A cosmetic product produced at a certain factory has a content weight that follows a normal distribution with mean 201.5 g and standard deviation 1.8 g. Using the standard normal distribution table on the right, what is the probability that the sample mean of 9 randomly selected cosmetic products from this factory is at least 200 g? [3 points]

$z$	$\mathrm { P } ( 0 \leq Z \leq z )$
1.0	0.3413
1.5	0.4332
2.0	0.4772
2.5	0.4938

(1) 0.7745
(2) 0.8413
(3) 0.9332
(4) 0.9772
(5) 0.9938

Recall that a random variable $X$, taking values in $\mathbb{N}$, follows the Poisson distribution $\mathcal{P}(\lambda)$ with parameter $\lambda$ if, for all $n \in \mathbb{N}$: $$\mathrm{P}(X = n) = \frac{\lambda^{n}}{n!} \mathrm{e}^{-\lambda}$$
Let $\left(X_{n}\right)_{n \geqslant 1}$ be a sequence of mutually independent random variables, with distribution $\mathcal{P}(\lambda)$, $S_{n} = X_{1} + X_{2} + \cdots + X_{n}$, and $T_{n} = \frac{S_{n} - n\lambda}{\sqrt{n\lambda}}$. Show that, for all $\varepsilon > 0$, there exists a real number $c(\varepsilon)$ such that, if $c \geqslant c(\varepsilon)$ and $n \in \mathbb{N}^{*}$, we have $\mathrm{P}\left(\left|T_{n}\right| \geqslant c\right) \leqslant \varepsilon$.

We fix two real numbers $a$ and $b$ such that $a < b$. For all integers $n \geqslant 1$ such that $a + \sqrt{n\lambda} > 0$, we define $$I_{n} = \{k \in \mathbb{N} \mid n\lambda + a\sqrt{n\lambda} \leqslant k \leqslant n\lambda + b\sqrt{n\lambda}\}$$ For $k \in \mathbb{Z}$, we define $x_{k,n} = \frac{k - n\lambda}{\sqrt{n\lambda}}$. We consider the function $f : \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = \mathrm{e}^{-\frac{1}{2}x^{2}}$ for all $x \in \mathbb{R}$, which is $M$-Lipschitz for some $M > 0$.
a) Show that, if $x, h \in \mathbb{R}$ and $h > 0$, then $\left| hf(x) - \int_{x}^{x+h} f(t) \mathrm{d}t \right| \leqslant M \frac{h^{2}}{2}$.
b) Deduce from this, when $I_{n}$ is non-empty, an upper bound for $$\left| \frac{1}{\sqrt{n\lambda}} \sum_{k \in I_{n}} f\left(x_{k,n}\right) - \int_{x_{p,n}}^{x_{q+1,n}} f(t) \mathrm{d}t \right|$$ where $p$ is the smallest element of $I_{n}$ and $q$ is the largest.
c) Show that $$\lim_{n \rightarrow +\infty} \frac{1}{\sqrt{n\lambda}} \sum_{k \in I_{n}} f\left(x_{k,n}\right) = \int_{a}^{b} f(x) \mathrm{d}x$$

