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 $k \in \{ 1 , \ldots , n \}$, $$P \left[ X _ { k } = 1 \right] = P \left[ X _ { k } = - 1 \right] = \frac { 1 } { 2 }$$ For each $\lambda \geqslant 0$, we set $$m ( \lambda ) = \frac { E \left[ X _ { 1 } \exp \left( \lambda X _ { 1 } \right) \right] } { E \left[ \exp \left( \lambda X _ { 1 } \right) \right] }$$ Show that the function $m$ is strictly increasing on $\mathbb { R } _ { + }$, and that for all $t \in [ 0,1 [$, there exists a unique $\lambda \geqslant 0$ such that $m ( \lambda ) = t$.

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 $\lambda \in \mathbb{R}$, $$\psi(\lambda) = \log\left(\frac{1}{2}e^{\lambda} + \frac{1}{2}e^{-\lambda}\right)$$ For each $\lambda \geqslant 0$, we set $$m(\lambda) = \frac{E[X_1 \exp(\lambda X_1)]}{E[\exp(\lambda X_1)]}$$ as well as $$D_n(\lambda) = \exp(\lambda n S_n - n \psi(\lambda))$$
(a) For $n \geqslant 2$ and $\lambda \geqslant 0$, show that $$E[(X_1 - m(\lambda))(X_2 - m(\lambda)) D_n(\lambda)] = 0$$
(b) Deduce that, for $n \geqslant 1$ and $\lambda \geqslant 0$, $$E[(S_n - m(\lambda))^2 D_n(\lambda)] \leqslant \frac{4}{n}.$$

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 $$S _ { n } = \frac { 1 } { n } \sum _ { k = 1 } ^ { n } X _ { k }$$ as well as, for all $\lambda \in \mathbb { R }$, $$\psi ( \lambda ) = \log \left( \frac { 1 } { 2 } e ^ { \lambda } + \frac { 1 } { 2 } e ^ { - \lambda } \right)$$ For each $\lambda \geqslant 0$, we set $$m ( \lambda ) = \frac { E \left[ X _ { 1 } \exp \left( \lambda X _ { 1 } \right) \right] } { E \left[ \exp \left( \lambda X _ { 1 } \right) \right] }$$ as well as $$D _ { n } ( \lambda ) = \exp \left( \lambda n S _ { n } - n \psi ( \lambda ) \right)$$
(a) For $n \geqslant 2$ and $\lambda \geqslant 0$, show that $$E \left[ \left( X _ { 1 } - m ( \lambda ) \right) \left( X _ { 2 } - m ( \lambda ) \right) D _ { n } ( \lambda ) \right] = 0$$
(b) Deduce that, for $n \geqslant 1$ and $\lambda \geqslant 0$, $$E \left[ \left( S _ { n } - m ( \lambda ) \right) ^ { 2 } D _ { n } ( \lambda ) \right] \leqslant \frac { 4 } { n }$$

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 $$S_n = \frac{1}{n} \sum_{k=1}^{n} X_k$$ For each $\lambda \geqslant 0$, we set $$m(\lambda) = \frac{E[X_1 \exp(\lambda X_1)]}{E[\exp(\lambda X_1)]}$$ For all $n \geqslant 1$, $\lambda \geqslant 0$ and $\varepsilon > 0$, we denote by $I_n(\lambda, \varepsilon)$ the random variable defined by $$I_n(\lambda, \varepsilon) = \begin{cases} 1 & \text{if } |S_n - m(\lambda)| \leqslant \varepsilon \\ 0 & \text{otherwise.} \end{cases}$$ Show that $$P[|S_n - m(\lambda)| \leqslant \varepsilon] \geqslant E[I_n(\lambda, \varepsilon) \exp(\lambda n(S_n - m(\lambda) - \varepsilon))],$$

