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 $n \in \mathbb{N}^*$, $$n \geqslant n_0 \Longrightarrow \ln\left(E\left(\mathrm{e}^{tS_n}\right)\right) \leqslant n(\lambda(t) + \varepsilon).$$

Let $(E, \langle \cdot, \cdot \rangle)$ be a reproducing kernel Hilbert space on an interval $I$, with reproducing kernel $K$. For all $(x,y) \in I^2$, we set $k_x(y) = K(x,y)$. Let $x \in I$ and $V_x$ defined on $E$ by $V_x(f) = f(x)$. We set $$N(V_x) = \sup_{\|f\|=1} |f(x)|$$ Prove that $$N(V_x) = \sqrt{\langle k_x, k_x \rangle}.$$

Let $(E, \langle \cdot, \cdot \rangle)$ be a reproducing kernel Hilbert space on an interval $I$, with reproducing kernel $K$. For all $(x,y) \in I^2$, we set $k_x(y) = K(x,y)$. Let $x \in I$ and $V_x$ defined on $E$ by $V_x(f) = f(x)$. We set $$N(V_x) = \sup_{\|f\|=1} |f(x)|$$ Prove that $$N(V_x) = \sqrt{\langle k_x, k_x \rangle}.$$

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 $t \geqslant 0$, $$P\left(S_n \geqslant nam\right) \leqslant \mathrm{e}^{-ntam} \mathrm{e}^{n(\lambda(t) + \varepsilon)}.$$

Let $(E, \langle \cdot, \cdot \rangle)$ be a reproducing kernel Hilbert space on an interval $I$, with reproducing kernel $K$. For all $(x,y) \in I^2$, we set $k_x(y) = K(x,y)$. Suppose that $K$ is continuous on $I \times I$. Prove that all functions in $E$ are continuous.

Let $(E, \langle \cdot, \cdot \rangle)$ be a reproducing kernel Hilbert space on an interval $I$, with reproducing kernel $K$. For all $(x,y) \in I^2$, we set $k_x(y) = K(x,y)$. Suppose that $K$ is continuous on $I \times I$. Prove that all functions in $E$ are continuous.

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 $n \geqslant n_0$, $$P\left(S_n \geqslant nam\right) \leqslant \mathrm{e}^{-n\left(\lambda^*(am) - \varepsilon\right)}.$$

We denote by $E$ the vector space of continuous functions defined on $[0,1]$ and taking values in $\mathbb{R}$ equipped with the inner product defined by $$\langle f, g \rangle = \int_0^1 f(t) g(t)\,\mathrm{d}t$$ We consider a function $A:[0,1] \times [0,1] \rightarrow \mathbb{R}$ continuous. We are interested in the application $T: E \rightarrow E$ defined by $$T(f)(x) = \int_0^1 A(x,t) f(t)\,\mathrm{d}t$$ We suppose that $\ker T$ is finite-dimensional. Justify that $T$ induces an isomorphism from $(\ker T)^\perp$ onto $\operatorname{Im} T$.

We denote by $E$ the vector space of continuous functions defined on $[0,1]$ and taking values in $\mathbb{R}$ equipped with the inner product defined by $$\langle f, g \rangle = \int_0^1 f(t) g(t) \, \mathrm{d}t$$ We consider a function $A : [0,1] \times [0,1] \rightarrow \mathbb{R}$ continuous. We are interested in the application $T : E \rightarrow E$ defined by $$T(f)(x) = \int_0^1 A(x,t) f(t) \, \mathrm{d}t$$ We suppose that $\ker T$ is finite-dimensional. Justify that $T$ induces an isomorphism from $(\ker T)^\perp$ onto $\operatorname{Im} T$.

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 set, for all $t \in I$, $$f ( t ) = \sum _ { n = 0 } ^ { + \infty } C _ { n } t ^ { n } \quad \text { and } \quad g ( t ) = 2 t f ( t ) .$$ Deduce that if $p \neq \frac { 1 } { 2 }$, then $T$ admits an expectation.

In this subsection, $I = ]{-1,1}[$ and $w(x) = \frac{1}{\sqrt{1-x^2}}$.
Show that, for every integer $k \in \mathbb{N}$, the function $x \mapsto x^k w(x)$ is integrable on $I$.

Let $A > 2$. Deduce that $$\lim_{n \rightarrow +\infty} \frac{1}{n} \mathbb{E}\left(\sum_{\substack{1 \leqslant i \leqslant n \\ |\Lambda_{i,n}| \geqslant A}} |f - P|\left(\Lambda_{i,n}\right)\right) = 0.$$

Let $A > 2$. Deduce that $$\lim_{n \rightarrow +\infty} \frac{1}{n} \mathbb{E}\left(\sum_{\substack{1 \leqslant i \leqslant n \\ |\Lambda_{i,n}| \geqslant A}} |f - P|\left(\Lambda_{i,n}\right)\right) = 0.$$

Let $f$ be a continuous and bounded function from $\mathbb{R}$ to $\mathbb{R}$. Show that $$\lim_{n \rightarrow +\infty} \mathbb{E}\left(\frac{1}{n} \sum_{i=1}^{n} f\left(\Lambda_{i,n}\right)\right) = \frac{1}{2\pi} \int_{-2}^{2} f(x) \sqrt{4 - x^{2}} \, \mathrm{d}x$$

Let $f$ be a continuous and bounded function from $\mathbb{R}$ to $\mathbb{R}$. Show that $$\lim_{n \rightarrow +\infty} \mathbb{E}\left(\frac{1}{n} \sum_{i=1}^{n} f\left(\Lambda_{i,n}\right)\right) = \frac{1}{2\pi} \int_{-2}^{2} f(x) \sqrt{4 - x^{2}} \, \mathrm{d}x$$

