Show that for all $\theta \in ] - \pi ; \pi [$, the function $f$ defined by $$\begin{aligned}
f : ] 0 ; + \infty [ & \longrightarrow \mathbf { C } \\
t & \longmapsto \frac { t ^ { x - 1 } } { 1 + t e ^ { \mathrm { i } \theta } }
\end{aligned}$$ is defined and integrable on $] 0 ; + \infty [$, where $x$ is a fixed element of $]0;1[$.
Let $g$ be the function defined by $$\begin{aligned}
g : ] - \pi ; \pi [ & \longrightarrow \mathbf { C } \\
\theta & \longmapsto e ^ { \mathrm { i } x \theta } \int _ { 0 } ^ { + \infty } \frac { t ^ { x - 1 } } { 1 + t e ^ { \mathrm { i } \theta } } \mathrm {~d} t
\end{aligned}$$ where $x$ is a fixed element of $]0;1[$. Show that the function $g$ is of class $\mathcal { C } ^ { 1 }$ on $] - \pi ; \pi [$ and that for all $\theta \in ] - \pi ; \pi [$, $$g ^ { \prime } ( \theta ) = \mathrm { i } e ^ { \mathrm { i } x \theta } \int _ { 0 } ^ { + \infty } h ^ { \prime } ( t ) \mathrm { d } t$$ where $h$ is the function defined by $$\begin{aligned}
h : ] 0 ; + \infty [ & \longrightarrow \mathbf { C } \\
t & \longmapsto \frac { t ^ { x } } { 1 + t e ^ { \mathrm { i } \theta } } .
\end{aligned}$$ Calculate $h ( 0 )$ and $$\lim _ { t \rightarrow + \infty } h ( t ) .$$ Deduce that the function $g$ is constant on $] - \pi ; \pi [$.
Let $g$ be the function defined by $$\begin{aligned}
g : ] - \pi ; \pi [ & \longrightarrow \mathbf { C } \\
\theta & \longmapsto e ^ { \mathrm { i } x \theta } \int _ { 0 } ^ { + \infty } \frac { t ^ { x - 1 } } { 1 + t e ^ { \mathrm { i } \theta } } \mathrm {~d} t
\end{aligned}$$ where $x$ is a fixed element of $]0;1[$. Show that for all $\theta \in ] 0 ; \pi [$, $$g ( \theta ) \sin ( x \theta ) = \frac { 1 } { 2 \mathrm { i } } \left( g ( - \theta ) e ^ { \mathrm { i } x \theta } - g ( \theta ) e ^ { - \mathrm { i } x \theta } \right) = \sin ( \theta ) \int _ { 0 } ^ { + \infty } \frac { t ^ { x } } { t ^ { 2 } + 2 t \cos ( \theta ) + 1 } \mathrm {~d} t$$
Recall that $x$ is a fixed element of $]0;1[$. Show that: $$\int _ { 0 } ^ { 1 } \frac { t ^ { x - 1 } } { 1 + t } \mathrm {~d} t = \sum _ { k = 0 } ^ { + \infty } \frac { ( - 1 ) ^ { k } } { k + x }$$
Show that: $$( \cos ( t ) ) ^ { 2 p } = \frac { 1 } { 2 ^ { 2 p } } \left( \binom { 2 p } { p } + 2 \sum _ { k = 0 } ^ { p - 1 } \binom { 2 p } { k } \cos ( 2 ( p - k ) t ) \right)$$ Hint: One may develop $\left( \frac { e ^ { \mathrm{i} t } + e ^ { - \mathrm{i} t } } { 2 } \right) ^ { 2 p }$.
Let $\left( X _ { k } \right) _ { k \in \mathbf{N} ^ { * } }$ be independent random variables with the same distribution given by: $$P \left( X _ { 1 } = - 1 \right) = P \left( X _ { 1 } = 1 \right) = \frac { 1 } { 2 }$$ For all $n \in \mathbf { N } ^ { * }$, we denote $S _ { n } = \sum _ { k = 1 } ^ { n } X _ { k }$. Determine, for all $n \in \mathbf { N } ^ { * }$, $E \left( S _ { n } \right)$ and $V \left( S _ { n } \right)$.
Let $S$ and $T$ be two independent random variables each taking a finite number of real values. Assume that $T$ and $- T$ follow the same distribution. Show that: $$E ( \cos ( S + T ) ) = E ( \cos ( S ) ) E ( \cos ( T ) )$$
Let $\left( X _ { k } \right) _ { k \in \mathbf{N} ^ { * } }$ be independent random variables with the same distribution given by: $$P \left( X _ { 1 } = - 1 \right) = P \left( X _ { 1 } = 1 \right) = \frac { 1 } { 2 }$$ For all $n \in \mathbf { N } ^ { * }$, we denote $S _ { n } = \sum _ { k = 1 } ^ { n } X _ { k }$. Deduce that for all $n \in \mathbf { N } ^ { * }$, and for all $t \in \mathbf { R }$: $$E \left( \cos \left( t S _ { n } \right) \right) = ( \cos ( t ) ) ^ { n } .$$
Let $\left( X _ { k } \right) _ { k \in \mathbf{N} ^ { * } }$ be independent random variables with the same distribution given by: $$P \left( X _ { 1 } = - 1 \right) = P \left( X _ { 1 } = 1 \right) = \frac { 1 } { 2 }$$ For all $n \in \mathbf { N } ^ { * }$, we denote $S _ { n } = \sum _ { k = 1 } ^ { n } X _ { k }$. Let $a , b \in \mathbf { R }$ such that $a \neq 0$ and $| b | \leq | a |$. Show that $$| a + b | = | a | + \operatorname { sign } ( a ) b$$ where $\operatorname { sign } ( x ) = x / | x |$ for nonzero real $x$. Deduce that: $$\forall n \in \mathbf { N } ^ { * } , \quad E \left( \left| S _ { 2 n } \right| \right) = E \left( \left| S _ { 2 n - 1 } \right| \right)$$
Let $\left( X _ { k } \right) _ { k \in \mathbf{N} ^ { * } }$ be independent random variables with the same distribution given by: $$P \left( X _ { 1 } = - 1 \right) = P \left( X _ { 1 } = 1 \right) = \frac { 1 } { 2 }$$ For all $n \in \mathbf { N } ^ { * }$, we denote $S _ { n } = \sum _ { k = 1 } ^ { n } X _ { k }$. Conclude that: $$\forall n \in \mathbf { N } ^ { * } , \quad E \left( \left| S _ { 2 n } \right| \right) = E \left( \left| S _ { 2 n - 1 } \right| \right) = \frac { ( 2 n - 1 ) ! } { 2 ^ { 2 n - 2 } ( ( n - 1 ) ! ) ^ { 2 } }$$