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 $$S _ { n } = \frac { 1 } { n } \sum _ { k = 1 } ^ { n } X _ { k }$$ For each $\lambda \geqslant 0$, we set $$m ( \lambda ) = \frac { E \left[ X _ { 1 } \exp \left( \lambda X _ { 1 } \right) \right] } { E \left[ \exp \left( \lambda X _ { 1 } \right) \right] }$$ For all $n \geqslant 1 , \lambda \geqslant 0$ and $\varepsilon > 0$, we denote by $I _ { n } ( \lambda , \varepsilon )$ the random variable defined by $$I _ { n } ( \lambda , \varepsilon ) = \begin{cases} 1 & \text { if } \left| S _ { n } - m ( \lambda ) \right| \leqslant \varepsilon \\ 0 & \text { otherwise } \end{cases}$$ Show that $$P \left[ \left| S _ { n } - m ( \lambda ) \right| \leqslant \varepsilon \right] \geqslant E \left[ I _ { n } ( \lambda , \varepsilon ) \exp \left( \lambda n \left( S _ { n } - m ( \lambda ) - \varepsilon \right) \right] , \right.$$

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 $\lambda \in \mathbb{R}$, $$\psi(\lambda) = \log\left(\frac{1}{2}e^{\lambda} + \frac{1}{2}e^{-\lambda}\right)$$ For each $\lambda \geqslant 0$, we set $$m(\lambda) = \frac{E[X_1 \exp(\lambda X_1)]}{E[\exp(\lambda X_1)]}$$ as well as $$D_n(\lambda) = \exp(\lambda n S_n - n \psi(\lambda))$$ For all $n \geqslant 1$, $\lambda \geqslant 0$ and $\varepsilon > 0$, we denote by $I_n(\lambda, \varepsilon)$ the random variable defined by $$I_n(\lambda, \varepsilon) = \begin{cases} 1 & \text{if } |S_n - m(\lambda)| \leqslant \varepsilon \\ 0 & \text{otherwise.} \end{cases}$$ Show that $$E[I_n(\lambda, \varepsilon) D_n(\lambda)] \geqslant 1 - \frac{4}{n\varepsilon^2}$$

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 $$S _ { n } = \frac { 1 } { n } \sum _ { k = 1 } ^ { n } X _ { k }$$ as well as, for all $\lambda \in \mathbb { R }$, $$\psi ( \lambda ) = \log \left( \frac { 1 } { 2 } e ^ { \lambda } + \frac { 1 } { 2 } e ^ { - \lambda } \right)$$ For each $\lambda \geqslant 0$, we set $$m ( \lambda ) = \frac { E \left[ X _ { 1 } \exp \left( \lambda X _ { 1 } \right) \right] } { E \left[ \exp \left( \lambda X _ { 1 } \right) \right] }$$ as well as $$D _ { n } ( \lambda ) = \exp \left( \lambda n S _ { n } - n \psi ( \lambda ) \right)$$ For all $n \geqslant 1 , \lambda \geqslant 0$ and $\varepsilon > 0$, we denote by $I _ { n } ( \lambda , \varepsilon )$ the random variable defined by $$I _ { n } ( \lambda , \varepsilon ) = \begin{cases} 1 & \text { if } \left| S _ { n } - m ( \lambda ) \right| \leqslant \varepsilon \\ 0 & \text { otherwise } \end{cases}$$ Show that $$E \left[ I _ { n } ( \lambda , \varepsilon ) D _ { n } ( \lambda ) \right] \geqslant 1 - \frac { 4 } { n \varepsilon ^ { 2 } }$$

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 $\lambda \in \mathbb{R}$, $$\psi(\lambda) = \log\left(\frac{1}{2}e^{\lambda} + \frac{1}{2}e^{-\lambda}\right)$$ For each $\lambda \geqslant 0$, we set $$m(\lambda) = \frac{E[X_1 \exp(\lambda X_1)]}{E[\exp(\lambda X_1)]}$$ as well as $$D_n(\lambda) = \exp(\lambda n S_n - n \psi(\lambda))$$ For all $n \geqslant 1$, $\lambda \geqslant 0$ and $\varepsilon > 0$, we denote by $I_n(\lambda, \varepsilon)$ the random variable defined by $$I_n(\lambda, \varepsilon) = \begin{cases} 1 & \text{if } |S_n - m(\lambda)| \leqslant \varepsilon \\ 0 & \text{otherwise.} \end{cases}$$
(a) Deduce, for each $\lambda \geqslant 0$ and $\varepsilon > 0$, the existence of a sequence $(u_n(\varepsilon))_{n \geqslant 1}$ that tends to 0 as $n$ tends to infinity and such that $$\frac{1}{n} \log P[S_n \geqslant m(\lambda) - \varepsilon] \geqslant \psi(\lambda) - \lambda m(\lambda) - \lambda \varepsilon + u_n(\varepsilon)$$
(b) Conclude that for all $t \in [0,1[$, $$\lim_{n \rightarrow \infty} \frac{1}{n} \log P[S_n \geqslant t] = \inf_{\lambda \geqslant 0} (\psi(\lambda) - \lambda t).$$
(c) Is the preceding formula still valid for $t = 1$?