We fix two real numbers $a$ and $b$ such that $a < b$. For all integers $n \geqslant 1$ such that $a + \sqrt{n\lambda} > 0$, we define $$I_{n} = \{k \in \mathbb{N} \mid n\lambda + a\sqrt{n\lambda} \leqslant k \leqslant n\lambda + b\sqrt{n\lambda}\}$$ For $k \in \mathbb{Z}$, we define $x_{k,n} = \frac{k - n\lambda}{\sqrt{n\lambda}}$. We consider the function $f(x) = \mathrm{e}^{-\frac{1}{2}x^{2}}$.
For all $k \in I_{n}$, we denote $y_{k,n} = \left(1 - \frac{x_{k,n}}{k}\sqrt{n\lambda}\right)^{k} \exp\left(x_{k,n}\sqrt{n\lambda}\right)$.
Let $\varepsilon > 0$. Prove the existence of an integer $N(\varepsilon)$ such that, for all $n \geqslant N(\varepsilon)$ and all $k \in I_{n}$, the following inequalities are satisfied:
a) $\frac{1-\varepsilon}{\sqrt{2\pi}} \frac{1}{\sqrt{n\lambda}} y_{k,n} \leqslant \mathrm{e}^{-n\lambda} \frac{(n\lambda)^{k}}{k!} \leqslant \frac{1+\varepsilon}{\sqrt{2\pi}} \frac{1}{\sqrt{n\lambda}} y_{k,n}$;
We will use Stirling's formula $n! \underset{n \rightarrow +\infty}{\sim} \sqrt{2\pi n} \left(\frac{n}{\mathrm{e}}\right)^{n}$.
b) $(1-\varepsilon) f\left(x_{k,n}\right) \leqslant y_{k,n} \leqslant (1+\varepsilon) f\left(x_{k,n}\right)$.

We fix two real numbers $a$ and $b$ such that $a < b$. For all integers $n \geqslant 1$ such that $a + \sqrt{n\lambda} > 0$, we define $$I_{n} = \{k \in \mathbb{N} \mid n\lambda + a\sqrt{n\lambda} \leqslant k \leqslant n\lambda + b\sqrt{n\lambda}\}$$
Express, in the form of an integral, $\lim_{n \rightarrow +\infty} \sum_{k \in I_{n}} \frac{(n\lambda)^{k}}{k!} \mathrm{e}^{-n\lambda}$.

We fix two real numbers $a$ and $b$ such that $a < b$. For all integers $n \geqslant 1$ such that $a + \sqrt{n\lambda} > 0$, we define $$I_{n} = \{k \in \mathbb{N} \mid n\lambda + a\sqrt{n\lambda} \leqslant k \leqslant n\lambda + b\sqrt{n\lambda}\}$$ Let $\left(X_{n}\right)_{n \geqslant 1}$ be a sequence of mutually independent random variables with distribution $\mathcal{P}(\lambda)$, $S_{n} = X_{1} + \cdots + X_{n}$, and $T_{n} = \frac{S_{n} - n\lambda}{\sqrt{n\lambda}}$.
Compare $\mathrm{P}\left(a \leqslant T_{n} \leqslant b\right)$ and $\sum_{k \in I_{n}} \mathrm{P}\left(S_{n} = k\right)$.

Let $\left(X_{n}\right)_{n \geqslant 1}$ be a sequence of mutually independent random variables with distribution $\mathcal{P}(\lambda)$, $S_{n} = X_{1} + \cdots + X_{n}$, and $T_{n} = \frac{S_{n} - n\lambda}{\sqrt{n\lambda}}$.
Determine the limits, when $n \rightarrow +\infty$, of $$\mathrm{P}\left(T_{n} \geqslant a\right), \quad \mathrm{P}\left(T_{n} = a\right), \quad \mathrm{P}\left(T_{n} > a\right) \quad \text{and} \quad \mathrm{P}\left(T_{n} \leqslant b\right)$$

Recall that a random variable $X$ follows the Poisson distribution $\mathcal{P}(\lambda)$ if $\mathrm{P}(X = n) = \frac{\lambda^{n}}{n!} \mathrm{e}^{-\lambda}$ for all $n \in \mathbb{N}$. Let $\left(X_{n}\right)_{n \geqslant 1}$ be a sequence of mutually independent random variables with distribution $\mathcal{P}(\lambda)$, $S_{n} = X_{1} + \cdots + X_{n}$, and $T_{n} = \frac{S_{n} - n\lambda}{\sqrt{n\lambda}}$.
Determine an equivalent, when $n \rightarrow +\infty$, of $$A_{n} = \sum_{k=0}^{\lfloor n\lambda \rfloor} \frac{(n\lambda)^{k}}{k!} \quad \text{and} \quad B_{n} = \sum_{k=\lfloor n\lambda \rfloor + 1}^{+\infty} \frac{(n\lambda)^{k}}{k!}$$ where $\lfloor t \rfloor$ denotes the integer part of the real number $t$.
We will interpret $\mathrm{e}^{-n\lambda} A_{n}$ as the probability of an event related to $S_{n}$ and thus to $T_{n}$.

