Let $X _ { 1 } , \ldots , X _ { n } , Y _ { 1 } , \ldots , Y _ { n }$ be mutually independent random variables with the same distribution $\mathcal { R }$ (where $X(\Omega) = \{-1,1\}$, $\mathbb{P}(X=-1)=\mathbb{P}(X=1)=\frac{1}{2}$). We define the random vectors $X = \frac { 1 } { \sqrt { n } } \left( X _ { 1 } , \ldots , X _ { n } \right) ^ { \top }$ and $Y = \frac { 1 } { \sqrt { n } } \left( Y _ { 1 } , \ldots , Y _ { n } \right) ^ { \top }$ taking values in $\mathcal { M } _ { n , 1 } ( \mathbb { R } )$. Prove that, for every real number $t$, $$\mathbb { E } ( \exp ( t \langle X \mid Y \rangle ) ) = \left( \operatorname { ch } \left( \frac { t } { n } \right) \right) ^ { n }$$
Let $X _ { 1 } , \ldots , X _ { n } , Y _ { 1 } , \ldots , Y _ { n }$ be mutually independent random variables with the same distribution $\mathcal { R }$ (where $X(\Omega) = \{-1,1\}$, $\mathbb{P}(X=-1)=\mathbb{P}(X=1)=\frac{1}{2}$). We define the random vectors $X = \frac { 1 } { \sqrt { n } } \left( X _ { 1 } , \ldots , X _ { n } \right) ^ { \top }$ and $Y = \frac { 1 } { \sqrt { n } } \left( Y _ { 1 } , \ldots , Y _ { n } \right) ^ { \top }$ taking values in $\mathcal { M } _ { n , 1 } ( \mathbb { R } )$.
Prove that, for every real number $t$,
$$\mathbb { E } ( \exp ( t \langle X \mid Y \rangle ) ) = \left( \operatorname { ch } \left( \frac { t } { n } \right) \right) ^ { n }$$