Let $X$ be an infinitely divisible random variable taking values in $\mathbb{N}$ and such that $\mathbb{P}(X = 0) > 0$. For all $n \in \mathbb{N}^{*}$, there exist $n$ independent random variables $X_{n,1}, \ldots, X_{n,n}$ with the same distribution such that the random variable $X_{n,1} + \cdots + X_{n,n}$ follows the distribution of $X$.
For all $k \in \mathbb{N}^{*}$, show that the sequence $\left(n\mathbb{P}\left(X_{n,1} = k\right)\right)_{n \in \mathbb{N}^{*}}$ converges to $\lambda_{k}$. Deduce that $X$ is $\lambda$-positive.

a) Show the result announced at the beginning of subsection III.C, namely that the following three assertions are equivalent: (i) $X$ is infinitely divisible; (ii) $X$ is $\lambda$-positive; (iii) there exists a sequence $\left(X_{i}\right)_{i \geqslant 1}$ of independent Poisson variables, as in II.C.3, such that $X \sim \sum_{i=1}^{\infty} i X_{i}$.
b) How to adapt this result to random variables taking values in $\mathbb{N}^{*}$?
c) Let $X$ be a random variable following the geometric distribution $\mathcal{G}(p)$, where $p \in ]0,1[$: $$\forall k \in \mathbb{N}^{*} \quad \mathbb{P}(X = k) = (1-p)^{k-1}p$$ Is the random variable $X$ infinitely divisible?

Deduce that, when $n$ tends to $+\infty$, $$P\left(S_n > n\right) \sim \frac{\mathrm{e}^{-n/2}}{n!} \left(\frac{n}{2}\right)^n.$$

Deduce, using Stirling's formula, that there exists a real $\alpha \in ]0,1[$ such that $P\left(S_n > n\right) = O\left(\alpha^n\right)$.

We define, for all $x \in \mathbb{R}$, $\lambda^*(x) = \sup_{t \geqslant 0}(tx - \lambda(t))$. We admit that the convergence of the sequence of functions $\left(t \mapsto \frac{\ln\left(E\left(\mathrm{e}^{tS_n}\right)\right)}{n}\right)_{n \in \mathbb{N}^*}$ towards $t \mapsto \ln(\gamma(t))$ is uniform on $\mathbb{R}^+$. Let $\varepsilon > 0$. Show that there exists a rank $n_0 \in \mathbb{N}^*$ such that, for all $t \in \mathbb{R}^+$ and for all $n \in \mathbb{N}^*$, $$n \geqslant n_0 \Longrightarrow \ln\left(E\left(\mathrm{e}^{tS_n}\right)\right) \leqslant n(\lambda(t) + \varepsilon).$$

We define, for all $x \in \mathbb{R}$, $\lambda^*(x) = \sup_{t \geqslant 0}(tx - \lambda(t))$. Let $\varepsilon > 0$ and $n_0$ as in Q35. Using Markov's inequality applied to the random variable $e^{tS_n}$, show that for $a > 1$, $n \geqslant n_0$ and $t \geqslant 0$, $$P\left(S_n \geqslant nam\right) \leqslant \mathrm{e}^{-ntam} \mathrm{e}^{n(\lambda(t) + \varepsilon)}.$$

We define, for all $x \in \mathbb{R}$, $\lambda^*(x) = \sup_{t \geqslant 0}(tx - \lambda(t))$. Let $\varepsilon > 0$ and $n_0$ as in Q35. Deduce that for $n \geqslant n_0$, $$P\left(S_n \geqslant nam\right) \leqslant \mathrm{e}^{-n\left(\lambda^*(am) - \varepsilon\right)}.$$

Give a concrete meaning to $m = \lim_{n \rightarrow +\infty} \frac{1}{n} E(S_n)$ in relation to the industrial process studied and interpret the inequality $P\left(S_n \geqslant nam\right) \leqslant \mathrm{e}^{-n(\lambda^*(am) - \varepsilon)}$. One may establish an intuitive link with the law of large numbers.

We have two finite sequences of real numbers $0 = t_1 < t_2 < \cdots < t_K$ ($K \geqslant 2$) and $x_1 < x_2 < \cdots < x_L$ ($L \geqslant 2$). The formula from question 32 applied at $t_i$ with $n$ sufficiently large allows us to estimate $\lambda(t_i)$ by an approximate value $\hat{\lambda}(t_i)$. Justify that for all $i \in \{1, \ldots, L\}$, $$\hat{\lambda}^*\left(x_i\right) = \max_{1 \leqslant j \leqslant K} \left(t_j x_i - \hat{\lambda}\left(t_j\right)\right)$$ constitutes a reasonable approximate value of $\lambda^*\left(x_i\right)$.