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 $$S _ { n } = \frac { 1 } { n } \sum _ { k = 1 } ^ { n } X _ { k }$$ as well as, for all $\lambda \in \mathbb { R }$, $$\psi ( \lambda ) = \log \left( \frac { 1 } { 2 } e ^ { \lambda } + \frac { 1 } { 2 } e ^ { - \lambda } \right)$$ For each $\lambda \geqslant 0$, we set $$m ( \lambda ) = \frac { E \left[ X _ { 1 } \exp \left( \lambda X _ { 1 } \right) \right] } { E \left[ \exp \left( \lambda X _ { 1 } \right) \right] }$$ as well as $$D _ { n } ( \lambda ) = \exp \left( \lambda n S _ { n } - n \psi ( \lambda ) \right)$$ For all $n \geqslant 1 , \lambda \geqslant 0$ and $\varepsilon > 0$, we denote by $I _ { n } ( \lambda , \varepsilon )$ the random variable defined by $$I _ { n } ( \lambda , \varepsilon ) = \begin{cases} 1 & \text { if } \left| S _ { n } - m ( \lambda ) \right| \leqslant \varepsilon \\ 0 & \text { otherwise } \end{cases}$$
(a) Deduce, for each $\lambda \geqslant 0$ and $\varepsilon > 0$, the existence of a sequence $\left( u _ { n } ( \varepsilon ) \right) _ { n \geqslant 1 }$ that tends to 0 as $n$ tends to infinity and such that $$\frac { 1 } { n } \log P \left[ S _ { n } \geqslant m ( \lambda ) - \varepsilon \right] \geqslant \psi ( \lambda ) - \lambda m ( \lambda ) - \lambda \varepsilon + u _ { n } ( \varepsilon )$$
(b) Conclude that for all $t \in [ 0,1 [$, $$\lim _ { n \rightarrow \infty } \frac { 1 } { n } \log P \left[ S _ { n } \geqslant t \right] = \inf _ { \lambda \geqslant 0 } ( \psi ( \lambda ) - \lambda t )$$
(c) Is the preceding formula still valid for $t = 1$ ?

Show that for all $t \in \mathbb { R } , \left| \phi _ { X } ( t ) \right| \leqslant 1$.

Show that, if there exist $a \in \mathbb { R }$ and $t _ { 0 } \in \mathbb { R } ^ { * }$ such that $X ( \Omega ) \subset a + \frac { 2 \pi } { t _ { 0 } } \mathbb { Z }$, then $\left| \phi _ { X } \left( t _ { 0 } \right) \right| = 1$.

We assume that there exists $t _ { 0 } \in \mathbb { R } ^ { * }$ such that $\left| \phi _ { X } \left( t _ { 0 } \right) \right| = 1$. We assume further that $X ( \Omega )$ is countable and we use the notations of question 2 (i.e. $X ( \Omega ) = \left\{ x _ { n } , n \in \mathbb { N } \right\}$ with $a _ { n } = \mathbb { P } \left( X = x _ { n } \right)$). Show that there exists $a \in \mathbb { R }$ such that $\sum _ { n = 0 } ^ { + \infty } a _ { n } \exp \left( \mathrm { i } \left( t _ { 0 } x _ { n } - t _ { 0 } a \right) \right) = 1$.

We assume that there exists $t _ { 0 } \in \mathbb { R } ^ { * }$ such that $\left| \phi _ { X } \left( t _ { 0 } \right) \right| = 1$. We assume further that $X ( \Omega )$ is countable and we use the notations of question 2 (i.e. $X ( \Omega ) = \left\{ x _ { n } , n \in \mathbb { N } \right\}$ with $a _ { n } = \mathbb { P } \left( X = x _ { n } \right)$). Using the result of Q11, deduce that $\sum _ { n = 0 } ^ { + \infty } a _ { n } \left( 1 - \cos \left( t _ { 0 } x _ { n } - t _ { 0 } a \right) \right) = 0$.

