$:

$$\begin{aligned} f_{X_1,X_3}(x_1,x_3) &= \int_0^1 2x_2(x_1+x_3)\,dx_2\\ &= (x_1+x_3) \int_0^1 2x_2\,dx_2\\ &= x_1+x_3. \end{aligned}$$

Hence

$$\boxed{ f_{X_1,X_3}(x_1,x_3) = x_1+x_3, \qquad 0<x_1,x_3<1. }$$

The general reflex is: to obtain a marginal density, integrate the joint density over the variables that are no longer of interest.

2. Expectation and variance of $X_1$

First obtain the marginal density of $X_1$ by integrating out $X_3$:

$$\begin{aligned} f_{X_1}(x_1) &= \int_0^1 f_{X_1,X_3}(x_1,x_3)\,dx_3\\ &= \int_0^1(x_1+x_3)\,dx_3\\ &= x_1+\frac12, \qquad 0<x_1<1. \end{aligned}$$

Then

$$\begin{aligned} E(X_1) &= \int_0^1 x_1 f_{X_1}(x_1)\,dx_1\\ &= \int_0^1 x_1\left(x_1+\frac12\right)dx_1\\ &= \frac13+\frac14 = \boxed{\frac7{12}}. \end{aligned}$$

For the variance, first compute the second moment:

$$\begin{aligned} E(X_1^2) &= \int_0^1 x_1^2\left(x_1+\frac12\right)dx_1\\ &= \frac14+\frac16 = \frac5{12}. \end{aligned}$$

Therefore

$$\begin{aligned} \operatorname{Var}(X_1) &= E(X_1^2)-E(X_1)^2\\ &= \frac5{12} - \left(\frac7{12}\right)^2\\ &= \boxed{\frac{11}{144}}. \end{aligned}$$

3. Covariance and independence of $X_1$ and $X_2$

First obtain their joint density by integrating out $X_3$:

$$\begin{aligned} f_{X_1,X_2}(x_1,x_2) &= \int_0^1 2x_2(x_1+x_3)\,dx_3\\ &= 2x_2\left(x_1+\frac12\right)\\ &= x_2(2x_1+1). \end{aligned}$$

We already know

$$f_{X_1}(x_1)=x_1+\frac12.$$

The marginal density of $X_2$ is

$$\begin{aligned} f_{X_2}(x_2) &= \int_0^1 x_2(2x_1+1)\,dx_1\\ &= 2x_2, \qquad 0<x_2<1. \end{aligned}$$

Now compute the moments needed for the covariance:

$$E(X_2) = \int_0^1 x_2(2x_2)\,dx_2 = \frac23,$$

and

$$\begin{aligned} E(X_1X_2) &= \int_0^1\int_0^1 x_1x_2\,f_{X_1,X_2}(x_1,x_2) \,dx_1dx_2\\ &= \int_0^1\int_0^1 x_1x_2^2(2x_1+1) \,dx_1dx_2\\ &= \frac7{18}. \end{aligned}$$

Hence

$$\begin{aligned} \operatorname{Cov}(X_1,X_2) &= E(X_1X_2)-E(X_1)E(X_2)\\ &= \frac7{18} - \frac7{12}\frac23\\ &= \boxed{0}. \end{aligned}$$

A zero covariance alone would not prove independence in general. We must check the densities:

$$\begin{aligned} f_{X_1}(x_1)f_{X_2}(x_2) &= \left(x_1+\frac12\right)(2x_2)\\ &= x_2(2x_1+1)\\ &= f_{X_1,X_2}(x_1,x_2). \end{aligned}$$

Therefore

$$\boxed{X_1\perp X_2}.$$

So here the covariance is zero and the variables are independent.

4. Conditional pdf given $X_3=x_3$

The conditional density is

$$f_{X_1,X_2\mid X_3}(x_1,x_2\mid x_3) = \frac{ f_{X_1,X_2,X_3}(x_1,x_2,x_3) }{ f_{X_3}(x_3) }.$$

First find the marginal density of $X_3$:

$$\begin{aligned} f_{X_3}(x_3) &= \int_0^1\int_0^1 2x_2(x_1+x_3)\,dx_1dx_2\\ &= x_3+\frac12. \end{aligned}$$

Hence, for $0<x_1,x_2,x_3<1$,

$$\begin{aligned} f_{X_1,X_2\mid X_3}(x_1,x_2\mid x_3) &= \frac{2x_2(x_1+x_3)} {x_3+\frac12}\\ &= \boxed{ \frac{4x_2(x_1+x_3)} {1+2x_3} }. \end{aligned}$$

Now calculate each component of the conditional expectation vector separately.

For $X_1$,

$$\begin{aligned} E(X_1\mid X_3=x_3) &= \int_0^1\int_0^1 x_1 \frac{4x_2(x_1+x_3)} {1+2x_3} \,dx_1dx_2\\ &= \boxed{ \frac{2+3x_3} {3(1+2x_3)} }. \end{aligned}$$

For $X_2$,

$$\begin{aligned} E(X_2\mid X_3=x_3) &= \int_0^1\int_0^1 x_2 \frac{4x_2(x_1+x_3)} {1+2x_3} \,dx_1dx_2\\ &= \boxed{\frac23}. \end{aligned}$$

Therefore

$$\boxed{ E\!\left( \begin{pmatrix} X_1\\ X_2 \end{pmatrix} \middle|X_3=x_3 \right) = \begin{pmatrix} \dfrac{2+3x_3}{3(1+2x_3)}\\[8pt] \dfrac23 \end{pmatrix}. }$$

↑ Retour à l’exercice 13

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Solution — Exercise 16

We have

$$X= \begin{pmatrix} X_1\\ X_2 \end{pmatrix} \sim N_2\!\left( \begin{pmatrix} 2\\ 3 \end{pmatrix}, \begin{pmatrix} 4&2\\ 2&3 \end{pmatrix} \right).$$

Thus

$$E(X_1)=2, \qquad E(X_2)=3,$$

and

$$\operatorname{Var}(X_1)=4, \qquad \operatorname{Var}(X_2)=3, \qquad \operatorname{Cov}(X_1,X_2)=2.$$

1. Distribution of $Y=3X_1+X_2-3$

A linear combination of a multivariate Normal vector is again Normal.