Using Table 1 below, give an approximate bound for the value of $m$ and the value of a real number $h > 0$ such that there exists a rank $n_0 \in \mathbb{N}^*$ satisfying for all $n \geqslant n_0$, $$P\left(S_n > 1{,}1 \times nm\right) \leqslant \mathrm{e}^{-nh}.$$

$x_i$	4,50	4,55	4,60	4,65	4,70
$\hat{\lambda}^*(x_i)$	$4{,}1 \times 10^{-12}$	$4{,}1 \times 10^{-12}$	$4{,}1 \times 10^{-12}$	$4{,}1 \times 10^{-12}$	$4{,}1 \times 10^{-12}$
$x_i$	4,75	4,80	4,85	4,90	4,95
$\hat{\lambda}^*(x_i)$	$5{,}1 \times 10^{-4}$	$5{,}5 \times 10^{-3}$	$1{,}1 \times 10^{-2}$	$1{,}6 \times 10^{-2}$	$2{,}1 \times 10^{-2}$
$x_i$	5,00	5,05	5,10	5,15	5,20
$\hat{\lambda}^*(x_i)$	$2{,}6 \times 10^{-2}$	$3{,}1 \times 10^{-2}$	$3{,}6 \times 10^{-2}$	$4{,}1 \times 10^{-2}$	$4{,}6 \times 10^{-2}$
$x_i$	5,25	5,30	5,35	5,40	5,45
$\hat{\lambda}^*(x_i)$	$5{,}1 \times 10^{-2}$	$5{,}6 \times 10^{-2}$	$6{,}1 \times 10^{-2}$	$6{,}6 \times 10^{-2}$	$7{,}1 \times 10^{-2}$

What relation exists between $S _ { n }$ and $Y _ { n }$ ? Deduce the expectation and variance of $S _ { n }$. Justify that $S _ { n }$ and $n$ have the same parity.

Let $n \in \mathbb { N } ^ { * }$ and let $\gamma = \left( \gamma _ { 1 } , \ldots , \gamma _ { 2 n } \right)$ be a Dyck path of length $2 n$. For $t \in \mathbb { N }$, express $\mathbb { P } \left( A _ { t , \gamma } \right)$ as a function of $n$ and $p$, where $A _ { t , \gamma } = \bigcap _ { k = 1 } ^ { m } \left( X _ { t + k } = \gamma _ { k } \right)$.

Let $T$ be the random variable, defined on $\Omega$ and taking values in $\mathbb { N }$, equal to the first instant when the particle returns to the origin, if this instant exists, and equal to 0 if the particle never returns to the origin: $$\forall \omega \in \Omega , \quad T ( \omega ) = \begin{cases} 0 & \text { if } \forall k \in \mathbb { N } ^ { * } , S _ { k } ( \omega ) \neq 0 \\ \min \left\{ k \in \mathbb { N } ^ { * } \mid S _ { k } ( \omega ) = 0 \right\} & \text { otherwise } \end{cases}$$ Show that $T$ takes even values and that, for all $n \in \mathbb { N } , \mathbb { P } ( T = 2 n + 2 ) = 2 C _ { n } p ^ { n + 1 } ( 1 - p ) ^ { n + 1 }$.

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 $\mathscr{B}(\mathscr{P}(\mathbb{N}^*), \mathbb{R})$.

For every integer $k \geqslant 2$, we set $\zeta(k) = \sum_{n=1}^{+\infty} n^{-k}$.
Let $s \in \mathbb{N}^*$. For $n \in \mathbb{N}$, let $X_n^{(1)}, X_n^{(2)}, \ldots, X_n^{(s)}$ be $s$ mutually independent random variables all following the uniform distribution on $\{1, 2, \ldots, n\}$.
We denote $Z_n^{(s)} = X_n^{(1)} \wedge \ldots \wedge X_n^{(s)}$ the gcd of $X_n^{(1)}, \ldots, X_n^{(s)}$.
For $r \in \mathbb{N}^*$ and $i \in \{1, 2, \ldots, s\}$, calculate $P(r \mid X_n^{(i)})$ and show that $P(r \mid X_n^{(i)}) \leqslant \frac{1}{r}$. Deduce that $$\lim_{n \rightarrow +\infty} P\left(r \mid Z_n^{(s)}\right) = \frac{1}{r^s}.$$

For every integer $k \geqslant 2$, we set $\zeta(k) = \sum_{n=1}^{+\infty} n^{-k}$.
For $s > 1$ fixed, we define a probability distribution $\mu_s$ on $\mathbb{N}^*$ by setting, for $n \in \mathbb{N}^*$, $$\mu_s(\{n\}) = \frac{1}{\zeta(s) n^s}.$$
Let $Z$ be a random variable defined on $(\Omega, \mathscr{A}, P)$ that follows the distribution $\mu_s$. Calculate $P(k \mid Z)$ for $k \in \mathbb{N}^*$.

