A continuous random variable $X$ defined on the interval $[ 0,1 ]$ has probability density function $f ( x )$. The mean of $X$ is $\frac { 1 } { 4 }$ and $\int _ { 0 } ^ { 1 } ( a x + 5 ) f ( x ) d x = 10$. Find the value of the constant $a$. [4 points]

Throughout this part, $\lambda$ is a real number belonging to the interval $]0,1[$ and $f, g, h$ are functions in $C^{0}(\mathbb{R}, \mathbb{R}_{+})$ that are integrable and satisfy the following inequality $$\forall x \in \mathbb{R}, \forall y \in \mathbb{R}, \quad h(\lambda x + (1-\lambda) y) \geq f(x)^{\lambda} g(y)^{1-\lambda}.$$ In questions 3), 4) and 5) we additionally assume that $f$ and $g$ are strictly positive, that is, for all real $x$, $f(x) > 0$ and $g(x) > 0$.
We denote $F = \int_{-\infty}^{+\infty} f(x)\,dx$ and $G = \int_{-\infty}^{+\infty} g(x)\,dx$. Show that for all $t$ in the interval $]0,1[$ there exists a unique real number denoted $u(t)$ and a unique real number denoted $v(t)$ such that $$\frac{1}{F} \int_{-\infty}^{u(t)} f(x)\,dx = t, \quad \frac{1}{G} \int_{-\infty}^{v(t)} g(x)\,dx = t$$ (One may study the variations of the function: $u \mapsto \frac{1}{F} \int_{-\infty}^{u} f(x)\,dx$).

We consider the function $f$ defined by: $\forall (x,y) \in \mathbb{R}^2,\ f(x,y) = \frac{1}{1+x^2+y^2}$.
Show that $\hat{f}$ is defined on $\mathbb{R}^2$ with $\hat{f}(q,\theta) = \frac{\pi}{\sqrt{1+q^2}}$.

We consider the function $f$ defined by: $\forall (x,y) \in \mathbb{R}^2,\ f(x,y) = \frac{1}{1+x^2+y^2}$.
We set $R(q) = \frac{1}{2\pi} \int_0^{2\pi} \hat{f}(q,\theta)\,\mathrm{d}\theta$. Prove that $q \mapsto \frac{R^{\prime}(q)}{q}$ is integrable on $]0, +\infty[$ and that $$f(0,0) = -\frac{1}{\pi} \int_0^{+\infty} \frac{R^{\prime}(q)}{q}\,\mathrm{d}q$$ One may, to compute this last integral, use the change of variable $q = \operatorname{sh}(u)$.

We assume that there exists a function $\varphi$ from $\mathbb{R}^+$ to $\mathbb{R}$, continuous and integrable on $\mathbb{R}^+$, such that: $\forall (x,y) \in \mathbb{R}^2,\ f(x,y) = \varphi\left(\sqrt{x^2+y^2}\right)$.
For $r \in \mathbb{R}^+$, compute $\bar{f}(r) = \frac{1}{2\pi} \int_0^{2\pi} f(r\cos t, r\sin t)\,\mathrm{d}t$.

We assume that there exists a function $\varphi$ from $\mathbb{R}^+$ to $\mathbb{R}$, continuous and integrable on $\mathbb{R}^+$, such that: $\forall (x,y) \in \mathbb{R}^2,\ f(x,y) = \varphi\left(\sqrt{x^2+y^2}\right)$.
Prove that the Radon transform of $f$ is defined on $\mathbb{R}^2$ and that $$\forall q \in \mathbb{R}^+,\quad \forall \theta \in \mathbb{R} \quad \hat{f}(q,\theta) = 2\int_q^{+\infty} \frac{r\varphi(r)}{\sqrt{r^2-q^2}}\,\mathrm{d}r$$

We assume that there exists a function $\varphi$ from $\mathbb{R}^+$ to $\mathbb{R}$, continuous and integrable on $\mathbb{R}^+$, such that: $\forall (x,y) \in \mathbb{R}^2,\ f(x,y) = \varphi\left(\sqrt{x^2+y^2}\right)$.
Deduce that $\forall q \in \mathbb{R}^+,\ \frac{1}{2\pi} \int_0^{2\pi} \hat{f}(q,\theta)\,\mathrm{d}\theta = 2\int_q^{+\infty} \frac{r\bar{f}(r)}{\sqrt{r^2-q^2}}\,\mathrm{d}r$.