Write

$$Y=a^\top X-3, \qquad a= \begin{pmatrix} 3\\ 1 \end{pmatrix}.$$

Its expectation is

$$\begin{aligned} E(Y) &= 3E(X_1)+E(X_2)-3\\ &= 3(2)+3-3\\ &= \boxed{6}. \end{aligned}$$

Its variance is

$$\begin{aligned} \operatorname{Var}(Y) &= 3^2\operatorname{Var}(X_1) + \operatorname{Var}(X_2) + 2(3)(1)\operatorname{Cov}(X_1,X_2)\\ &= 9(4)+3+6(2)\\ &= \boxed{51}. \end{aligned}$$

Equivalently, in matrix form,

$$\operatorname{Var}(Y) = a^\top\Sigma a = \begin{pmatrix} 3&1 \end{pmatrix} \begin{pmatrix} 4&2\\ 2&3 \end{pmatrix} \begin{pmatrix} 3\\ 1 \end{pmatrix} = 51.$$

Therefore

$$\boxed{ Y\sim N(6,51). }$$

2. Conditional distribution of $X_2$ given $X_1=x_1$

For a bivariate Normal vector, the conditional distribution is again Normal.

Here we condition on $X_1$, so identify

$$\mu_1=2, \qquad \mu_2=3, \qquad \sigma_{11}=4, \qquad \sigma_{22}=3, \qquad \sigma_{12}=\sigma_{21}=2.$$

The conditional mean is

$$\begin{aligned} E(X_2\mid X_1=x_1) &= \mu_2 + \frac{\sigma_{21}}{\sigma_{11}} (x_1-\mu_1)\\ &= 3+\frac24(x_1-2)\\ &= \boxed{ 3+\frac12(x_1-2) }. \end{aligned}$$

The conditional variance is

$$\begin{aligned} \operatorname{Var}(X_2\mid X_1=x_1) &= \sigma_{22} - \frac{\sigma_{21}\sigma_{12}} {\sigma_{11}}\\ &= 3-\frac{2\cdot2}{4}\\ &= \boxed{2}. \end{aligned}$$

Hence

$$\boxed{ X_2\mid X_1=x_1 \sim N\!\left( 3+\frac12(x_1-2), \;2 \right). }$$

A useful way to reconstruct the formula is:

$$\text{conditional mean} = \text{old mean} + \text{link with the known variable} \times \text{observed deviation},$$

while

$$\text{conditional variance} = \text{old variance} - \text{uncertainty explained by the known variable}.$$

↑ Retour à l’exercice 16

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Solutions — TP2

Solution — TP2 Exercise 2

a) Expectation and method of moments

The unknown quantity is the model parameter $\theta$. The observed information is the sample $X_1,\ldots,X_n$.

Because the density has two pieces,

$$E(X) = \int_0^\theta x\frac{1}{3\theta}\,dx + \int_\theta^{21}x\frac{2}{3(21-\theta)}\,dx.$$

The first term is

$$\frac{1}{3\theta} \left[\frac{x^2}{2}\right]_0^\theta = \frac{\theta}{6}.$$

For the second term,

$$\frac{2}{3(21-\theta)} \left[\frac{x^2}{2}\right]_\theta^{21} = \frac{441-\theta^2}{3(21-\theta)} = \frac{21+\theta}{3}.$$

Hence

$$\boxed{ E(X)=7+\frac{\theta}{2} }.$$

With one unknown parameter, the first-moment equation is

$$E(X)=\bar X.$$

Therefore

$$7+\frac{\hat\theta_{\mathrm{MM}}}{2} = \bar X,$$

so

$$\boxed{ \hat\theta_{\mathrm{MM}} = 2(\bar X-7) }.$$

b) Bias

By definition,

$$B(\hat\theta_{\mathrm{MM}}) = E(\hat\theta_{\mathrm{MM}})-\theta.$$

Using $E(\bar X)=E(X)$,

$$\begin{aligned} E(\hat\theta_{\mathrm{MM}}) &= 2(E(\bar X)-7)\\ &= 2\left(7+\frac{\theta}{2}-7\right)\\ &= \theta. \end{aligned}$$

Thus

$$\boxed{ B(\hat\theta_{\mathrm{MM}})=0 }.$$

The method-of-moments estimator is unbiased.

c) Likelihood and why there is no simple closed form

Let

$$n_1(\theta) = \#\{i:x_i\in[0,\theta]\},$$

and

$$n_2(\theta) = \#\{i:x_i\in(\theta,21]\}.$$

Then

$$n_1(\theta)+n_2(\theta)=n.$$

The likelihood is

$$\boxed{ L(\theta) = \left(\frac{1}{3\theta}\right)^{n_1(\theta)} \left(\frac{2}{3(21-\theta)}\right)^{n_2(\theta)} }.$$

Hence

$$\boxed{ \ell(\theta) = -n_1(\theta)\ln(3\theta) - n_2(\theta) \ln\!\left(\frac32(21-\theta)\right) }.$$

The important point is that $n_1(\theta)$ and $n_2(\theta)$ change whenever $\theta$ crosses one of the observations $x_i$. Therefore the log-likelihood is not one smooth expression over the whole parameter domain.

So the usual shortcut

$$\ell'(\theta)=0$$

does not produce a single analytical closed-form estimator. The maximum can nevertheless be found numerically.

Réflexe de l'exercice 2

A maximum-likelihood problem is not automatically “differentiate and set equal to zero”. First check whether the support or the branch of the density depends on the unknown parameter.

↑ Retour à l’exercice 2

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Solution — TP2 Exercise 3

a) Moments and method of moments

The first moment is

$$E(X) = \int_0^1 x(\alpha+1)x^\alpha\,dx = (\alpha+1)\int_0^1x^{\alpha+1}\,dx = \boxed{ \frac{\alpha+1}{\alpha+2} }.$$

The second moment is

$$E(X^2) = (\alpha+1)\int_0^1x^{\alpha+2}\,dx = \boxed{ \frac{\alpha+1}{\alpha+3} }.$$

Thus

$$\boxed{ V(X) = \frac{\alpha+1}{\alpha+3} - \left(\frac{\alpha+1}{\alpha+2}\right)^2 }.$$

For the method of moments,

$$\frac{\hat\alpha_{\mathrm{MM}}+1} {\hat\alpha_{\mathrm{MM}}+2} = \bar X.$$

Solving for $\hat\alpha_{\mathrm{MM}}$,

$$\hat\alpha_{\mathrm{MM}}(1-\bar X) = 2\bar X-1,$$

so

$$\boxed{ \hat\alpha_{\mathrm{MM}} = \frac{2\bar X-1}{1-\bar X} }.$$

b) Maximum likelihood

For an iid sample,

$$L(\alpha) = \prod_{i=1}^n(\alpha+1)x_i^\alpha = (\alpha+1)^n\prod_{i=1}^n x_i^\alpha.$$