Show that, if $f$ and $g$ are two functions in $E$, then the integral $\int _ { 0 } ^ { + \infty } f ( t ) g ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$ is absolutely convergent, where $E$ is the set of continuous functions $f$ from $\mathbb { R } _ { + } ^ { * }$ to $\mathbb { R }$ such that the integral $\int _ { 0 } ^ { + \infty } f ^ { 2 } ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$ 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 $E$. 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 $\mathscr{M}(E)$ and let $\mu$ be an element of $\mathscr{M}(E)$. Show that if the sequence $(\mu_n)_{n \in \mathbb{N}}$ converges to $\mu$ in the normed vector space $\mathscr{B}(\mathscr{P}(E), \mathbb{R})$, then $$\forall x \in E, \quad \lim_{n \rightarrow +\infty} \mu_n(x) = \mu(x).$$

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 let $\mu$ be an element of $\mathscr{M}(E)$. Show that if the sequence $(\mu_n)_{n \in \mathbb{N}}$ converges to $\mu$ in the normed vector space $\mathscr{B}(\mathscr{P}(E), \mathbb{R})$, then $$\forall x \in E, \quad \lim_{n \rightarrow +\infty} \mu_n(x) = \mu(x)$$

Deduce that $E$ is a vector subspace of the vector space $\mathcal { C } \left( \mathbb { R } _ { + } ^ { * } , \mathbb { R } \right)$ of continuous functions on $\mathbb { R } _ { + } ^ { * }$ with values in $\mathbb { R }$, where $E$ is the set of continuous functions $f$ from $\mathbb { R } _ { + } ^ { * }$ to $\mathbb { R }$ such that the integral $\int _ { 0 } ^ { + \infty } f ^ { 2 } ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$ 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 $E$. 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 $\|f\| = \sup\{|f(A)|, \quad A \in \mathscr{P}(E)\}$.
We fix a sequence $(\mu_n)_{n \in \mathbb{N}}$ of elements of $\mathscr{M}(E)$ and $\mu \in \mathscr{M}(E)$ satisfying $$\forall x \in E, \quad \lim_{n \rightarrow +\infty} \mu_n(x) = \mu(x). \tag{1}$$ We also fix a real number $\varepsilon > 0$.
Show that there exists a finite subset $F_\varepsilon$ of $E$ and an integer $N_\varepsilon \geqslant 0$ such that $\mu(F_\varepsilon) > 1 - \varepsilon$ and for all integer $n \geqslant N_\varepsilon$ $$\sum_{x \in F_\varepsilon} |\mu_n(x) - \mu(x)| < \varepsilon.$$

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 $\|f\| = \sup\{|f(A)|, \quad A \in \mathscr{P}(E)\}$.
We fix a sequence $(\mu_n)_{n \in \mathbb{N}}$ of elements of $\mathscr{M}(E)$ and $\mu \in \mathscr{M}(E)$ satisfying condition (1). We also fix a real number $\varepsilon > 0$ and a finite subset $F_\varepsilon$ of $E$ and integer $N_\varepsilon \geqslant 0$ as in 10a.
Show that for every subset $A$ of $E$: $$|\mu_n(A) - \mu(A)| \leqslant |\mu_n(A \cap F_\varepsilon) - \mu(A \cap F_\varepsilon)| + \mu(E \backslash F_\varepsilon) + \mu_n(E \backslash F_\varepsilon)$$ and deduce that if $n \geqslant N_\varepsilon$, then $|\mu_n(A) - \mu(A)| < 4\varepsilon$.

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 $\|f\| = \sup\{|f(A)|, \quad A \in \mathscr{P}(E)\}$.
Deduce that the sequence $(\mu_n)_{n \in \mathbb{N}}$ converges to $\mu$ in $\mathscr{B}(\mathscr{P}(E), \mathbb{R})$ if and only if it satisfies condition $$\forall x \in E, \quad \lim_{n \rightarrow +\infty} \mu_n(x) = \mu(x). \tag{1}$$

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 $E$. We fix a sequence $(\mu_n)_{n \in \mathbb{N}}$ of elements of $\mathscr{M}(E)$ and $\mu \in \mathscr{M}(E)$ satisfying $$\forall x \in E, \quad \lim_{n \rightarrow +\infty} \mu_n(x) = \mu(x) \tag{1}$$ We also fix a real number $\varepsilon > 0$. Show that there exists a finite subset $F_\varepsilon$ of $E$ and an integer $N_\varepsilon \geqslant 0$ such that $\mu(F_\varepsilon) > 1 - \varepsilon$ and for all integer $n \geqslant N_\varepsilon$ $$\sum_{x \in F_\varepsilon} |\mu_n(x) - \mu(x)| < \varepsilon$$

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 $E$. We fix a sequence $(\mu_n)_{n \in \mathbb{N}}$ of elements of $\mathscr{M}(E)$ and $\mu \in \mathscr{M}(E)$ satisfying condition (1): $\forall x \in E, \lim_{n \to +\infty} \mu_n(x) = \mu(x)$. We fix $\varepsilon > 0$ and a finite subset $F_\varepsilon$ of $E$ and integer $N_\varepsilon$ as in 10a. Show that for every subset $A$ of $E$: $$\left|\mu_n(A) - \mu(A)\right| \leqslant \left|\mu_n(A \cap F_\varepsilon) - \mu(A \cap F_\varepsilon)\right| + \mu(E \backslash F_\varepsilon) + \mu_n(E \backslash F_\varepsilon)$$ and deduce that if $n \geqslant N_\varepsilon$, then $\left|\mu_n(A) - \mu(A)\right| < 4\varepsilon$.