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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.

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    $\begingroup$ 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 $\operatorname{Cov}(X_1,X_3)=\operatorname{Var}(X_1)=\operatorname{Var}(X_3).$ $\endgroup$
    – whuber ♦
    Dec 4, 2019 at 15:40
  • $\begingroup$ @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.. $\endgroup$
    – independentvariable
    Dec 4, 2019 at 15:43
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    $\begingroup$ 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. $\endgroup$
    – whuber ♦
    Dec 4, 2019 at 16:05
  • $\begingroup$ @whuber thanks for your answer and help! $\endgroup$
    – independentvariable
    Dec 4, 2019 at 16:10

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