Therefore

$$\boxed{ \ell(\alpha) = n\ln(\alpha+1) + \alpha\sum_{i=1}^n\ln x_i }.$$

Differentiate:

$$\ell'(\alpha) = \frac{n}{\alpha+1} + \sum_{i=1}^n\ln x_i.$$

Set the derivative equal to zero:

$$\frac{n}{\hat\alpha+1} = -\sum_{i=1}^n\ln x_i.$$

Hence

$$\boxed{ \hat\alpha_{\mathrm{MLE}} = -1- \frac{n}{\sum_{i=1}^n\ln X_i} }.$$

Moreover,

$$\ell''(\alpha) = -\frac{n}{(\alpha+1)^2}<0,$$

so the stationary point is indeed a maximum.

c) Bias of the MLE

We want

$$B(\hat\alpha_{\mathrm{MLE}}) = E(\hat\alpha_{\mathrm{MLE}})-\alpha.$$

The logarithms in the estimator suggest the transformation

$$Y=-\ln X.$$

For $y\ge0$,

$$\begin{aligned} F_Y(y) &= P(Y\le y)\\ &= P(-\ln X\le y)\\ &= P(X\ge e^{-y})\\ &= \int_{e^{-y}}^1(\alpha+1)x^\alpha\,dx\\ &= 1-e^{-(\alpha+1)y}. \end{aligned}$$

Therefore

$$f_Y(y) = (\alpha+1)e^{-(\alpha+1)y}, \qquad y\ge0.$$

Using the scale parametrization of the course,

$$\boxed{ Y\sim Expo\left(\beta=\frac{1}{\alpha+1}\right) }.$$

Now define

$$U = \sum_{i=1}^nY_i = -\sum_{i=1}^n\ln X_i.$$

A sum of $n$ iid exponentials with scale $\beta$ is Gamma distributed:

$$\boxed{ U \sim Gamma\left(n,\frac{1}{\alpha+1}\right) }.$$

The MLE becomes

$$\hat\alpha_{\mathrm{MLE}} = -1+\frac{n}{U}.$$

For $U\sim Gamma(n,\beta)$, $n>1$,

$$E\left(\frac1U\right) = \frac{1}{(n-1)\beta}.$$

Since

$$\beta=\frac1{\alpha+1},$$

we get

$$E\left(\frac1U\right) = \frac{\alpha+1}{n-1}.$$

Hence

$$\begin{aligned} E(\hat\alpha_{\mathrm{MLE}}) &= -1+ n\frac{\alpha+1}{n-1}\\ &= \frac{n\alpha+1}{n-1}. \end{aligned}$$

Finally,

$$\boxed{ B(\hat\alpha_{\mathrm{MLE}}) = \frac{\alpha+1}{n-1} }.$$

The estimator is biased for finite $n$, but

$$B(\hat\alpha_{\mathrm{MLE}}) \longrightarrow0 \qquad\text{when }n\to\infty.$$

So it is asymptotically unbiased.

Réflexe de l'exercice 3

The method of moments uses a theoretical moment such as $E(X)$. MLE uses the full density through the likelihood. The transformation $Y=-\ln X$ is introduced only because it turns the awkward sum of logarithms in the MLE into a Gamma-distributed sum whose expectation can be handled.

↑ Retour à l’exercice 3

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Solution — TP2 Exercise 4

a) Log-likelihood

For one observation,

$$f(x_i;\mu,\sigma) = \frac{1}{\sigma\sqrt{2\pi}} \exp\!\left( -\frac{(x_i-\mu)^2}{2\sigma^2} \right).$$

By independence,

$$L(\mu,\sigma^2) = \left(\frac{1}{\sigma\sqrt{2\pi}}\right)^n \exp\!\left[ -\frac{1}{2\sigma^2} \sum_{i=1}^n(x_i-\mu)^2 \right].$$

Thus

$$\boxed{ \ell(\mu,\sigma^2) = -\frac n2\ln(2\pi\sigma^2) - \frac{1}{2\sigma^2} \sum_{i=1}^n(x_i-\mu)^2 }.$$

b) MLEs and their biases

Differentiate with respect to $\mu$:

$$\frac{\partial\ell}{\partial\mu} = \frac{1}{\sigma^2} \sum_{i=1}^n(x_i-\mu).$$

Setting this equal to zero gives

$$\boxed{ \hat\mu_{\mathrm{MLE}} = \bar X }.$$

Since

$$E(\bar X)=\mu,$$

we have

$$\boxed{ B(\hat\mu)=0 }.$$

Now differentiate with respect to $\sigma^2$:

$$\frac{\partial\ell}{\partial\sigma^2} = -\frac{n}{2\sigma^2} + \frac{1}{2\sigma^4} \sum_{i=1}^n(x_i-\mu)^2.$$

After substituting $\hat\mu=\bar X$, the solution is

$$\boxed{ \hat\sigma^2_{\mathrm{MLE}} = \frac1n \sum_{i=1}^n(X_i-\bar X)^2 }.$$

Compare this with the unbiased sample variance

$$S^2 = \frac1{n-1} \sum_{i=1}^n(X_i-\bar X)^2.$$

Therefore

$$\hat\sigma^2_{\mathrm{MLE}} = \frac{n-1}{n}S^2.$$

Because $E(S^2)=\sigma^2$,

$$E(\hat\sigma^2_{\mathrm{MLE}}) = \frac{n-1}{n}\sigma^2.$$

Hence

$$\boxed{ B(\hat\sigma^2_{\mathrm{MLE}}) = -\frac{\sigma^2}{n} }.$$

This explains why the usual unbiased sample variance uses $1/(n-1)$ rather than the MLE factor $1/n$.

c) Distribution, variance and MSE of $\hat\mu$

This part concerns the estimator of the mean, not the estimator of the variance.

Since

$$X_1,\ldots,X_n \overset{iid}{\sim} N(\mu,\sigma^2),$$

their average is exactly Normal:

$$\boxed{ \hat\mu=\bar X \sim N\left(\mu,\frac{\sigma^2}{n}\right) }.$$

Therefore

$$\boxed{ V(\hat\mu) = \frac{\sigma^2}{n} }.$$

From part (b),

$$B(\hat\mu)=0.$$

Using

$$\operatorname{MSE}(\hat\mu) = V(\hat\mu)+B(\hat\mu)^2,$$

we obtain

$$\boxed{ \operatorname{MSE}(\hat\mu) = \frac{\sigma^2}{n} }.$$

The $1/n$ factor quantifies the gain from increasing the sample size: the sampling variability of $\bar X$ decreases as $n$ grows.

