For all $(i,j) \in (\mathbb{N}^{\star})^{2}$ and for all $C > 0$, we set $\sigma_{ij}(C) = \sqrt{\mathbb{V}\left(X_{ij} \mathbb{1}_{|X_{ij}| \leqslant C}\right)}$. If $\sigma_{ij}(C) \neq 0$, we set $$\widehat{X}_{ij}(C) = \frac{1}{\sigma_{ij}(C)} \left(X_{ij} \mathbb{1}_{|X_{ij}| \leqslant C} - \mathbb{E}\left(X_{ij} \mathbb{1}_{|X_{ij}| \leqslant C}\right)\right).$$
Justify that, for $C$ sufficiently large, the variables $\widehat{X}_{ij}(C)$ are well defined and that they are then bounded, centered, of variance 1 and that they are mutually independent for $1 \leqslant i \leqslant j$.

For every $(i,j) \in (\mathbb{N}^{\star})^{2}$ and for every $C > 0$, we set $\sigma_{ij}(C) = \sqrt{\mathbb{V}\left(X_{ij} \mathbb{1}_{|X_{ij}| \leqslant C}\right)}$. If $\sigma_{ij}(C) \neq 0$, we set $$\widehat{X}_{ij}(C) = \frac{1}{\sigma_{ij}(C)}\left(X_{ij} \mathbb{1}_{|X_{ij}| \leqslant C} - \mathbb{E}\left(X_{ij} \mathbb{1}_{|X_{ij}| \leqslant C}\right)\right).$$
Show that $$X_{ij} - \widehat{X}_{ij}(C) = \left(1 - \frac{1}{\sigma_{ij}(C)}\right) X_{ij} + \frac{1}{\sigma_{ij}(C)}\left(X_{ij} \mathbb{1}_{|X_{ij}| > C} - \mathbb{E}\left(X_{ij} \mathbb{1}_{|X_{ij}| > C}\right)\right).$$

For all $(i,j) \in (\mathbb{N}^{\star})^{2}$ and for all $C > 0$, we set $\sigma_{ij}(C) = \sqrt{\mathbb{V}\left(X_{ij} \mathbb{1}_{|X_{ij}| \leqslant C}\right)}$. If $\sigma_{ij}(C) \neq 0$, we set $$\widehat{X}_{ij}(C) = \frac{1}{\sigma_{ij}(C)} \left(X_{ij} \mathbb{1}_{|X_{ij}| \leqslant C} - \mathbb{E}\left(X_{ij} \mathbb{1}_{|X_{ij}| \leqslant C}\right)\right).$$
Show that $$X_{ij} - \widehat{X}_{ij}(C) = \left(1 - \frac{1}{\sigma_{ij}(C)}\right) X_{ij} + \frac{1}{\sigma_{ij}(C)} \left(X_{ij} \mathbb{1}_{|X_{ij}| > C} - \mathbb{E}\left(X_{ij} \mathbb{1}_{|X_{ij}| > C}\right)\right).$$

For every $(i,j) \in (\mathbb{N}^{\star})^{2}$ and for every $C > 0$, we set $\sigma_{ij}(C) = \sqrt{\mathbb{V}\left(X_{ij} \mathbb{1}_{|X_{ij}| \leqslant C}\right)}$. If $\sigma_{ij}(C) \neq 0$, we set $$\widehat{X}_{ij}(C) = \frac{1}{\sigma_{ij}(C)}\left(X_{ij} \mathbb{1}_{|X_{ij}| \leqslant C} - \mathbb{E}\left(X_{ij} \mathbb{1}_{|X_{ij}| \leqslant C}\right)\right).$$
Show that $$\lim_{C \rightarrow +\infty} \mathbb{E}\left(\left(X_{ij} - \widehat{X}_{ij}(C)\right)^{2}\right) = 0.$$

For all $(i,j) \in (\mathbb{N}^{\star})^{2}$ and for all $C > 0$, we set $\sigma_{ij}(C) = \sqrt{\mathbb{V}\left(X_{ij} \mathbb{1}_{|X_{ij}| \leqslant C}\right)}$. If $\sigma_{ij}(C) \neq 0$, we set $$\widehat{X}_{ij}(C) = \frac{1}{\sigma_{ij}(C)} \left(X_{ij} \mathbb{1}_{|X_{ij}| \leqslant C} - \mathbb{E}\left(X_{ij} \mathbb{1}_{|X_{ij}| \leqslant C}\right)\right).$$
Show that $$\lim_{C \rightarrow +\infty} \mathbb{E}\left(\left(X_{ij} - \widehat{X}_{ij}(C)\right)^{2}\right) = 0.$$