We consider a function $f$ belonging to $\mathcal{B}_1$ and we recall that $$\hat{f}(q,\theta) = \int_{-\infty}^{+\infty} f(q\cos\theta - t\sin\theta,\, q\sin\theta + t\cos\theta)\,\mathrm{d}t$$
Justify that for all $q$ and all $\theta$ we have $\hat{f}(-q, \theta+\pi) = \hat{f}(q,\theta)$.

We consider a function $f$ in $\mathcal{B}_1$ whose partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are in $\mathcal{B}_2$. We set: $$\forall q \in \mathbb{R}^+,\quad F(q) = \frac{1}{2\pi}\int_0^{2\pi} \hat{f}(q,\theta)\,\mathrm{d}\theta = 2\int_q^{+\infty} \frac{r\bar{f}(r)}{\sqrt{r^2-q^2}}\,\mathrm{d}r$$
Prove: $\forall \varepsilon > 0,\ \int_\varepsilon^{+\infty} \frac{F^{\prime}(q)}{q}\,\mathrm{d}q = -\frac{F(\varepsilon)}{\varepsilon} + 2\int_\varepsilon^{+\infty} \frac{1}{q^2}\left(\int_q^{+\infty} \frac{r\bar{f}(r)}{\sqrt{r^2-q^2}}\,\mathrm{d}r\right)\mathrm{d}q$.

We consider a function $f$ in $\mathcal{B}_1$ whose partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are in $\mathcal{B}_2$. We set: $$\forall q \in \mathbb{R}^+,\quad F(q) = \frac{1}{2\pi}\int_0^{2\pi} \hat{f}(q,\theta)\,\mathrm{d}\theta = 2\int_q^{+\infty} \frac{r\bar{f}(r)}{\sqrt{r^2-q^2}}\,\mathrm{d}r$$
We admit that we can interchange the two integrals and therefore that $$\forall \varepsilon > 0 \quad \int_\varepsilon^{+\infty} \left(\frac{1}{q^2}\int_q^{+\infty} \frac{r\bar{f}(r)}{\sqrt{r^2-q^2}}\,\mathrm{d}r\right)\mathrm{d}q = \int_\varepsilon^{+\infty} \left(\int_\varepsilon^r \frac{r\bar{f}(r)}{q^2\sqrt{r^2-q^2}}\,\mathrm{d}q\right)\mathrm{d}r$$
Deduce that $\forall \varepsilon > 0,\ \int_\varepsilon^{+\infty} \frac{F^{\prime}(q)}{q}\,\mathrm{d}q = -2\varepsilon \int_\varepsilon^{+\infty} \frac{\bar{f}(r)}{r\sqrt{r^2-\varepsilon^2}}\,\mathrm{d}r$.

We consider a function $f$ in $\mathcal{B}_1$ such that $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are in $\mathcal{B}_2$. The Radon inversion formula states: $\forall (x,y) \in \mathbb{R}^2,\ f(x,y) = \frac{-1}{\pi} \int_0^{+\infty} \frac{R_{x,y}^{\prime}(q)}{q}\,\mathrm{d}q$, where $R_{x,y}(q) = \frac{1}{2\pi}\int_0^{2\pi} \hat{f}(x\cos\theta + y\sin\theta + q, \theta)\,\mathrm{d}\theta$.
Establish the Radon inversion formula for this function $f$ at the point $(x,y) = (0,0)$.

We consider a function $f$ in $\mathcal{B}_1$ such that $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are in $\mathcal{B}_2$. The Radon inversion formula states: $\forall (x,y) \in \mathbb{R}^2,\ f(x,y) = \frac{-1}{\pi} \int_0^{+\infty} \frac{R_{x,y}^{\prime}(q)}{q}\,\mathrm{d}q$.
Are the hypotheses made on $f$ necessary for the Radon inversion formula to be verified at the point $(x,y) = (0,0)$?