Réflexe de l'exercice 4

For a Normal population with both parameters unknown,

$$\hat\mu_{\mathrm{MLE}}=\bar X$$

is unbiased, whereas

$$\hat\sigma^2_{\mathrm{MLE}} = \frac1n\sum_i(X_i-\bar X)^2$$

is biased. Do not confuse this MLE with the unbiased sample variance $S^2$, which uses $1/(n-1)$.

↑ Retour à l’exercice 4

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Solution — TP2 Exercise 6

The density is that of a Gamma random variable with shape $2$ and scale $\theta$:

$$X_i\sim Gamma(2,\theta).$$

For this parametrization,

$$E(X_i)=2\theta, \qquad V(X_i)=2\theta^2.$$

a) Method of moments

With one unknown parameter, match the first theoretical moment with the sample mean:

$$E(X)=\bar X.$$

Since

$$E(X)=2\theta,$$

we obtain

$$2\hat\theta_{\mathrm{MM}}=\bar X,$$

hence

$$\boxed{ \hat\theta_{\mathrm{MM}} = \frac{\bar X}{2} }.$$

b) Maximum likelihood

For an iid sample,

$$L(\theta) = \prod_{i=1}^n \left[ \frac{1}{\Gamma(2)} \frac{x_i}{\theta^2} e^{-x_i/\theta} \right].$$

Equivalently,

$$L(\theta) = \left(\frac{1}{\Gamma(2)}\right)^n \left(\prod_{i=1}^n x_i\right) \theta^{-2n} \exp\!\left( -\frac{1}{\theta}\sum_{i=1}^n x_i \right).$$

Therefore,

$$\ell(\theta) = -n\ln\Gamma(2) + \sum_{i=1}^n\ln x_i - 2n\ln\theta - \frac{1}{\theta}\sum_{i=1}^n x_i.$$

Differentiate with respect to $\theta$:

$$\ell'(\theta) = -\frac{2n}{\theta} + \frac{1}{\theta^2} \sum_{i=1}^n x_i.$$

Setting $\ell'(\theta)=0$ gives

$$-2n\theta+\sum_{i=1}^n x_i=0,$$

so

$$\boxed{ \hat\theta_{\mathrm{MLE}} = \frac{1}{2n}\sum_{i=1}^n X_i = \frac{\bar X}{2} }.$$

Thus,

$$\boxed{ \hat\theta_{\mathrm{MM}} = \hat\theta_{\mathrm{MLE}} }.$$

c) Distribution, expectation, variance and MSE

Because the $X_i$ are independent and

$$X_i\sim Gamma(2,\theta),$$

their sum satisfies

$$\boxed{ \sum_{i=1}^nX_i \sim Gamma(2n,\theta) }.$$

Since

$$\hat\theta = \frac{1}{2n}\sum_{i=1}^nX_i,$$

the estimator has distribution

$$\boxed{ \hat\theta \sim Gamma\left(2n,\frac{\theta}{2n}\right) }.$$

Also,

$$E\left(\sum_{i=1}^nX_i\right) = 2n\theta, \qquad V\left(\sum_{i=1}^nX_i\right) = 2n\theta^2.$$

Therefore,

$$E(\hat\theta) = \frac{1}{2n}(2n\theta) = \boxed{\theta},$$

so $\hat\theta$ is unbiased, and

$$V(\hat\theta) = \frac{1}{(2n)^2}(2n\theta^2) = \boxed{ \frac{\theta^2}{2n} }.$$

Finally,

$$\operatorname{MSE}(\hat\theta) = V(\hat\theta)+Bias(\hat\theta)^2.$$

Since the bias is zero,

$$\boxed{ \operatorname{MSE}(\hat\theta) = \frac{\theta^2}{2n} }.$$

d) Normal approximation and the central 95% interval

For sufficiently large $n$,

$$\hat\theta \approx N\left( \theta, \frac{\theta^2}{2n} \right).$$

Hence

$$Z = \frac{\hat\theta-\theta} {\theta/\sqrt{2n}} \approx N(0,1).$$

The central $95\%$ of a standard Normal distribution satisfies

$$P(-1.96\le Z\le1.96) \approx 0.95.$$

Thus,

$$P\left( -1.96 \le \frac{\hat\theta-\theta}{\theta/\sqrt{2n}} \le 1.96 \right) \approx 0.95,$$

which gives

$$\boxed{ P\left( \theta- 1.96\frac{\theta}{\sqrt{2n}} \le \hat\theta \le \theta+ 1.96\frac{\theta}{\sqrt{2n}} \right) \approx 0.95 }.$$

Therefore the interval is

$$\boxed{ \left[ \theta- 1.96\frac{\theta}{\sqrt{2n}}, \; \theta+ 1.96\frac{\theta}{\sqrt{2n}} \right] }.$$

This interval cannot be computed numerically in practice, because it still contains the unknown parameter $\theta$.

Réflexe de l'exercice 6

$$X_i\sim Gamma(2,\theta) \Rightarrow \hat\theta=\frac{\bar X}{2}$$

Here MM and MLE give the same estimator. Then study $\hat\theta$ through the distribution of $\sum_i X_i$. For part (d), $95\%$ corresponds to the central Normal interval $[-1.96,1.96]$.

↑ Retour à l’exercice 6

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Solution — TP2 Exercise 9

a) MLE and method-of-moments estimators

For

$$X\sim U(0,\theta),$$

the density is

$$f(x;\theta) = \begin{cases} \dfrac1\theta, & 0\le x\le\theta,\\ 0, & \text{otherwise}. \end{cases}$$

For an observed sample,

$$L(\theta) = \begin{cases} \dfrac1{\theta^n}, & \theta\ge \max(x_1,\ldots,x_n),\\[6pt] 0, & \theta<\max(x_1,\ldots,x_n). \end{cases}$$

The likelihood is zero if $\theta$ is smaller than an observed value. On the admissible region, $1/\theta^n$ decreases as $\theta$ increases. Therefore the maximum occurs at the smallest admissible value:

$$\boxed{ \hat\theta_{\mathrm{MLE}} = \max(X_1,\ldots,X_n) }.$$

This is an example where the MLE occurs at a boundary, not at a point satisfying $\ell'(\theta)=0$.

For the method of moments,

$$E(X)=\frac{\theta}{2}.$$

Matching $E(X)$ with $\bar X$ gives

$$\boxed{ \tilde\theta_{\mathrm{MM}} = 2\bar X }.$$

b) Expectations and variances

Let

$$Y=\max(X_1,\ldots,X_n).$$

For $0\le y\le\theta$,

$$\begin{aligned} P(Y\le y) &= P(X_1\le y,\ldots,X_n\le y)\\ &= \prod_{i=1}^nP(X_i\le y)\\ &= \left(\frac y\theta\right)^n. \end{aligned}$$

Thus

$$f_Y(y) = \frac{n y^{n-1}}{\theta^n}, \qquad 0\le y\le\theta.$$

This yields

$$\boxed{ E(\hat\theta) = \frac{n}{n+1}\theta }$$

and

$$\boxed{ V(\hat\theta) = \frac{n}{(n+2)(n+1)^2}\theta^2 }.$$

Therefore

$$\boxed{ B(\hat\theta) = -\frac{\theta}{n+1} }.$$

For the method-of-moments estimator,

$$\tilde\theta=2\bar X.$$

Since a Uniform $U(0,\theta)$ variable satisfies

$$E(X)=\frac{\theta}{2}, \qquad V(X)=\frac{\theta^2}{12},$$

we obtain

$$\boxed{ E(\tilde\theta)=\theta }$$

and

$$\boxed{ V(\tilde\theta) = 4V(\bar X) = \frac{\theta^2}{3n} }.$$

Thus $\tilde\theta$ is unbiased.

c) Which estimator is preferable?

Bias alone is not enough. Compare

$$\operatorname{MSE}(\hat\theta) = V(\hat\theta)+B(\hat\theta)^2$$

with

$$\operatorname{MSE}(\tilde\theta) = V(\tilde\theta),$$

because $\tilde\theta$ is unbiased.

For the method-of-moments estimator,

$$\boxed{ \operatorname{MSE}(\tilde\theta) = \frac{\theta^2}{3n} }.$$

For the MLE,

$$\boxed{ \operatorname{MSE}(\hat\theta) = \frac{2\theta^2}{(n+1)(n+2)} }.$$

The official correction therefore prefers the maximum-likelihood estimator: despite its bias, its variance is small enough to give the better MSE comparison. More precisely, the two MSEs are equal for $n=1$ and $n=2$, and the MLE has the smaller MSE for integer $n\ge3$.

Réflexe de l'exercice 9

An unbiased estimator is not automatically the best estimator.

$$\boxed{ \text{compare bias + variance through the MSE} }$$

This exercise also shows that an MLE may lie on the boundary of the admissible parameter set.

↑ Retour à l’exercice 9

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Solution — TP2 Exercise 10

a) Maximum likelihood estimator

The observations are iid and follow an Exponential distribution parameterized by its scale $\lambda$, so

$$E(X_i)=\lambda,\qquad V(X_i)=\lambda^2.$$

For an observed sample $(x_1,\ldots,x_n)$, the likelihood is the product of densities:

$$\begin{aligned} L(\lambda) &=\prod_{i=1}^{n}\frac1\lambda e^{-x_i/\lambda}\\ &=\lambda^{-n}\exp\left(-\frac1\lambda\sum_{i=1}^n x_i\right). \end{aligned}$$

Taking the logarithm and differentiating with respect to $\lambda$:

$$\ell(\lambda)=-n\ln\lambda-\frac1\lambda\sum_{i=1}^n x_i,$$ $$\ell'(\lambda)=-\frac n\lambda+\frac{\sum_i x_i}{\lambda^2}.$$

Solve $\ell'(\lambda)=0$:

$$-n\lambda+\sum_{i=1}^n x_i=0.$$

This stationary point is a maximum, so the estimator (before observation) and observed estimate are respectively

$$\boxed{\hat\lambda_{\mathrm{MLE}}=\bar X=\frac1n\sum_{i=1}^nX_i}, \qquad \hat\lambda_{\mathrm{obs}}=\bar x.$$

b) Bias, variance and MSE

The bias evaluates whether the estimator is systematically shifted away from the unknown parameter:

$$\begin{aligned} \operatorname{Bias}(\hat\lambda) &=E(\hat\lambda)-\lambda\\ &=\frac1n\sum_{i=1}^nE(X_i)-\lambda\\ &=\lambda-\lambda=0. \end{aligned}$$

Thus $\hat\lambda$ is unbiased. Since the observations are independent,

$$\begin{aligned} V(\hat\lambda) &=V\left(\frac1n\sum_{i=1}^nX_i\right)\\ &=\frac1{n^2}\sum_{i=1}^nV(X_i)\\ &=\boxed{\frac{\lambda^2}{n}}. \end{aligned}$$

Finally,

$$\operatorname{MSE}(\hat\lambda) =V(\hat\lambda)+\operatorname{Bias}(\hat\lambda)^2 =\boxed{\frac{\lambda^2}{n}}.$$

The estimator is centered on $\lambda$, and its variance (hence its MSE) decreases as the sample size increases.

c) Approximate 95% confidence interval for $\lambda$

The central limit theorem gives, for sufficiently large $n$,

$$\hat\lambda=\bar X\approx N\left(\lambda,\frac{\lambda^2}{n}\right).$$

Standardize and keep the central $95\%$:

$$P\left( -1.96\le \frac{\hat\lambda-\lambda}{\lambda/\sqrt n} \le1.96 \right)\approx0.95.$$

Because $\lambda>0$ and $n=100$, the denominators below remain positive. Isolating the unknown parameter $\lambda$ gives the interval in the official solution:

$$\boxed{ CI_{95\%}(\lambda)= \left[ \frac{\bar x}{1+1.96/\sqrt n}, \frac{\bar x}{1-1.96/\sqrt n} \right] }.$$

For $n=100$,

$$\boxed{ CI_{95\%}(\lambda)= \left[ \frac{\bar x}{1.196}, \frac{\bar x}{0.804} \right] }.$$

About the hint in the statement. It asks to replace the unknown population variance $V(X)$ with the observed sample variance

$$s^2=\frac1{n-1}\sum_{i=1}^n(x_i-\bar x)^2.$$

That alternative, using a plug-in standard error $s/\sqrt n$, leads to the usual large-sample Wald interval

$$\boxed{ CI_{95\%,\,\mathrm{plug\!-\!in}}(\lambda) \approx \left[ \bar x-1.96\frac{s}{10}, \bar x+1.96\frac{s}{10} \right] }.$$

The official correction instead uses the theoretical variance $\lambda^2$ and solves for $\lambda$, producing the asymmetric interval above. These are two different approximate constructions; the first displayed interval reproduces the official solution, while the plug-in formula follows the wording of the hint. No numerical endpoints can be computed because neither $\bar x$ nor $s$ is provided.

