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Suppose $\{X_n\}$ and $\{Y_n\}$ are two sequences of random variables and we know that $X_n \overset{L^2}{\to} X$ and $Y_n \overset{L^2}{\to} Y$, where $\overset{L^2}{\to}$ means converge in mean square sense. In addition, we know that $X$ and $Y$ are independent random variables, and both have finite mean and variance. I'm interested in showing $$\lim_{n \to +\infty} Cov(X_n, Y_n) = 0$$

I know that the above is not always true (e.g., see this Math StackExchange thread). However, I wonder what other conditions can one impose that will make it true?

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  • $\begingroup$ I'm a bit confused: doesn't this just follow from Holder's inequality? Note that the linked question assumes convergence in distribution, which is much weaker. $\endgroup$
    – Leo Moos
    Apr 9, 2021 at 15:21
  • $\begingroup$ @LeoMoos, you are right that the L2 convergence here is much stronger than the convergence-in-distribution in the referenced link. That being said, would you mind elaborating a bit about how the limit of covariance can be proved (e.g., via Holder's inequality as you suggested)? Thanks! $\endgroup$ Apr 9, 2021 at 15:25
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    $\begingroup$ By Holder's inequality $\mathbf{E}[X_n Y_n] - \mathbf{E}[XY] = \mathbf{E}[X_n ( Y_n - Y)] + \mathbf{E}[(X_n - X) Y] \to 0$ as $n \to \infty$. $\endgroup$
    – Leo Moos
    Apr 9, 2021 at 16:14
  • $\begingroup$ @LeoMoos This makes sense. Does it matter that $X$ and $Y$ may not have zero mean? I.e., it is possible that $Cov(X,Y) = E(XY) - E(X)E(Y) \neq E(XY)$? $\endgroup$ Apr 9, 2021 at 16:33

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First off $\mathbf{E}[X_n] \to \mathbf{E}[X]$ and $\mathbf{E}[Y_n] \to \mathbf{E}[Y]$ as $n \to \infty$, whence also $\mathbf{E}[X_n] \mathbf{E}[Y_n] \to \mathbf{E}[X] \mathbf{E}[Y]$. Via Holder's inequality one additionally has $\mathbf{E}[X_n Y_n] - \mathbf{E}[XY] = \mathbf{E}[X_n (Y_n - Y)] + \mathbf{E}[Y(X_n - X)] \to 0$.

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  • $\begingroup$ Much appreciated! $\endgroup$ Apr 9, 2021 at 16:58

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