For every integer $k \geqslant 2$, we set $\zeta(k) = \sum_{n=1}^{+\infty} n^{-k}$.
For $s > 1$ fixed, we define a probability distribution $\mu_s$ on $\mathbb{N}^*$ by setting, for $n \in \mathbb{N}^*$, $$\mu_s(\{n\}) = \frac{1}{\zeta(s) n^s}.$$
Let $s \geqslant 2$ be an integer. Let $X_n^{(1)}, X_n^{(2)}, \ldots, X_n^{(s)}$ be $s$ mutually independent random variables all following the uniform distribution on $\{1, 2, \ldots, n\}$, and let $Z_n^{(s)} = X_n^{(1)} \wedge \ldots \wedge X_n^{(s)}$ be their gcd.
Deduce that the sequence $(\mu_{Z_n^{(s)}})_{n \in \mathbb{N}}$ converges in $\mathscr{B}(\mathscr{P}(\mathbb{N}^*), \mathbb{R})$ to $\mu_s$.

For every integer $k \geqslant 2$, we set $\zeta(k) = \sum_{n=1}^{+\infty} n^{-k}$.
Let $s, n \in \mathbb{N}^*$ with $2 \leqslant s \leqslant n$. We randomly draw $s$ numbers from $\{1, 2, \ldots, n\}$ and we denote $P_n(s)$ the probability that these numbers are coprime. Show that $$\lim_{n \rightarrow +\infty} P_n(s) = \frac{1}{\zeta(s)}$$ and give the value of $\lim_{n \rightarrow +\infty} P_n(s)$ in the case where $s = 2$, then $s = 4$, and finally $s = 6$.

Let $( \Omega , \mathcal { A } , \mathbb { P } )$ be a probability space and $X$ a discrete random variable such that $\mathbb { P } ( X = - 1 ) = 1 / 2$ and $\mathbb { P } ( X = 1 ) = 1 / 2$. Consider a sequence $\left( X _ { i } \right) _ { i \in \mathbb { N } ^ { * } }$ of mutually independent discrete random variables with the same distribution as $X$. Set $S _ { 0 } = 0$ and $S _ { n } = \sum _ { i = 1 } ^ { n } X _ { i }$. The functions $\varphi$ and $\Phi$ are defined by $\varphi ( x ) = \frac { 1 } { \sqrt { 2 \pi } } \mathrm { e } ^ { - x ^ { 2 } / 2 }$ and $\Phi ( x ) = \int _ { - \infty } ^ { x } \varphi ( t ) \mathrm { d } t$.
Deduce that we have $$\lim _ { n \rightarrow + \infty } \mathbb { P } \left( \left\{ u \leqslant \frac { S _ { n } } { \sqrt { n } } \leqslant v \right\} \right) = \int _ { u } ^ { v } \varphi ( x ) \mathrm { d } x$$ then that $$\lim _ { n \rightarrow + \infty } \mathbb { P } \left( \left\{ u \leqslant \frac { S _ { n } } { \sqrt { n } } \right\} \right) = 1 - \Phi ( u )$$

Let $( \Omega , \mathcal { A } , \mathbb { P } )$ be a probability space and $X$ a discrete random variable such that $\mathbb { P } ( X = - 1 ) = 1 / 2$ and $\mathbb { P } ( X = 1 ) = 1 / 2$. Consider a sequence $\left( X _ { i } \right) _ { i \in \mathbb { N } ^ { * } }$ of mutually independent discrete random variables with the same distribution as $X$. Set $S _ { 0 } = 0$ and $S _ { n } = \sum _ { i = 1 } ^ { n } X _ { i }$. Fix $\varepsilon \in ] 0 , 1 [$.
Show that there exists $x _ { 0 } \geqslant 1$ such that we have $$\forall x \geqslant x _ { 0 } , \quad \exists n _ { x } \in \mathbb { N } , \quad \forall n \geqslant n _ { x } , \quad x ^ { 2 } \mathbb { P } \left( \left\{ \left| S _ { n } \right| \geqslant x \sqrt { n } \right\} \right) \leqslant \varepsilon .$$

Let $( \Omega , \mathcal { A } , \mathbb { P } )$ be a probability space and $X$ a discrete random variable such that $\mathbb { P } ( X = - 1 ) = 1 / 2$ and $\mathbb { P } ( X = 1 ) = 1 / 2$. Consider a sequence $\left( X _ { i } \right) _ { i \in \mathbb { N } ^ { * } }$ of mutually independent discrete random variables with the same distribution as $X$. Set $S _ { 0 } = 0$ and $S _ { n } = \sum _ { i = 1 } ^ { n } X _ { i }$. Fix $\varepsilon \in ] 0 , 1 [$.
For $x _ { 0 }$ and $x$ as in the previous question, we fix $N \geqslant \frac { n _ { x } } { \varepsilon }$ and we choose $n \geqslant N$. Show that then $$x ^ { 2 } \mathbb { P } \left( \left\{ \max _ { 1 \leqslant p \leqslant n } \left| S _ { p } \right| \geqslant 3 x \sqrt { n } \right\} \right) \leqslant 3 \varepsilon$$