0
$\begingroup$

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 \begin{equation} E(Y) = E(X), \end{equation} and \begin{equation} Cov(X,H) = Cov(Y,H), \end{equation} 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

\begin{equation} \sigma(X) = \sigma(Y) \end{equation}

$\endgroup$
4
  • 1
    $\begingroup$ 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? $\endgroup$
    – Matt Samuel
    Commented Jul 7, 2015 at 18:42
  • $\begingroup$ 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. $\endgroup$
    – fabian
    Commented Jul 7, 2015 at 18:47
  • 3
    $\begingroup$ 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. $\endgroup$
    – Matt Samuel
    Commented Jul 7, 2015 at 18:50
  • $\begingroup$ Reformulating a question after some answer is posted is not a good choice. $\endgroup$
    – Did
    Commented Jul 18, 2015 at 9:59

1 Answer 1

Reset to default
2
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ Great remark! I change my question and assume that $X$ and $ Y$ are in general not independent of $ H$. $\endgroup$
    – fabian
    Commented Jul 7, 2015 at 18:36
  • 1
    $\begingroup$ @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. $\endgroup$
    – Dilip Sarwate
    Commented Jul 7, 2015 at 23:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .