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Let $X$ and $Y$ be independent random variables with finite variances, and let $U = X + Y$ and $V = XY$. Under what condition are $U$ and $V$ uncorrelated?

MY ATTEMPT

We say that two variables are uncorrelated if their covariance equals zero. Bearing this in mind, we have \begin{align*} \textbf{Cov}(U,V) & = \textbf{E}(UV) - \textbf{E}(U)\textbf{E}(V)\\\\ & = \textbf{E}(X^{2}Y + XY^{2}) - \textbf{E}(X+Y)\textbf{E}(XY)\\\\ & = \textbf{E}(X^{2})\textbf{E}(Y) + \textbf{E}(X)\textbf{ E}(Y^{2}) - \textbf{E}(X)^{2}\textbf{E}(Y) - \textbf{E}(X)\textbf{E}(Y)^{2}\\\\ & = \textbf{E}(Y)[\textbf{E}(X^{2}) - \textbf{E}(X)^{2}] + \textbf{E}(X)[\textbf{E}(Y^{2}) - \textbf{E}(Y)^{2}]\\\\ & = \textbf{E}(Y)\textbf{Var}(X) + \textbf{E}(X)\textbf{Var}(Y) = 0 \end{align*}

Is there a specific name for this last expression? Any contribution is appreciated. Thanks :)

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  • $\begingroup$ "Is there a specific name for this last expression?" No. $\endgroup$ – Did Jan 30 at 17:37
  • $\begingroup$ Looks like you've found the condition. There is no specific name for this. $\endgroup$ – Ben Jan 31 at 9:16
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First of all, there is no specific name of what you've got. Second of all, assuming r.v. $X$ is not degenerated we have the following condition $$ \mathbf{E} Y = - \mathbf{E} X \frac{\mathbf{Var} Y}{\mathbf{Var} X}. $$

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