I have problem figuring out the solution for this task:

X1 and X2 are independent random variables with normal distribution ~N(2,1). What is a covariance of $X_1 − 4X_2^2$ and $X_1 + X_2$.

So far I've managed to come up with this:

$cov(X_1 − 4X_2^2$ , $X_1 + X_2) = cov(X_1, X_1) - cov(X_1,X_2)-4cov(X_1, X_2^2) - 4cov(X_2, X_2^2) = Var(X_1) + 0 -4cov(X_1, X_2^2) - 4cov(X_2, X_2^2)$

But I don't know how to handle both $cov(X_1, X_2^2)$ and $cov(X_2, X_2^2)$. X1 and X2 are independent, but does it mean that $X_1$ and $X_2^2$ are independent too? What about $X_2$ and $X_2^2$?

  • $\begingroup$ Hints: If $X$ and $Y$ are independent random variables, then so are $g(X)$ and $h(Y)$ independent random variables for measurable functions $g(\cdot)$ and $h(\cdot)$. In particular, $E[g(X)h(Y)] = E[g(X)]E[h(Y)]$ and also $\operatorname{cov}(g(X), h(Y)) = 0$. Finally, $E[X^n]$ is readily computed (or looked up) for normal random variables. $\endgroup$ – Dilip Sarwate Apr 8 '15 at 20:07

\begin{align} cov(X_1,X_2^2)&=EX_1X_2^2-EX_1EX_2^2 \mbox{ (by definition)}\\ &=EX_1EX_2^2-EX_1EX_2^2 \mbox{ (by independence)}\\ &=0 \end{align}

\begin{align} cov(X_2,X_2^2)&=EX_2X_2^2-EX_2EX_2^2 \mbox{ (by definition)}\\ &=EX_2^3-EX_2EX_2^2\\ &=14-2\times5\\ &=4 \end{align}

Note if $X\sim N(\mu,\sigma^2)$ then \begin{align} EX^3&=\mu^3+3\mu\sigma^2\\ EX^2&=\mu^2+\sigma^2 \end{align}

  • $\begingroup$ Thank you. I understand what you did in covariances, but this normal distribution moments surprised me. I tried to calculate them using integral, but this would take ages. $\endgroup$ – Przemysław Robert Wilk Apr 8 '15 at 21:42

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