why is it equal to $E[S^2(t_1)]E[S^2(t_2)]+E^2[S(t_1)S(t_2)]+E[N^2(t_1)]E[N^2(t_2)]+E^2[N(t_1)N(t_2)]$

Question

$Y(t)=a[S(t)+N(t)]^2$,$S(t)$ and $N(t)$ are both Gaussian random process and WSS with zero mean,and $S(t)$ is independent of $N(t)$

\begin{align} R_Y(t_1,t_2) & =E[Y(t_1)Y^*(t_2)] \\ & =a^2E[(S(t_1)+N(t_1))^2(S(t_2)+N(t_2))^2]\\ &=a^2(E[S^2(t_1)S^2(t_2)]+E[N^2(t_1)N^2(t_2)]+E[S^2(t_1)N^2(t_2)]+E[S^2(t_2)N^2(t_1)])\\ &=a^2(E[S^2(t_1)]E[S^2(t_2)]+E^2[S(t_1)S(t_2)]+E[N^2(t_1)]E[N^2(t_2)]+E^2[N(t_1)N(t_2)]) \end{align}

Can anyone tell me why is \begin{align} E[S^2(t_1)S^2(t_2)]+E[N^2(t_1)N^2(t_2)]+E[S^2(t_1)N^2(t_2)]+E[S^2(t_2)N^2(t_1)] = E[S^2(t_1)]E[S^2(t_2)]+E^2[S(t_1)S(t_2)]+E[N^2(t_1)]E[N^2(t_2)]+E^2[N(t_1)N(t_2)]\end{align},it seems a little wierd

Answer 1

Seems to be false. Take $S(t)=X,N(t)=Y$ for all $t$ where $\{X,Y\}$ is i.i.d with standard normal distribution. Then the identity you have written does not hold.