Interpretation: over repeated iid samples of this size, the chosen confidence-interval procedure covers the fixed parameter $\lambda$ approximately $95\%$ of the time.

↑ Retour à l’exercice 10

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Solutions — TP3

Solution — TP3 Exercise 2

Model: $X_1,\ldots,X_{100}$ are assumed iid $N(\mu,\sigma^2)$. Both population parameters are unknown, while $\bar x=174$ cm and $s^2=11.7^2=136.89$ cm$^2$ are observed.

a) 90% confidence interval for $\mu$

We estimate the unknown population variance with $S^2$. Because the population is Normal, the standardized sample mean follows Student’s distribution exactly:

$$T=\frac{\bar X-\mu}{S/\sqrt n}\sim t_{n-1}=t_{99}.$$

For $\alpha=0.10$, the two-sided critical quantile is $t_{99,0.95}\approx1.6604$. Therefore

$$\begin{aligned} CI_{90\%}(\mu) &=\left[\bar x-t_{99,0.95}\frac{s}{\sqrt n}, \bar x+t_{99,0.95}\frac{s}{\sqrt n}\right]\\ &=\left[174-1.6604\frac{11.7}{10}, 174+1.6604\frac{11.7}{10}\right]\\ &=\boxed{[172.06,\;175.94]\ \mathrm{cm}}. \end{aligned}$$

The official correction uses the approximation $1.645$ (Normal critical quantile); the exact Student quantile with $99$ degrees of freedom is approximately $1.6604$. Using a table row close to $99$ degrees of freedom also gives about $1.66$.

b) 90% confidence interval for $\sigma^2$

Here the unknown parameter changes: we estimate the population variance $\sigma^2$, not the mean. Under the Normal assumption,

$$Q=\frac{(n-1)S^2}{\sigma^2}\sim\chi^2_{n-1}=\chi^2_{99}.$$

Using the $5\%$ and $95\%$ quantiles and isolating $\sigma^2$ (note that the order reverses when taking reciprocals),

$$\boxed{ CI_{90\%}(\sigma^2)= \left[ \frac{(n-1)s^2}{\chi^2_{99,0.95}}, \frac{(n-1)s^2}{\chi^2_{99,0.05}} \right] }.$$

With $\chi^2_{99,0.05}\approx77.0463$ and $\chi^2_{99,0.95}\approx123.2252$,

$$\begin{aligned} CI_{90\%}(\sigma^2) &=\left[ \frac{99(136.89)}{123.2252}, \frac{99(136.89)}{77.0463} \right]\\ &=\boxed{[109.98,\;175.90]\ \mathrm{cm}^2}. \end{aligned}$$

The official correction replaces the unavailable table row $99$ by row $100$ and obtains approximately $[108.99,173.90]$ cm$^2$. The interval above uses the actual $99$ degrees of freedom.

Interpretation: the interval procedures have respectively $90\%$ coverage for the fixed mean $\mu$ and for the fixed variance $\sigma^2$, under the model assumptions.

↑ Retour à l’exercice 2

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Solution — TP3 Exercise 3

We assume independent Normal samples with sizes $n_f=n_m=100$ and one common unknown population variance

$$\sigma_m^2=\sigma_f^2=\sigma^2.$$

The observed means differ by

$$\bar x_m-\bar x_f=177-173=4\ \mathrm{cm}.$$

First calculate the pooled sample variance:

$$\begin{aligned} s_{\mathrm{pool}}^2 &=\frac{(n_m-1)s_m^2+(n_f-1)s_f^2}{n_m+n_f-2}\\ &=\frac{99(9.3)^2+99(8.7)^2}{198}\\ &=\boxed{81.09\ \mathrm{cm}^2}. \end{aligned}$$

Therefore $s_{\mathrm{pool}}=\sqrt{81.09}\approx9.005$ cm. Keep these two quantities separate: $S_{\mathrm{pool}}$ is a standard deviation; $S_{\mathrm{pool}}^2$ is a variance.

a) 95% confidence interval for $\mu_m-\mu_f$

Since the common variance is unknown and estimated with $S_{\mathrm{pool}}^2$, the correct statistic follows a Student distribution:

$$T= \frac{(\bar X_m-\bar X_f)-(\mu_m-\mu_f)} {S_{\mathrm{pool}}\sqrt{1/n_m+1/n_f}} \sim t_{n_m+n_f-2}=t_{198}.$$

The observed standard error is

$$s_{\mathrm{pool}}\sqrt{\frac1{n_m}+\frac1{n_f}} =\sqrt{81.09}\sqrt{\frac1{100}+\frac1{100}} \approx1.2735\ \mathrm{cm}.$$

With $t_{198,0.975}\approx1.9720$,

$$\begin{aligned} CI_{95\%}(\mu_m-\mu_f) &=\left[(\bar x_m-\bar x_f)\pm t_{198,0.975}\,s_{\mathrm{pool}} \sqrt{\frac1{n_m}+\frac1{n_f}}\right]\\ &=\left[4-1.9720(1.2735), 4+1.9720(1.2735)\right]\\ &=\boxed{[1.49,\;6.51]\ \mathrm{cm}}. \end{aligned}$$

Interpretation: the interval is entirely positive, suggesting that the population mean male height exceeds the population mean female height under the assumptions. The official correction uses $1.96$ as a large-df approximation; the exact $t_{198}$ quantile yields the interval above.

b) 95% confidence interval for the common variance $\sigma^2$

Since two independent Normal populations share the same unknown variance, the pooled variance gives the chi-square statistic

$$\boxed{ Q=\frac{(n_m+n_f-2)S_{\mathrm{pool}}^2}{\sigma^2} \sim\chi^2_{n_m+n_f-2}=\chi^2_{198} }.$$

Important: the numerator contains $S_{\mathrm{pool}}^2$, not $S_{\mathrm{pool}}$. We want a confidence interval for a variance.

