Variables indépendantes normales

Cher Mohammed, voici un petit texte qui peut t'interesser. Ici $\Sigma_r=[1,r;r,1]$ Pour l'appliquer a ton probleme il faudrait pouvoir dire quelque chose en termes des $a_k$ et des $b_k$ des $f_1$ et $f_2$ tels que $f_1(Y_1)$ et $f_1(Y_1)$ sont normales. L'independance est donnee dans ce cas par$ \sum_ka_kb_kr^k=0.$




Let $(Y_1,Y_2)\sim N(0,\Sigma_r).$ Let $f_1$ and $f_2$ be two real measurable functions such that $\mathbb{E}(f_i(Y_i)^2)$ are finite for $i=1,2.$ Consider the Hermite polynomials $(H_k)_{k=0}^{\infty}$ defined by the generating function
$$e^{xt-\frac{t^2}{2}}=\sum_{k=0}^{\infty}H_k(x)\frac{t^k}{k!}$$ and the expansion
$$f_1(x)=\sum_{k=1}^{\infty}a_k\frac{H_k(x)}{\sqrt{k!}},\ f_2(x)=\sum_{k=1}^{\infty}b_k\frac{H_k(x)}{\sqrt{k!}}.$$ Then for all $-1\leq r\leq 1$
$$\mathbb{E}(f_1(Y_1)f_2(Y_2))=\sum_{k=1}^{\infty}a_kb_kr^k.$$

{Proof.} Let us compute
$$\mathbb{E}(e^{Y_1t-\frac{t^2}{2}}e^{Y_2s-\frac{s^2}{2}})= \sum_{k=0}^{\infty}\sum_{m=0}^{\infty}\frac{t^k}{k!}\frac{s^m}{m!}\mathbb{E}(H_k(Y_1)H_m(Y_2))$$
For this, write $r=\cos \theta$ with $0\leq \theta\leq \pi.$ If $Y_1,Y_3$ are independent centered real Gaussian random variables with variance 1, then $Y_2=Y_1\cos \theta +Y_3\sin \theta $ is centered with variance 1, $(Y_1,Y_2)$ is Gaussian and $\mathbb{E}(Y_1Y_2)=\cos \theta.$
Furthermore a simple calculation using the definition of $Y_2$ gives $$\mathbb{E}(e^{Y_1t-\frac{t^2}{2}}e^{Y_2\; s-\frac{s^2}{2}})=e^{ts\cos \theta}$$
This shows that
$\mathbb{E}(H_k(Y_1)H_m(Y_2))=0$ if $k\neq m$ and that $\mathbb{E}(H_k(Y_1)H_k(Y_2))=k!\cos^k \theta.$ From this we get the result. $\square$

$\Sigma_r=\left[\begin{array}{cc}1&r\\r&1\end{array}\right].$