# Expected value of multiplication of four zero-mean normal random variables

I am given the following statement:

For zero-mean Gaussian random variables $$X_k$$ (where nothing about variance - covariances is assumed) we have: $$\mathbb{E}[X_1X_2X_3 X_4] =\mathbb{E}[X_1X_2]\mathbb{E}[X_3X_4] + \mathbb{E}[X_1X_3]\mathbb{E}[X_2X_4] + \mathbb{E}[X_1X_4]\mathbb{E}[X_2X_3]$$

My question is, can I assume $$X_1 = X_3, \ X_2 = X_4$$ and re-write the above expression as: $$\mathbb{E}[X_1^2 X_2^2] = \mathbb{E}[X_1^2]\mathbb{E}[X_2^2] + 2\mathbb{E}[X_1X_2]^2$$

My concern is $$X_1X_3$$ is multiplication of two Gaussian variables. However, setting $$X_3= X_1$$ gives us $$X_1^2$$ which is the square of one Gaussian variable, and I am not sure if/how the above equality follows.

• In what sense do you mean "rewrite"? Certainly the case $X_1=X_3$ is a special case of the formula (and similarly when $X_2=X_4$): this occurs when $\operatorname{Cov}(X_1,X_3)=\operatorname{Var}(X_1)=\operatorname{Var}(X_3).$
– whuber
Dec 4, 2019 at 15:40
• @whuber does the equation follow for $X_1 = X_3$? Because when I impose two of these variables to be the same, maybe I change the definition.. Dec 4, 2019 at 15:43
• When $X_1=X_3$ simply apply the equation. There's nothing special to do, because the tuple $(X_1,X_2,X_1,X_4)$ satisfies all the assumptions: its components are zero-mean Gaussian variables.
– whuber
Dec 4, 2019 at 16:05
• @whuber thanks for your answer and help! Dec 4, 2019 at 16:10