For $\alpha=0.05$, use the chi-square quantiles

$$\chi^2_{198,0.025}\approx160.9246, \qquad \chi^2_{198,0.975}\approx238.8612.$$

Isolate $\sigma^2$:

$$\boxed{ CI_{95\%}(\sigma^2)= \left[ \frac{(n_m+n_f-2)s_{\mathrm{pool}}^2}{\chi^2_{198,0.975}}, \frac{(n_m+n_f-2)s_{\mathrm{pool}}^2}{\chi^2_{198,0.025}} \right] }.$$

Numerically, the numerator is

$$198\times81.09=16055.82,$$

so

$$\begin{aligned} CI_{95\%}(\sigma^2) &=\left[ \frac{16055.82}{238.8612}, \frac{16055.82}{160.9246} \right]\\ &=\boxed{[67.22,\;99.77]\ \mathrm{cm}^2}. \end{aligned}$$

Check against the official correction: the printed result $[67.88,100.77]$ cm$^2$ is inconsistent with $s_{\mathrm{pool}}^2=81.09$ and the stated chi-square quantiles. In particular, $(n_m+n_f-2)s_{\mathrm{pool}}^2=16055.82$, not $16216.2$. The interval displayed here is the result of the stated statistic and data.

↑ Retour à l’exercice 3

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Solutions — TP4

Solution — TP4 Exercise 1

Observed: $n=9$, $\bar x=60.62$ V and $s^2=1.31$ V$^2$. Unknown: the true mean $\mu$ and variance $\sigma^2$ of all measurements made under the same conditions. For the exact tests and confidence intervals below, assume independent measurements from a Normal population.

a) Is the voltmeter distorted? Test on one mean

An undistorted voltmeter should read 60 V on average. The question is not about one measurement being exactly 60 V.

$$H_0:\mu=60,\qquad H_1:\mu\ne60.$$

Since $\sigma^2$ is unknown, use the Single mean test from page 1 of the official formula sheet:

$$T=\frac{\bar X-\mu_0}{S/\sqrt n}\sim t_{n-1} \quad\text{under }H_0.$$ $$t_{\mathrm{obs}}= \frac{60.62-60}{\sqrt{1.31}/\sqrt9} \approx\boxed{1.625}, \qquad df=8.$$

At $\alpha=0.05$, the two-sided critical values are $t_{8,0.025}=-2.306$ and $t_{8,0.975}=2.306$.

$$-2.306<1.625<2.306.$$

Therefore we do not reject $H_0$. As this is a two-sided test, both unusually low and unusually high measurements count:

$$\begin{aligned} p&=P(T_8\le-1.625)+P(T_8\ge1.625)\\ &=2P(T_8\ge1.625) \approx\boxed{0.1428=14.28\%}. \end{aligned}$$

Interpretation: if the true mean were 60 V, a statistic at least this extreme in either direction would occur about 14.28% of the time under repeated samples. This is not sufficient evidence of systematic distortion at 5%; it does not prove the voltmeter is accurate.

b) Manufacturer’s claim: variance greater than $0.81\ \mathrm V^2$

The parameter is now the population variance. A right-tailed test addresses the manufacturer’s claim:

$$H_0:\sigma^2\le0.81,\qquad H_1:\sigma^2>0.81.$$

Evaluate the statistic at the null boundary $\sigma_0^2=0.81$, using the Single variance test on page 1 of the formula sheet:

$$Q=\frac{(n-1)S^2}{\sigma_0^2} \sim\chi^2_{n-1} \quad\text{under }\sigma^2=\sigma_0^2.$$ $$q_{\mathrm{obs}}=\frac{8(1.31)}{0.81} \approx\boxed{12.94}, \qquad df=8.$$

The upper critical value is $\chi^2_{8,0.95}\approx15.51$. Thus $12.94<15.51$: we do not reject $H_0$.

Large chi-square values support $H_1$, so the p-value is the right-tail area only:

$$p=P(\chi^2_8\ge12.94) \approx\boxed{0.1139=11.39\%}.$$

Interpretation: there is insufficient evidence at the 5% level that the population variance exceeds $0.81\ \mathrm V^2$.

c) 95% confidence interval for the mean $\mu$

Unknown population variance implies a Student confidence interval (Normal-population assumption):

$$CI_{95\%}(\mu)= \left[ \bar x-t_{8,0.975}\frac{s}{\sqrt n}, \bar x+t_{8,0.975}\frac{s}{\sqrt n} \right].$$

The two endpoints come from subtracting and adding the margin of error:

$$\begin{aligned} CI_{95\%}(\mu) &=60.62\pm2.306\frac{\sqrt{1.31}}{\sqrt9}\\ &=60.62\pm0.88\\ &=\boxed{[59.74,\;61.50]\ \mathrm V}. \end{aligned}$$

Because 60 V is inside the interval, the data are consistent with the non-rejection in (a). A frequentist 95% confidence interval has 95% long-run coverage over repeated samples; it does not assign a probability of 95% to the fixed parameter $\mu$.

d) 95% confidence interval for the variance $\sigma^2$

From the Properties of $\bar X$ and $S^2$, 1 population section of the formula sheet:

$$\frac{(n-1)S^2}{\sigma^2}\sim\chi^2_{n-1}.$$

For $\alpha=0.05$, use the chi-square quantiles at $0.025$ and $0.975$:

$$\chi^2_{8,0.025}\approx2.180,\qquad \chi^2_{8,0.975}\approx17.535.$$

When isolating $\sigma^2$, the quantiles reverse order because $\sigma^2$ appears in the denominator:

$$CI_{95\%}(\sigma^2)= \left[ \frac{(n-1)s^2}{\chi^2_{n-1,0.975}}, \frac{(n-1)s^2}{\chi^2_{n-1,0.025}} \right].$$ $$\begin{aligned} CI_{95\%}(\sigma^2) &=\left[\frac{8(1.31)}{17.535},\; \frac{8(1.31)}{2.180}\right]\\ &\approx\boxed{[0.598,\;4.808]\ \mathrm V^2}. \end{aligned}$$

This interval estimates the variance (V$^2$), not the standard deviation (V).

↑ Retour à l’exercice 1

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Solution — TP4 Exercise 2 (a–b)

Populations: dust emissions of boilers of types A and B under comparable operating conditions. Samples: 8 A-boilers and 6 B-boilers. Unknown: $\mu_A,\mu_B,\sigma_A^2,\sigma_B^2$. Assume the samples are independent and normally distributed for the exact Fisher and Student tests.