For every integer $n$ such that $n \geqslant 1$, we denote by $\widehat{M}_{n}(C) = \left(\widehat{X}_{ij}(C)\right)_{1 \leqslant i,j \leqslant n}$. For every $\omega \in \Omega$, we denote by $\widehat{\Lambda}_{1,n}(\omega) \geqslant \ldots \geqslant \widehat{\Lambda}_{n,n}(\omega)$ the eigenvalues of $\frac{1}{\sqrt{n}} \widehat{M}_{n}(C)(\omega)$ arranged in decreasing order. Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a $K$-Lipschitz function.
Show that $$\left|\mathbb{E}\left(\frac{1}{n} \sum_{i=1}^{n} f\left(\Lambda_{i,n}\right)\right) - \mathbb{E}\left(\frac{1}{n} \sum_{i=1}^{n} f\left(\widehat{\Lambda}_{i,n}\right)\right)\right| \leqslant \frac{K}{n} \mathbb{E}\left(\left\|M_{n} - \widehat{M}_{n}(C)\right\|_{F}\right).$$

For every integer $n$ such that $n \geqslant 1$, we denote by $\widehat{M}_{n}(C) = (\widehat{X}_{ij}(C))_{1 \leqslant i,j \leqslant n}$. For all $\omega \in \Omega$, we denote by $\widehat{\Lambda}_{1,n}(\omega) \geqslant \ldots \geqslant \widehat{\Lambda}_{n,n}(\omega)$ the eigenvalues of $\frac{1}{\sqrt{n}} \widehat{M}_{n}(C)(\omega)$ arranged in decreasing order. Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a $K$-Lipschitz function.
Show that $$\left|\mathbb{E}\left(\frac{1}{n} \sum_{i=1}^{n} f\left(\Lambda_{i,n}\right)\right) - \mathbb{E}\left(\frac{1}{n} \sum_{i=1}^{n} f\left(\widehat{\Lambda}_{i,n}\right)\right)\right| \leqslant \frac{K}{n} \mathbb{E}\left(\left\|M_{n} - \widehat{M}_{n}(C)\right\|_{F}\right).$$

For every integer $n$ such that $n \geqslant 1$, we denote by $\widehat{M}_{n}(C) = \left(\widehat{X}_{ij}(C)\right)_{1 \leqslant i,j \leqslant n}$. For every $\omega \in \Omega$, we denote by $\widehat{\Lambda}_{1,n}(\omega) \geqslant \ldots \geqslant \widehat{\Lambda}_{n,n}(\omega)$ the eigenvalues of $\frac{1}{\sqrt{n}} \widehat{M}_{n}(C)(\omega)$ arranged in decreasing order. Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a $K$-Lipschitz function.
Assume furthermore that $f$ is bounded. Show $$\mathbb{E}\left(\frac{1}{n} \sum_{i=1}^{n} f\left(\Lambda_{i,n}\right)\right) \xrightarrow{n \rightarrow +\infty} \frac{1}{2\pi} \int_{-2}^{2} f(x) \sqrt{4 - x^{2}} \, \mathrm{d}x.$$

For every integer $n$ such that $n \geqslant 1$, we denote by $\widehat{M}_{n}(C) = (\widehat{X}_{ij}(C))_{1 \leqslant i,j \leqslant n}$. For all $\omega \in \Omega$, we denote by $\widehat{\Lambda}_{1,n}(\omega) \geqslant \ldots \geqslant \widehat{\Lambda}_{n,n}(\omega)$ the eigenvalues of $\frac{1}{\sqrt{n}} \widehat{M}_{n}(C)(\omega)$ arranged in decreasing order. Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a $K$-Lipschitz function.
Assume further that $f$ is bounded. Show $$\mathbb{E}\left(\frac{1}{n} \sum_{i=1}^{n} f\left(\Lambda_{i,n}\right)\right) \xrightarrow{n \rightarrow +\infty} \frac{1}{2\pi} \int_{-2}^{2} f(x) \sqrt{4 - x^{2}} \, \mathrm{d}x.$$

Show the semicircle law in the general case:
For every function $f : \mathbb{R} \rightarrow \mathbb{R}$, continuous and bounded, $$\mathbb{E}\left(\frac{1}{n} \sum_{i=1}^{n} f\left(\Lambda_{i,n}\right)\right) \xrightarrow{n \rightarrow +\infty} \frac{1}{2\pi} \int_{-2}^{2} f(x) \sqrt{4 - x^{2}} \, \mathrm{d}x$$

Show the semicircle law in the general case:
For every function $f : \mathbb{R} \rightarrow \mathbb{R}$, continuous and bounded, $$\mathbb{E}\left(\frac{1}{n} \sum_{i=1}^{n} f\left(\Lambda_{i,n}\right)\right) \xrightarrow{n \rightarrow +\infty} \frac{1}{2\pi} \int_{-2}^{2} f(x) \sqrt{4 - x^{2}} \, \mathrm{d}x$$

Let $E = \{x_1, x_2, \ldots, x_n, \ldots\}$ be a countably infinite set where the $x_i$ are pairwise distinct elements. We denote by $\mathscr{M}(E)$ the set of probability measures on $E$.
Let $(\mu_n)_{n \in \mathbb{N}}$ be a sequence of elements of $\mathscr{M}(E)$.
Show that there exists a sequence $(\varphi_k)_{k \in \mathbb{N}^*}$ of strictly increasing maps from $\mathbb{N}^*$ to $\mathbb{N}^*$ such that, for all $k \in \mathbb{N}^*$ and for all integer $1 \leqslant i \leqslant k$, the sequence $\left(\mu_{\varphi_1 \circ \varphi_2 \circ \ldots \circ \varphi_k(n)}(x_i)\right)_{n \in \mathbb{N}^*}$ converges.

Let $E = \{x_1, x_2, \ldots, x_n, \ldots\}$ be a countably infinite set where the $x_i$ are pairwise distinct elements. We denote by $\mathscr{M}(E)$ the set of probability measures on $E$.
Let $(\mu_n)_{n \in \mathbb{N}}$ be a sequence of elements of $\mathscr{M}(E)$. Let $(\varphi_k)_{k \in \mathbb{N}^*}$ be the sequence of strictly increasing maps from 12a.
Show that for all $i \in \mathbb{N}^*$ and all integer $k \geqslant i$, the limit of the sequence $\left(\mu_{\varphi_1 \circ \varphi_2 \circ \ldots \circ \varphi_k(n)}(x_i)\right)_{n \in \mathbb{N}^*}$ depends only on $i$ and not on $k$. We denote this limit by $\mu_\infty(x_i)$.

Let $E = \{x_1, x_2, \ldots, x_n, \ldots\}$ be a countably infinite set where the $x_i$ are pairwise distinct elements. We denote by $\mathscr{M}(E)$ the set of probability measures on $E$.
Let $(\mu_n)_{n \in \mathbb{N}}$ be a sequence of elements of $\mathscr{M}(E)$. Let $(\varphi_k)_{k \in \mathbb{N}^*}$ be the sequence of strictly increasing maps from 12a, and let $\mu_\infty(x_i)$ be the limits defined in 12b.
Show that the map $$\begin{array}{rcl} \psi : \mathbb{N}^* & \longrightarrow & \mathbb{N}^* \\ k & \longmapsto & \varphi_1 \circ \varphi_2 \circ \ldots \circ \varphi_k(k) \end{array}$$ is strictly increasing, and that, for all integer $i \in \mathbb{N}^*$, the sequence $\left(\mu_{\psi(k)}(x_i)\right)_{k \in \mathbb{N}^*}$ converges to $\mu_\infty(x_i)$.

Let $E = \{x_1, x_2, \ldots, x_n, \ldots\}$ be a countably infinite set where the $x_i$ are pairwise distinct elements. We denote by $\mathscr{M}(E)$ the set of probability measures on $E$.
Let $(\mu_n)_{n \in \mathbb{N}}$ be a sequence of elements of $\mathscr{M}(E)$. Let $\mu_\infty(x_i)$ be the limits defined in 12b.
Show that $\mu_\infty(x_i) \geqslant 0$ for all $i$ in $\mathbb{N}^*$, and that $\sum_{i=1}^{\infty} \mu_\infty(x_i) \leqslant 1$.

