# Equivalence of the second moment of two random variables when their first moments and covariance with a third random variable are equal

I'm trying to check under which conditions the standard deviation of two random variables is identical when I know some properties about other moments of these random variables. I suppose that their their first moments and their covariance with a third random variable are equal. In more detail, suppose $X$ and $Y$ and $H$ a three random variables. Suppose furthermore $$E(Y) = E(X),$$ and $$Cov(X,H) = Cov(Y,H),$$ the moments are taken with respect to the same measure $P$. The random variable $H$ is not constructed such that it is Independent of either $X$ or $Y$. Finally, Suppose also that both random variables $X$ and $Y$ are always positive. Then, what conditions are needed so that

$$\sigma(X) = \sigma(Y)$$

• It's unclear what you're looking for. The independence counterexample should have resolved the question in my opinion. Are you trying to find cases where it's true? Commented Jul 7, 2015 at 18:42
• Thank you very much , Matt. Yes, you're right independence resolves my question. You're also right in guessing that I'm looking for cases where the standard deviations are identical. I reformulate my question. Commented Jul 7, 2015 at 18:47
• I don't think it will work for any $H$ unless it changes depending on $X$ and $Y$. An evil adversary could always construct a counterexample for any $H$, and I wouldn't be surprised if it they could always be chosen not to be independent. Commented Jul 7, 2015 at 18:50
• Reformulating a question after some answer is posted is not a good choice.
– Did
Commented Jul 18, 2015 at 9:59

## 1 Answer

Hint: What would happen if $H$ was chosen to be independent of both $X$ and $Y$?

• Great remark! I change my question and assume that $X$ and $Y$ are in general not independent of $H$. Commented Jul 7, 2015 at 18:36
• @fabian Better insist on the identical covariances being nonzero; else if we choose $H$ so that it is uncorrelated with both $X$ and $Y$ (but not independent of $X$ and $Y$), we have the same problem again. Commented Jul 7, 2015 at 23:13