We assume that there exists $t _ { 0 } \in \mathbb { R } ^ { * }$ such that $\left| \phi _ { X } \left( t _ { 0 } \right) \right| = 1$. We assume further that $X ( \Omega )$ is countable and we use the notations of question 2 (i.e. $X ( \Omega ) = \left\{ x _ { n } , n \in \mathbb { N } \right\}$ with $a _ { n } = \mathbb { P } \left( X = x _ { n } \right)$). Show that for all $n \in \mathbb { N }$, if $a _ { n } \neq 0$, then $x _ { n } \in a + \frac { 2 \pi } { t _ { 0 } } \mathbb { Z }$.

We assume that there exists $t _ { 0 } \in \mathbb { R } ^ { * }$ such that $\left| \phi _ { X } \left( t _ { 0 } \right) \right| = 1$. We assume further that $X ( \Omega )$ is countable and we use the notations of question 2 (i.e. $X ( \Omega ) = \left\{ x _ { n } , n \in \mathbb { N } \right\}$ with $a _ { n } = \mathbb { P } \left( X = x _ { n } \right)$). Deduce that $\mathbb { P } \left( X \in a + \frac { 2 \pi } { t _ { 0 } } \mathbb { Z } \right) = 1$.

Let $X$ be a real and discrete random variable and $m \in \mathbb { R }$. For $T \in \mathbb { R } _ { + } ^ { * }$, we set $V _ { m } ( T ) = \frac { 1 } { 2 T } \int _ { - T } ^ { T } \phi _ { X } ( t ) \mathrm { e } ^ { - \mathrm { i } m t } \mathrm {~d} t$. We assume that $X ( \Omega )$ is finite and we use the notations of question 1: $X ( \Omega ) = \left\{ x _ { 1 } , \ldots , x _ { r } \right\}$ with $a _ { k } = \mathbb { P } \left( X = x _ { k } \right)$. Show that, for all $T \in \mathbb { R } _ { + } ^ { * }$, we have $V _ { m } ( T ) = \sum _ { n = 1 } ^ { r } \operatorname { sinc } \left( T \left( x _ { n } - m \right) \right) \mathbb { P } \left( X = x _ { n } \right)$.

Let $X$ be a real and discrete random variable and $m \in \mathbb { R }$. For $T \in \mathbb { R } _ { + } ^ { * }$, we set $V _ { m } ( T ) = \frac { 1 } { 2 T } \int _ { - T } ^ { T } \phi _ { X } ( t ) \mathrm { e } ^ { - \mathrm { i } m t } \mathrm {~d} t$. We assume that $X ( \Omega )$ is countable and we use the notations of question 2: $X ( \Omega ) = \left\{ x _ { n } , n \in \mathbb { N } \right\}$ with $a _ { n } = \mathbb { P } \left( X = x _ { n } \right)$. For $n \in \mathbb { N }$ and $h \in \mathbb { R } _ { + } ^ { * }$, we set $g _ { n } ( h ) = \operatorname { sinc } \left( \frac { x _ { n } - m } { h } \right) \mathbb { P } \left( X = x _ { n } \right)$. Show that for all $T \in \mathbb { R } _ { + } ^ { * }$, we have $V _ { m } ( T ) = \sum _ { n = 0 } ^ { + \infty } g _ { n } \left( \frac { 1 } { T } \right)$.

Let $X$ be a real and discrete random variable and $m \in \mathbb { R }$. For $n \in \mathbb { N }$ and $h \in \mathbb { R } _ { + } ^ { * }$, we set $g _ { n } ( h ) = \operatorname { sinc } \left( \frac { x _ { n } - m } { h } \right) \mathbb { P } \left( X = x _ { n } \right)$. Show that the function $g _ { n }$ extends to a function $\tilde { g } _ { n }$ defined and continuous on $\mathbb { R } ^ { + }$.

Let $X$ be a real and discrete random variable and $m \in \mathbb { R }$. For $n \in \mathbb { N }$ and $h \in \mathbb { R } _ { + } ^ { * }$, we set $g _ { n } ( h ) = \operatorname { sinc } \left( \frac { x _ { n } - m } { h } \right) \mathbb { P } \left( X = x _ { n } \right)$, and $\tilde{g}_n$ denotes its continuous extension to $\mathbb{R}^+$. Show that the function $G = \sum _ { n = 0 } ^ { + \infty } \tilde { g } _ { n }$ is defined and continuous on $\mathbb { R } ^ { + }$.

