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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.

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

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 $E[XY]$.

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    $\begingroup$ +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

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