Let $E = \{x_1, x_2, \ldots, x_n, \ldots\}$ be a countably infinite set where the $x_i$ are pairwise distinct elements. We denote by $\mathscr{M}(E)$ the set of probability measures on $E$. We denote by $\mathscr{P}(E)$ the set of subsets of $E$. Let $\mathscr{B}(\mathscr{P}(E), \mathbb{R})$ be the $\mathbb{R}$-vector space of bounded functions from $\mathscr{P}(E)$ to $\mathbb{R}$ with norm $\|f\| = \sup\{|f(A)|, \quad A \in \mathscr{P}(E)\}$.
Let $(\mu_n)_{n \in \mathbb{N}}$ be a sequence of elements of $\mathscr{M}(E)$. Let $\mu_\infty$ be defined as in 12b and 12d.
We say that the sequence $(\mu_n)_{n \in \mathbb{N}}$ of elements of $\mathscr{M}(E)$ is tight if for every real number $\varepsilon > 0$, there exists a finite subset $F_\varepsilon$ of $E$ such that $\mu_n(F_\varepsilon) \geqslant 1 - \varepsilon$ for all natural integer $n$.
Suppose further that the sequence $(\mu_n)_{n \in \mathbb{N}}$ is tight. Show then that $\mu_\infty$ defines an element of $\mathscr{M}(E)$ which is a cluster point of the sequence $(\mu_n)_{n \in \mathbb{N}}$ in $\mathscr{B}(\mathscr{P}(E), \mathbb{R})$.

Let $E = \{x_1, x_2, \ldots, x_n, \ldots\}$ be an infinite countable set. We denote $\mathscr{M}(E)$ the set of probability measures on $E$. Let $(\mu_n)_{n \in \mathbb{N}}$ be a sequence of elements of $\mathscr{M}(E)$. Show that there exists a sequence $(\varphi_k)_{k \in \mathbb{N}^*}$ of strictly increasing applications from $\mathbb{N}^*$ to $\mathbb{N}^*$ such that, for all $k \in \mathbb{N}^*$ and for all integer $1 \leqslant i \leqslant k$, the sequence $\left(\mu_{\varphi_1 \circ \varphi_2 \circ \ldots \circ \varphi_k(n)}(x_i)\right)_{n \in \mathbb{N}^*}$ converges.

Let $E = \{x_1, x_2, \ldots, x_n, \ldots\}$ be an infinite countable set. We denote $\mathscr{M}(E)$ the set of probability measures on $E$. Let $(\mu_n)_{n \in \mathbb{N}}$ be a sequence of elements of $\mathscr{M}(E)$ and $(\varphi_k)_{k \in \mathbb{N}^*}$ the sequence of strictly increasing applications from 12a. Show that for all $i \in \mathbb{N}^*$ and all integer $k \geqslant i$, the limit of the sequence $\left(\mu_{\varphi_1 \circ \varphi_2 \circ \ldots \circ \varphi_k(n)}(x_i)\right)_{n \in \mathbb{N}^*}$ depends only on $i$ and not on $k$. We denote this limit $\mu_\infty(x_i)$.

Let $E = \{x_1, x_2, \ldots, x_n, \ldots\}$ be an infinite countable set. We denote $\mathscr{M}(E)$ the set of probability measures on $E$. Let $(\mu_n)_{n \in \mathbb{N}}$ be a sequence of elements of $\mathscr{M}(E)$ and $(\varphi_k)_{k \in \mathbb{N}^*}$ the sequence of strictly increasing applications from 12a, with limits $\mu_\infty(x_i)$ as defined in 12b. Show that the application $$\begin{aligned} \psi : \mathbb{N}^* &\longrightarrow \mathbb{N}^* \\ k &\longmapsto \varphi_1 \circ \varphi_2 \circ \ldots \circ \varphi_k(k) \end{aligned}$$ is strictly increasing, and that, for all integer $i \in \mathbb{N}^*$, the sequence $\left(\mu_{\psi(k)}(x_i)\right)_{k \in \mathbb{N}^*}$ converges to $\mu_\infty(x_i)$.

Let $E = \{x_1, x_2, \ldots, x_n, \ldots\}$ be an infinite countable set. We denote $\mathscr{M}(E)$ the set of probability measures on $E$. Let $(\mu_n)_{n \in \mathbb{N}}$ be a sequence of elements of $\mathscr{M}(E)$ and $\mu_\infty$ as defined in 12b. Show that $\mu_\infty(x_i) \geqslant 0$ for all $i$ in $\mathbb{N}^*$, and that $\sum_{i=1}^{\infty} \mu_\infty(x_i) \leqslant 1$.

