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Let $A$, $B$ and $B'$ be random variables, and consider

$\text{Var}(AB) \text{ and} \text{ Var}(AB')$,

such that $\text{Cov}(A,B)\ne 0$, $\text{Cov}(A,B')\ne 0$ and $\text{Var}(B)\geq \text{Var}(B')$.

Then, we might conclude that $\text{Var}(AB)\geq \text{Var}(AB')$.

I have been trying to verify this, but this is pretty difficult.

Does anyone have a nice clue of how to do this?

kurtkim
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    Start with looking at [this answer](http://stats.stackexchange.com/a/15986/6633) which gives an explicit formula for the variance of the product of two non-independent random variables. It might give a clue as to why you are having difficulty proving what you want to prove, such as the result does not hold in general..... – Dilip Sarwate Oct 30 '15 at 12:47
  • Yes, I've already seen that answer, so complicated. But in my question, there is a helpful hint that $Var(B)\geq Var(B')$. With this, it might lead us to my potential conclusion. – kurtkim Oct 30 '15 at 13:11
  • But the correct answer, which you might have seen but apparently have not understood, includes terms such as $\operatorname{cov}(A^2, B^2)$ whose value is unknown and cannot be determined from the additional assumptions that you make. So, go to it, and find a proof that leads to your desired conclusion using just the additional assumptions that you have already made. But, don't expect me to believe your proof. – Dilip Sarwate Oct 30 '15 at 13:18
  • Well I know what you point out, however the additional codition would help us to at least \it {approximate} my desired inequality – kurtkim Oct 31 '15 at 16:09

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