# If there are two geometric random variables $X_1$ and $X_2$, how to prove $E[X_1^2]E[X_2^2] \geq (E[X_{1}X_{2}])^2$?

If there are two geometric random variables $X_1$ and $X_2$, how to prove $E[X_1^2]E[X_2^2] \geq(E[X_{1}X_{2}])^2$? In addition, under what condition will the equality $E[X_1^2]E[X_2^2]=(E[X_{1}X_{2}])^2$ hold?

From my point of view, the left part of $E[X_1^2]E[X_2^2] \geq (E[X_{1}X_{2}])^2$ can be considered to be self-correlation. The right part is cross-correlation. Thus, it seems to be reasonable that self-correlation is larger than cross-correlation. But, I have no idea how to prove it.

Thanks.

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What you have written down is true for any random variables, not just for 2 geometric random variables. It is the Cauchy Schwarz inequality.

The equality holds iff $X_1=cX_2$ for Cauchy Schwarz. but since they are both geometric distributions, it must be the case $X_1=X_2$

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Thanks a lot. It is easy to prove it by the the Cauchy Schwarz inequality. – Stanely Apr 10 '13 at 16:43