Minimize the expected value of the product of 2 normally distributed variables So, there are 2 variables, $X$ and $Y$, both are normally distributed. We are given that $E(X)=E(Y)=0$ and $Var(X)=2$, while $Var(Y)=8$. Additionally, $Corr(X,Y)=-\frac{1}{2}$. The question is to find the smallest value of $E(X^5Y^3)$.
My first instinct was to somehow use the definition of covariance:
$$Cov(X^5,Y^3)=E(X^5Y^3)-E(X^5)E(Y^3)$$
$$E(X^5Y^3)=Cov(X^5,Y^3)+E(X^5)E(Y^3)$$
I knew that I could find the $E(Y^3)$ using the moment generating function. Since we are given that $Y\sim N(0,8)$, the moment generating function is
$$M_Y(t)=e^{4t^2}$$
So, $$E(Y^3)=\frac{d^3}{dt^3}M_Y(0)=0\implies E(X^5Y^3)=Cov(X^5,Y^3)$$ The same goes for $E(X^5)=\frac{d^5}{dt^5}M_X(0)=0$. Another idea is to now play with the definition:
$$Cov(X,X^4Y^3)=E(X^5Y^3)-E(X)E(X^4Y^3)=E(X^5Y^3)$$
So that $$Cov(X,X^4Y^3)=Cov(X^5,Y^3)$$
Because we are given the correlation for a reason, I got $Corr(X,Y)=\frac{Cov(X,Y)}{\sqrt{Var(X)Var(Y)}}=-\frac{1}{2}=\frac{Cov(X,Y)}{4} \implies Cov(X,Y)=-2$.
However, I do not see the connection with my previous result. Perhaps I need to use some different method and not covariance. Can anyone point me in the right direction?
$\pmb{Edit:}$ I have used a suspicious formula from the paper by Kan, URL: https://www-2.rotman.utoronto.ca/~kan/papers/moment.pdf on page 5 and got the following result for my particular case:
$$E(X^5Y^3)=\frac{45}{128}\sum_{j=0}^{1}{\frac{-1}{(2-j)!(1-j)!(2j+1)!}}=\frac{45}{128}\left(-\frac{1}{2}-\frac{1}{6}\right)=-\frac{15}{64}=-0.234375$$
But this result requires verification. Can anyone tell me if the formula is legit and if yes, then did I use it correctly?$\pmb{Edit\space №2:}$ The question about the formula is resolved. I have also found that $E(X^5Y^3)=5760\sqrt{7}Corr(X^5,Y^3)$ from the definition of correlation coefficient, so the question is: is it possible to minimize the correlation coefficient now?

The formula is correct, but your application of the formula is not.  The correct expectation using this formula is
$$\begin{align}
\operatorname{E}[X^5 Y^3] 
&= \sigma_1^5 \sigma_2^3 \frac{5! 3!}{2^{(5+3)/2}} \sum_{j=0}^{\lfloor \min(5,3)/2 \rfloor} \frac{(2\rho)^{2j+1}}{(\frac{5-1}{2} - j)! (\frac{3-1}{2} - j)! (2j+1)!} \\
&= (\sqrt{2})^5 (\sqrt{8})^3 (45) \sum_{j=0}^1 \frac{(-1)^{2j+1}}{(2-j)!(1-j)!(2j+1)!} \\ 
&= 5760 \left( \frac{-1}{2! 1! 1!} + \frac{-1}{1! 0! 3!}\right) \\
&= -5760 \left( \frac{1}{2} + \frac{1}{6} \right) \\
&= -3840.
\end{align}$$
That said, there is no guarantee that $(X,Y)$ is bivariate normal.  As such, the formula is not applicable because you are interested in the minimum value of this expectation under all distributions that satisfy the given criteria, and it is not assured that the minimum is attained for the case when $(X,Y)$ is in fact bivariate normal.
We can use Cauchy-Schwarz in the form $$|\operatorname{E}[X_1 X_2]|^2 \le \operatorname{E}[X_1^2]\operatorname{E}[X_2^2]$$ to show that $\operatorname{E}[X^5 Y^3] \ge - 5760\sqrt{7}$, but this lower bound is not necessarily attainable.
