Following up on this question, how would you derive the expectation and variance of the sum of two normally distributed random variables that aren't necessarily independent?

For example, if $$X \sim N(\mu, 3\sigma^2)$$ and $$Y \sim N(\mu + 9, \sigma^2)$$ is there a way to calculate the expectation and variance? What would the resulting X + Y distribution be in concrete terms? We aren't given the covariance.


$$E[X+Y]=E[X]+E[Y]=\mu+\mu+9=2 \mu +9$$ $$\begin{array} VVar[X+Y]&=Var[X]+Var[Y]+2Cov[X,Y]\\ &=3 \sigma^2+\sigma^2+E[XY]-E[X]E[Y]\\ &=4 \sigma^2+E[XY]-\mu(\mu+9) \end{array}$$

We can't say that much about the distribution of $XY$ without some more assumptions. If we know that $X$ and $Y$ are independent then $XY$ has a chi-squared distribution and we can compute $E[XY]$.

  • 1
    $\begingroup$ +1, However, if we know that $X$ and $Y$ are independent, then $E(XY) = E(X)E(Y)$ or equivalently, $Cov(X,Y) = 0$. No need to use the chi-square relationship here. $\endgroup$ – knrumsey Oct 13 '17 at 20:12
  • $\begingroup$ In other words, the variance would just be $4\sigma^2$? $\endgroup$ – username Oct 13 '17 at 20:18
  • $\begingroup$ Yes, if your independent then $Cov[X,Y]=0$. $\endgroup$ – Wintermute Oct 13 '17 at 20:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.