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This is a fairly straight forward problem but I don't know how to calculate the covariance given just the variance of X and Y.

Suppose X and Y are independent random variables such that Var(X)=1 and Var(Y)=2. What is the variance of X+2Y-3?

So far, I have

(1^2)var(x) + (2^2)var(y) + 2(1)(2)cov(x, y)

Which comes out to

1 + 8 + 4cov(x, y)

How can I calculate 4cov(x, y)?

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  • $\begingroup$ Hint: If $X$ and $Y$ are independent, then $\mathrm{Var}(X+Y) = \mathrm{Var}(X) + \mathrm{Var}(Y)$. In particular, $\mathrm{Cov}(X,Y) = 0$. $\endgroup$ – snar Mar 10 '15 at 18:16
  • $\begingroup$ Thank you! Wow, that was very simple. $\endgroup$ – likeabbas Mar 10 '15 at 18:29
  • $\begingroup$ If $X$ and $Y$ are independent their covariance is $0$. $\endgroup$ – André Nicolas Mar 10 '15 at 18:40
  • $\begingroup$ Correct me if I'm wrong, but I think you just want $V(X + 2Y - 3)$ and you're imagining you need the covariance, which you don't. By independence, $$V(X + 2Y - 3) = V(X) + V(2Y) = V(X) + 4V(Y).$$ $\endgroup$ – BruceET Mar 13 '15 at 3:01

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