Suppose $X_1, \dots, X_n$ are Gaussian random variables with mean zero and variance $1$. Is it true that $|\mathbb{E}(X_1\dots X_n)|\le n!$? I am not quite sure about the $n!$ bound on the RHS, but it was suggested by my teacher, who guessed that should be it. Originally I was only trying to get a bound for the moment. Note the value of the expectation is zero if $n$ is odd. So we only consider cases when $n$ is even.
I was trying to apply the Wick Formula and Cauchy Schwartz inequality but it gives me $|\mathbb{E}(X_1\dots X_n)|\le \#(\text{pair partitions of }\{1,\dots, n\})$ But although it gives a bound I found the value of RHS is not $n!$, but $n!/2^{n/2}$ which is sharper. So is this correct or am I making a mistake?
Use generalized Holder's inequality which gives $E|X_1X_2...X_n| \leq (E|X_1|^{n})^{1/n}E|X_2|^{n})^{1/n}... E|X_n|^{n})^{1/n} =E|X_1|^{n}$. Use integration by parts for $E|X_1|^{n}=\frac 2 {\sqrt{2\pi}} \int_0^{\infty} x^{n}e^{-x^{2}/2}dx$