Suppose that $X$ admits a second moment. If $u$ and $v$ are two real numbers such that $u < E(X) < v$, determine the limit of the sequence $\left(\pi_{n}\right)_{n \in \mathbb{N}^{*}}$ defined by $$\forall n \in \mathbb{N}^{*}, \quad \pi_{n} = P\left(nu \leqslant S_{n} \leqslant nv\right)$$

In subsection II.D, we assume that there exists a strictly positive real number $c$ such that the discrete real random variable $X$ satisfies $\mathbb{E}(X)=0$ and $\forall \omega \in \Omega,|X(\omega)| \leqslant c$. For every strictly positive integer $n$, $S_{n}=\sum_{k=1}^{n} X_{k}$ where $\left(X_{k}\right)$ are mutually independent with the same distribution as $X$.
Show that $\forall n \in \mathbb{N}^{*}, \mathbb{P}\left(\left|\frac{S_{n}}{n}\right| \geqslant \varepsilon\right) \leqslant 2 \exp\left(-n \frac{\varepsilon^{2}}{2c^{2}}\right)$.

In subsection II.D, we assume that there exists a strictly positive real number $c$ such that the discrete real random variable $X$ satisfies $\mathbb{E}(X)=0$ and $\forall \omega \in \Omega,|X(\omega)| \leqslant c$. We have shown that $\forall n \in \mathbb{N}^{*}, \mathbb{P}\left(\left|\frac{S_{n}}{n}\right| \geqslant \varepsilon\right) \leqslant 2 \exp\left(-n \frac{\varepsilon^{2}}{2c^{2}}\right)$.
Let $n$ be a non-zero natural number, $p$ an element of the interval $]0,1[$ and $Z$ a random variable following a binomial distribution with parameter $(n, p)$. Using the previous question, bound $\mathbb{P}\left(\left|\frac{Z}{n}-p\right| \geqslant \varepsilon\right)$ as a function of $n, p$ and $\varepsilon$.

Let $n \geqslant 1$ be a natural integer, and let $\left( X _ { 1 } , \ldots , X _ { n } \right)$ be mutually independent discrete real random variables such that, for all $k \in \{ 1 , \ldots , n \}$, $$P \left[ X _ { k } = 1 \right] = P \left[ X _ { k } = - 1 \right] = \frac { 1 } { 2 }$$ We define $$S _ { n } = \frac { 1 } { n } \sum _ { k = 1 } ^ { n } X _ { k }$$ Show that $P \left[ S _ { n } \geqslant 0 \right] \geqslant \frac { 1 } { 2 }$.

Let $n \geqslant 1$ be a natural integer, and let $(X_1, \ldots, X_n)$ be discrete real random variables that are mutually independent such that, for all $k \in \{1, \ldots, n\}$, $$P[X_k = 1] = P[X_k = -1] = \frac{1}{2}$$ We define $$S_n = \frac{1}{n} \sum_{k=1}^{n} X_k$$ as well as, for all $\lambda \in \mathbb{R}$, $$\psi(\lambda) = \log\left(\frac{1}{2}e^{\lambda} + \frac{1}{2}e^{-\lambda}\right)$$ Show that for all $t \in \mathbb{R}$, we have $$\frac{1}{n} \log P[S_n \geqslant t] \leqslant \inf_{\lambda \geqslant 0} (\psi(\lambda) - \lambda t).$$