First compute the sample means and unbiased sample variances:

$$\bar x=\frac1n\sum_{i=1}^n x_i,\qquad s^2=\frac1{n-1}\sum_{i=1}^n(x_i-\bar x)^2.$$

Observed statistic	Boiler A	Boiler B
Sample size	$n_A=8$	$n_B=6$
Mean	$\bar x_A=61$ g	$\bar x_B=49$ g
Sample variance	$s_A^2=\frac{1550}{7}\approx221.43$ g$^2$	$s_B^2=\frac{690}{5}=138$ g$^2$

a) Are the variances different? Fisher test, $\alpha=0.02$

$$H_0:\sigma_A^2=\sigma_B^2,\qquad H_1:\sigma_A^2\ne\sigma_B^2.$$

From Test on 2 variances, page 2 of the formula sheet:

$$F=\frac{S_A^2}{S_B^2}\sim F_{n_A-1,n_B-1}=F_{7,5} \quad\text{under }H_0.$$ $$f_{\mathrm{obs}}=\frac{221.43}{138} \approx\boxed{1.605}.$$

With a two-sided $2\%$ significance level, each rejection tail has probability $0.01$:

$$F_{7,5,0.01}\approx0.134,\qquad F_{7,5,0.99}\approx10.46.$$

We reject $H_0$ if $F<0.134$ or $F>10.46$. Here:

$$0.134<1.605<10.46.$$

Decision: we do not reject $H_0$. The data do not give sufficient evidence of unequal variances at the 2% level. The two-sided p-value is approximately $0.623$ (62.3%).

Important: not rejecting equality is not proof that the variances are equal. The pooled-variance model in (b) is an additional modeling assumption, as in the official correction.

b) Is the mean emission from A more than 5 g higher? Student test, $\alpha=0.05$

Test the boundary of the statement “at least 5 g greater” against a strictly greater difference:

$$H_0:\mu_A-\mu_B\le5,\qquad H_1:\mu_A-\mu_B>5.$$

The test statistic is calculated at the boundary $\mu_A-\mu_B=5$. Under a common unknown population variance, use the Test on 2 sample means with pooled variance from page 2:

$$S_{\mathrm{pool}}^2= \frac{(n_A-1)S_A^2+(n_B-1)S_B^2}{n_A+n_B-2}.$$ $$\begin{aligned} s_{\mathrm{pool}}^2 &=\frac{7(221.43)+5(138)}{12}\\ &\approx\boxed{186.67\ \mathrm g^2}. \end{aligned}$$

(The official correction uses rounded $s_A^2=221.4$, giving $s_{\mathrm{pool}}^2=186.65$.)

$$T=\frac{(\bar X_A-\bar X_B)-5} {S_{\mathrm{pool}}\sqrt{\frac1{n_A}+\frac1{n_B}}} \sim t_{n_A+n_B-2}=t_{12} \quad\text{at the null boundary}.$$ $$\begin{aligned} t_{\mathrm{obs}} &=\frac{(61-49)-5} {\sqrt{186.67}\sqrt{\frac18+\frac16}}\\ &\approx\boxed{0.949}. \end{aligned}$$

Since $H_1$ is right-sided, we reject for large positive values. At $\alpha=0.05$, $t_{12,0.95}\approx1.782$ and

$$0.949<1.782.$$

Decision: we do not reject $H_0$. The sample does not establish that A’s population mean emission exceeds B’s by more than 5 g.

The one-sided p-value is

$$\boxed{p=P(T_{12}\ge0.949)\approx0.181=18.1\%.}$$

Official-correction note: the provided solutions report 18.14%; a direct calculation with $t_{12}$ and the stated statistic gives approximately 18.07%. This small discrepancy does not affect the decision.

↑ Retour à l’exercice 2

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Solution — TP4 Exercise 3

Real question: do two laboratory methods produce systematically different results on average? Population: phosphate measurements on ore samples under the stated conditions. Data: 10 paired measurements, because every ore sample is measured by A and B. Unknown: the mean within-sample difference $\mu_D$.

An independent two-population test would ignore the pairing. Instead, define the difference on sample $i$ as

$$D_i=A_i-B_i.$$

Sample	1	2	3	4	5	6	7	8	9	10
$d_i=A_i-B_i$ (g)	3.2	−3.9	−2.3	−0.6	1.1	−2.9	−1.8	−0.5	−2.3	−0.3

A paired t-test is a single-mean Student test applied to these differences. For an exact small-sample test, the paired differences are assumed iid Normal across ore samples; Normal marginals for A and B alone would not guarantee that their differences are Normal.

a) Select and carry out the test, $\alpha=0.05$

The null hypothesis states that the methods have the same population mean measurement:

$$H_0:\mu_D=0,\qquad H_1:\mu_D\ne0.$$

It is a two-sided test. Calculate the sample mean difference:

$$\bar d=\frac{3.2-3.9-2.3-0.6+1.1-2.9-1.8-0.5-2.3-0.3}{10} =\boxed{-1.03\ \mathrm g}.$$

For the unbiased variance of the differences, use the computational identity

$$s_D^2= \frac{\sum_{i=1}^{10}d_i^2-n\bar d^2}{n-1}.$$

Here $\sum d_i^2=49.59$:

$$s_D^2=\frac{49.59-10(-1.03)^2}{9} \approx\boxed{4.33\ \mathrm g^2}.$$

From the Single mean test, page 1 of the formula sheet, replace $\bar X,S$ with $\bar D,S_D$ and take $\mu_0=0$:

$$T=\frac{\bar D-0}{S_D/\sqrt n} \sim t_{n-1}=t_9 \quad\text{under }H_0.$$ $$\begin{aligned} t_{\mathrm{obs}} &=\frac{-1.03}{\sqrt{4.33}/\sqrt{10}}\\ &\approx\boxed{-1.565}. \end{aligned}$$

With $\alpha=0.05$, the rejection region from the two-sided line of the official formula is:

$$T<t_{9,0.025}=-2.262 \quad\text{or}\quad T>t_{9,0.975}=2.262.$$

But $-2.262<-1.565<2.262$, so we do not reject $H_0$.

Interpretation: A measured average difference of $-1.03$ g is not, on this sample, statistically significant at 5%. We cannot conclude that the two methods have different population mean measurements.

b) Calculate the p-value

Because the test is two-sided and Student’s t distribution is symmetric, both tails count:

$$\begin{aligned} p &=P(T_9\le-1.565)+P(T_9\ge1.565)\\ &=2P(T_9\ge1.565)\\ &\approx\boxed{0.1520=15.20\%}. \end{aligned}$$

Since $p=0.1520>0.05$, we do not reject $H_0$, consistent with (a). The p-value is not the probability that $H_0$ is true; it measures how unusual this result would be if $H_0$ held.

To verify the numerical result in Python:

from scipy.stats import t

t_obs = -1.565
p_value = 2 * t.sf(abs(t_obs), df=9)
print(p_value)  # approximately 0.1520

Exam method: same ore sample measured twice $\rightarrow$ compute paired differences $\rightarrow$ use the single-mean Student test on $D$ with $n-1$ degrees of freedom; do not use the independent two-population formula.

↑ Retour à l’exercice 3