Let $X : \Omega \rightarrow \mathbb { R }$ be a real-valued random variable. We assume that $X ( \Omega )$ is finite and we use the notation from question 1: $X ( \Omega ) = \left\{ x _ { 1 } , \ldots , x _ { r } \right\}$ with $a _ { k } = \mathbb { P } \left( X = x _ { k } \right)$. Show that $\phi _ { X }$ is expandable as a power series on $\mathbb { R }$ and, for all real $t , \phi _ { X } ( t ) = \sum _ { n = 0 } ^ { + \infty } \frac { ( \mathrm { i } t ) ^ { n } } { n ! } \mathbb { E } \left( X ^ { n } \right)$.

Let $X : \Omega \rightarrow \mathbb { R }$ be a real-valued random variable. We assume that $X ( \Omega )$ is countable and we use the notation from question 2: $X ( \Omega ) = \left\{ x _ { n } , n \in \mathbb { N } \right\}$ with $a _ { n } = \mathbb { P } \left( X = x _ { n } \right)$. We also assume that, for all integer $n \in \mathbb { N } , X$ admits a moment of order $n$ and that there exists a real $R > 0$ such that $$\mathbb { E } \left( | X | ^ { n } \right) = O \left( \frac { n ^ { n } } { R ^ { n } } \right) \quad \text { when } n \rightarrow + \infty$$ Show that for all $n \in \mathbb { N }$ and all $y \in \mathbb { R } , \left| \mathrm { e } ^ { \mathrm { i } y } - \sum _ { k = 0 } ^ { n } \frac { ( \mathrm { i } y ) ^ { k } } { k ! } \right| \leqslant \frac { | y | ^ { n + 1 } } { ( n + 1 ) ! }$.

Let $X : \Omega \rightarrow \mathbb { R }$ be a real-valued random variable. We assume that $X ( \Omega )$ is countable and we use the notation from question 2: $X ( \Omega ) = \left\{ x _ { n } , n \in \mathbb { N } \right\}$ with $a _ { n } = \mathbb { P } \left( X = x _ { n } \right)$. We also assume that, for all integer $n \in \mathbb { N } , X$ admits a moment of order $n$ and that there exists a real $R > 0$ such that $$\mathbb { E } \left( | X | ^ { n } \right) = O \left( \frac { n ^ { n } } { R ^ { n } } \right) \quad \text { when } n \rightarrow + \infty$$ Using the result of Q39, deduce that for all real $t \in \left[ - \frac { R } { \mathrm { e } } , \frac { R } { \mathrm { e } } \right]$, $$\phi _ { X } ( t ) = \sum _ { k = 0 } ^ { + \infty } \frac { ( \mathrm { i } t ) ^ { k } } { k ! } \mathbb { E } \left( X ^ { k } \right)$$

Let $s > 1$ be a real number and let $X$ be a random variable taking values in $\mathbb{N}^*$ following the zeta distribution with parameter $s$. If $n \in \mathbb{N}^*$, we set $g(n) = r_1(n) - r_3(n)$ where $r_i(n) = \operatorname{Card}\{d \in \mathbb{N} : d \equiv i [4] \text{ and } d \mid n\}$.
Show that $E(g(X)) = \lim_{n \rightarrow +\infty} \prod_{k=1}^{n} E\left(g\left(p_k^{\nu_{p_k}(X)}\right)\right)$.

Let $s > 1$ be a real number and let $X$ be a random variable taking values in $\mathbb{N}^*$ following the zeta distribution with parameter $s$.
Deduce $$E(g(X)) = \lim_{n \rightarrow +\infty} \prod_{k=1}^{n} \frac{1}{1 - \chi_4\left(p_k\right) p_k^{-s}}.$$

Let $s > 1$ be a real number and let $X$ be a random variable taking values in $\mathbb{N}^*$ following the zeta distribution with parameter $s$. We recall that $\chi_4(2n) = 0$ and $\chi_4(2n-1) = (-1)^{n-1}$ for $n \in \mathbb{N}^*$.
Show that, if $p$ is a prime number, $$E\left(\chi_4\left(p^{\nu_p(X)}\right)\right) = \frac{1 - p^{-s}}{1 - \chi_4(p) p^{-s}}.$$