Let $E = \{x_1, x_2, \ldots, x_n, \ldots\}$ be an infinite countable set. We denote $\mathscr{M}(E)$ the set of probability measures on $E$. Let $\mathscr{B}(\mathscr{P}(E), \mathbb{R})$ be the $\mathbb{R}$-vector space of bounded functions from $\mathscr{P}(E)$ to $\mathbb{R}$ with norm $\|f\| = \sup\{|f(A)|, \; A \in \mathscr{P}(E)\}$. Let $(\mu_n)_{n \in \mathbb{N}}$ be a sequence of elements of $\mathscr{M}(E)$ and $\mu_\infty$ as defined in 12b. We say that the sequence $(\mu_n)_{n \in \mathbb{N}}$ is tight if for all real $\varepsilon > 0$, there exists a finite subset $F_\varepsilon$ of $E$ such that $\mu_n(F_\varepsilon) \geqslant 1 - \varepsilon$ for all natural integer $n$. Suppose further that the sequence $(\mu_n)_{n \in \mathbb{N}}$ is tight. Show then that $\mu_\infty$ defines an element of $\mathscr{M}(E)$ which is a cluster point of the sequence $(\mu_n)_{n \in \mathbb{N}}$ in $\mathscr{B}(\mathscr{P}(E), \mathbb{R})$.

We consider $n$ discrete random variables $Y_1, \ldots, Y_n$ defined on $(\Omega, \mathcal{B}, \mathbb{P})$ with real values and define the random vector $Y(\omega) = \left(\begin{array}{c} Y_1(\omega) \\ \vdots \\ Y_n(\omega) \end{array}\right)$. The covariance matrix $\Sigma_Y$ has general term $\sigma_{i,j} = \operatorname{cov}(Y_i, Y_j)$.
Verify that $\Sigma_Y$ is a symmetric matrix, that $$\Sigma_Y = \mathbb{E}\left((Y - \mathbb{E}(Y))(Y - \mathbb{E}(Y))^\top\right)$$ and that, if $U$ is a constant vector in $\mathcal{M}_{n,1}(\mathbb{R})$, then $$\Sigma_{Y+U} = \Sigma_Y.$$

We consider $n$ discrete random variables $Y_1, \ldots, Y_n$ with random vector $Y$ and covariance matrix $\Sigma_Y$. Let $p \in \mathbb{N}^*$ and $M \in \mathcal{M}_{p,n}(\mathbb{R})$. We define the discrete random variable $Z = MY$, with values in $\mathcal{M}_{p,1}(\mathbb{R})$. Justify that $Z$ admits an expectation and express $\mathbb{E}(Z)$ in terms of $\mathbb{E}(Y)$. Show that $Z$ admits a covariance matrix $\Sigma_Z$ and that $$\Sigma_Z = M \Sigma_Y M^\top.$$

Let $(\Omega, \mathscr{A}, P)$ be a probability space. Let $(X_n)_{n \in \mathbb{N}}$ be a sequence of random variables defined on $(\Omega, \mathscr{A}, P)$ and taking values in $\mathbb{N}^*$ and let $X$ be a random variable defined on $(\Omega, \mathscr{A}, P)$ and taking values in $\mathbb{N}^*$. Assume that:

1. [i.] The sequence $(\mu_{X_n})_{n \in \mathbb{N}}$ is tight.
2. [ii.] For all $r \in \mathbb{N}^*$, $\lim_{n \rightarrow +\infty} P(r \mid X_n) = P(r \mid X)$.

Show that then the sequence $(\mu_{X_n})_{n \in \mathbb{N}}$ converges to $\mu_X$ in $\mathscr{B}(\mathscr{P}(\mathbb{N}^*), \mathbb{R})$.

Let $(X_n)_{n \in \mathbb{N}}$ be a sequence of random variables defined on $(\Omega, \mathscr{A}, P)$ and taking values in $\mathbb{N}^*$ and let $X$ be a random variable defined on $(\Omega, \mathscr{A}, P)$ and taking values in $\mathbb{N}^*$. Assume that:

1. [i.] The sequence $(\mu_{X_n})_{n \in \mathbb{N}}$ is tight.
2. [ii.] For all $r \in \mathbb{N}^*$, $\lim_{n \rightarrow +\infty} \mathbf{P}(r \mid X_n) = \mathbf{P}(r \mid X)$.

Show that the sequence $(\mu_{X_n})_{n \in \mathbb{N}}$ converges to $\mu_X$ in $\mathscr{B}(\mathscr{P}(\mathbb{N}^*), \mathbb{R})$.