We consider a function $f$ in $\mathcal{B}_1$ such that $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are in $\mathcal{B}_2$. The Radon inversion formula states: $\forall (x,y) \in \mathbb{R}^2,\ f(x,y) = \frac{-1}{\pi} \int_0^{+\infty} \frac{R_{x,y}^{\prime}(q)}{q}\,\mathrm{d}q$.
Propose a method to obtain the Radon inversion formula at any pair $(x,y)$ from the formula at $(0,0)$.

Let $p , q , r : \mathbb { R } \rightarrow \mathbb { R } _ { * } ^ { + }$ be three continuous functions, with strictly positive values and integrable on $\mathbb { R }$.
14a. Show that there exists a function $u : ] 0,1 [ \rightarrow \mathbb { R }$ of class $\mathscr { C } ^ { 1 }$ bijective such that $$\forall t \in ] 0,1 [ , \quad u ^ { \prime } ( t ) p ( u ( t ) ) = \int p ( x ) d x$$ Similarly, there exists an analogous function $v : ] 0,1 [ \rightarrow \mathbb { R }$ for $q$.
14b. We assume that $$\forall x , y \in \mathbb { R } , \quad p ( x ) q ( y ) \leqslant \left( r \left( \frac { x + y } { 2 } \right) \right) ^ { 2 } . \tag{4}$$ Show that $$\left( \int p ( x ) d x \right) \left( \int q ( x ) d x \right) \leqslant \left( \int r ( x ) d x \right) ^ { 2 } \tag{5}$$ You may use, after having justified its validity, the change of variable defined by $x = \frac { u ( t ) + v ( t ) } { 2 }$ in the right-hand side of inequality (5).

Assuming that $\int_{-\infty}^{+\infty} \exp\left(-x^{2}\right) \mathrm{d}x = \sqrt{\pi}$, give the value of $\int_{-\infty}^{+\infty} g_{\sigma}(x) \mathrm{d}x$.

Show that the function $I : \left\lvert\, \begin{aligned} & \mathbb{R}_{+}^{*} \rightarrow \mathbb{R} \\ & t \mapsto \int_{-\infty}^{+\infty} f(t, x) \mathrm{d}x \end{aligned}\right.$ is constant. One may use the result of question 17.

We define $\gamma_\lambda(y) = \exp(-y^2/\lambda)$ and for all $x \in \mathbf{R}$, $\tau_x(f)(y) = f(y-x)$. For all $f, g \in \mathcal{E}$, $(f \mid g) = \int_{-\infty}^{+\infty} f(y)g(y)\,\mathrm{d}y$.
(a) Let $a > 0$. Show that there exists $c \geq 0$ such that for all $x \in \mathbf{R}$ we have $$\int_{-\infty}^{+\infty} \exp\left(-\frac{(y-x)^2}{\lambda}\right) \exp\left(-\frac{y^2}{a}\right) \mathrm{d}y = c \exp\left(-\frac{x^2}{a+\lambda}\right).$$ Hint: One may show the equality $$\frac{(y-x)^2}{\lambda} + \frac{y^2}{a} = \frac{a+\lambda}{a\lambda}\left(y - \frac{ax}{a+\lambda}\right)^2 + \frac{x^2}{a+\lambda}.$$
(b) Let $g \in \mathcal{E}$. We consider $C(g) : \mathbf{R} \rightarrow \mathbf{R}$ defined for all $x \in \mathbf{R}$ by $$C(g)(x) = \left(\tau_x(\gamma_\lambda) \mid g\right)$$ Show that $C(g) \in \mathcal{E}$.
(c) Show that $C : \mathcal{E} \rightarrow \mathcal{E}$ defines an endomorphism of $\mathcal{E}$.