Let $X _ { 1 } , \ldots , X _ { n } , Y _ { 1 } , \ldots , Y _ { n }$ be mutually independent random variables with the same distribution $\mathcal { R }$. We define the random vectors $X = \frac { 1 } { \sqrt { n } } \left( X _ { 1 } , \ldots , X _ { n } \right) ^ { \top }$ and $Y = \frac { 1 } { \sqrt { n } } \left( Y _ { 1 } , \ldots , Y _ { n } \right) ^ { \top }$ taking values in $\mathcal { M } _ { n , 1 } ( \mathbb { R } )$.
Prove that $$\mathbb { P } ( | \langle X \mid Y \rangle | \geqslant \varepsilon ) \leqslant 2 \exp \left( - \frac { \varepsilon ^ { 2 } n } { 2 } \right)$$

$N$ being a non-zero natural integer, $\left( X _ { j } ^ { i } \right) _ { 1 \leqslant i \leqslant N , 1 \leqslant j \leqslant n }$ is a family of $n \times N$ mutually independent real random variables with the same distribution $\mathcal { R }$. For every $i \in \llbracket 1 , N \rrbracket$, we set $X ^ { i } = \frac { 1 } { \sqrt { n } } \left( X _ { 1 } ^ { i } , \ldots , X _ { n } ^ { i } \right) ^ { \top }$.
Deduce from the previous questions that $$\mathbb { P } \left( \bigcup _ { 1 \leqslant i < j \leqslant N } \left| \left\langle X ^ { i } \mid X ^ { j } \right\rangle \right| \geqslant \varepsilon \right) \leqslant N ( N - 1 ) \exp \left( - \frac { \varepsilon ^ { 2 } n } { 2 } \right) .$$

$N$ being a non-zero natural integer, $\left( X _ { j } ^ { i } \right) _ { 1 \leqslant i \leqslant N , 1 \leqslant j \leqslant n }$ is a family of $n \times N$ mutually independent real random variables with the same distribution $\mathcal { R }$. For every $i \in \llbracket 1 , N \rrbracket$, we set $X ^ { i } = \frac { 1 } { \sqrt { n } } \left( X _ { 1 } ^ { i } , \ldots , X _ { n } ^ { i } \right) ^ { \top }$.
We assume that $n \geqslant 4 \frac { \ln N } { \varepsilon ^ { 2 } }$. Prove that $$\mathbb { P } \left( \bigcup _ { 1 \leqslant i < j \leqslant N } \left| \left\langle X ^ { i } \mid X ^ { j } \right\rangle \right| \geqslant \varepsilon \right) < 1 .$$

$N$ being a non-zero natural integer, $\left( X _ { j } ^ { i } \right) _ { 1 \leqslant i \leqslant N , 1 \leqslant j \leqslant n }$ is a family of $n \times N$ mutually independent real random variables with the same distribution $\mathcal { R }$. For every $i \in \llbracket 1 , N \rrbracket$, we set $X ^ { i } = \frac { 1 } { \sqrt { n } } \left( X _ { 1 } ^ { i } , \ldots , X _ { n } ^ { i } \right) ^ { \top }$.
Deduce that, for every natural integer $N$ less than or equal to $\exp \left( \frac { \varepsilon ^ { 2 } n } { 4 } \right)$, there exists a family of $N$ unit vectors of $\mathbb { R } ^ { n }$ whose coherence parameter is bounded by $\varepsilon$.

In this subsection, we still assume that $J_n = J_n^{(\mathrm{C})}$. Moreover, we assume that $\beta = 1$ and $h = 0$.
We denote $Z_\infty = \int_{-\infty}^{+\infty} \exp\left(-\frac{x^4}{12}\right) \mathrm{d}x$ and we consider the function $$\varphi_\infty : x \longmapsto \frac{1}{Z_\infty} \exp\left(-\frac{x^4}{12}\right).$$
Show finally that $$\mathbb{P}\left(n^{1/4} M_n \leqslant x\right) \underset{n \rightarrow +\infty}{\longrightarrow} \int_{-\infty}^{x} \varphi_\infty(u) \mathrm{d}u$$