Let $\mathcal{C}^{1}$ be the space of functions of class $C^{1}$ from $[-\pi, \pi]$ to $\mathbf{C}$. For $f \in \mathcal{C}^{1}$, we set $$\|f\|_{\infty} = \max\{|f(t)|; t \in [-\pi, \pi]\} \quad \text{and} \quad V(f) = \int_{-\pi}^{\pi} |f^{\prime}|.$$
Let $f \in \mathcal{C}^{1}$ with real values. We assume that the set $C(f)$ of points in $]-\pi, \pi[$ where the function $f^{\prime}$ vanishes is finite. We denote by $\ell$ the cardinality of $C(f)$ and, if $\ell \geq 1$, we denote by $t_{1} < \cdots < t_{\ell}$ the elements of $C(f)$. We set $t_{0} = -\pi$ and $t_{\ell+1} = \pi$.
Show that $$V(f) = \sum_{j=0}^{\ell} \left|f\left(t_{j+1}\right) - f\left(t_{j}\right)\right|$$
For $0 \leq j \leq \ell$, let $\psi_{j}$ be the function from $\mathbf{R}$ to $\{0,1\}$ equal to 1 on $\left[f\left(t_{j}\right), f\left(t_{j+1}\right)\right[$ and to 0 on $\mathbf{R} \backslash \left[f\left(t_{j}\right), f\left(t_{j+1}\right)\right[$. Show that $$V(f) = \sum_{j=0}^{\ell} \int_{-\|f\|_{\infty}}^{\|f\|_{\infty}} \psi_{j}$$

For $\mu > 0$ and $\varphi \in \mathcal{C}_{c}(\mathbb{R})$, we define $T_{\mu} : \varphi \mapsto T_{\mu}\varphi$, where for all $x \in \mathbb{R}$,
$$T_{\mu}\varphi(x) = \frac{1}{2\mu} \int_{x-\mu}^{x+\mu} \varphi(t)\, dt$$
Show that if $\varphi \in \mathcal{C}_{c}(\mathbb{R})$ is a positive function, we have $\|T_{\mu}\varphi\|_{1} = \|\varphi\|_{1}$.

Show that for all $f \in C^{0}(\mathbf{R}) \cap CL(\mathbf{R})$ and all $x \in \mathbf{R}$,
$$\lim_{t \rightarrow +\infty} P_{t}(f)(x) = \int_{-\infty}^{+\infty} f(y) \varphi(y) \mathrm{d}y$$

Deduce that we have:
$$\forall t \in \mathbf{R}_{+}^{*}, \quad -S^{\prime}(t) \leq \mathrm{e}^{-2t} \int_{-\infty}^{+\infty} \frac{f^{\prime 2}(x)}{f(x)} \varphi(x) \mathrm{d}x$$

Show that for all $f \in C^0(\mathbf{R}) \cap CL(\mathbf{R})$ and all $x \in \mathbf{R}$, $$\lim_{t \rightarrow +\infty} P_t(f)(x) = \int_{-\infty}^{+\infty} f(y)\varphi(y)\,\mathrm{d}y,$$ where $P_t(f)(x) = \int_{-\infty}^{+\infty} f\!\left(\mathrm{e}^{-t}x + \sqrt{1-\mathrm{e}^{-2t}}\,y\right)\varphi(y)\,\mathrm{d}y$.

Deduce that we have: $$\forall t \in \mathbf{R}_+^*, \quad -S'(t) \leq \mathrm{e}^{-2t}\int_{-\infty}^{+\infty} \frac{f'^2(x)}{f(x)}\,\varphi(x)\,\mathrm{d}x.$$

Let $p \in \left[ 1 , + \infty \right[$. Let $X$ be a positive and finite real random variable. Let $F _ { X }$ be the function defined for all $t \geq 0$ by $$F _ { X } ( t ) = \mathbf { P } ( X > t ) .$$ Show that the integral $\int _ { 0 } ^ { + \infty } t ^ { p - 1 } F _ { X } ( t ) \mathrm { d } t$ converges, then that $$\mathbf { E } \left( X ^ { p } \right) = p \int _ { 0 } ^ { + \infty } t ^ { p - 1 } F _ { X } ( t ) \mathrm { d } t$$