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Suppose we have $3$ jointly normally distributed random variables $X_1, X_2, X_3$ with mean $0$ and variances $\sigma_1^2, \sigma_2^2, \sigma_3^2$. Suppose their correlations are $\rho_{12}, \rho_{23}, \rho_{13}$.

Now I know that $X_1 + X_2 \sim N(0, \sigma_1^2 + \sigma_2^2 + 2 \rho_{12} \sigma_1 \sigma_2)$, but I'm not sure how to extend this to get the distribution of $X_1 + X_2 + X_3$. I'm guessing that it is $X_1 + X_2 + X_3 \sim N(0, \sigma_1^2 + \sigma_2^2 + \sigma_3^2 + 2 \rho_{12} \sigma_1 \sigma_2 + 2 \rho_{13} \sigma_1 \sigma_3 + 2 \rho_{23} \sigma_2 \sigma_3)$ but I find it hard to actually prove it. I don't really know the correlation between $X_1 + X_2$ and $X_3$ so I can't just repeat this twice.

How would I do this?

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You can explicitly write $$\begin{align} \operatorname{Var}[X_1+X_2+X_3] &= \mathbb{E}[(X_1+X_2+X_3)^2] = \mathbb{E}[X_1^2+X_2^2+X_3^2 + 2X_1X_2 + 2X_1X_3 + 2X_2X_3] \\ &=\mathbb{E}[X_1^2]+\mathbb{E}[X_2^2]+\mathbb{E}[X_3^2] + 2\mathbb{E}[X_1X_2] + 2\mathbb{E}[X_1X_3] + 2\mathbb{E}[X_2X_3] \\ &= \sigma_1^2+\sigma_2^2+\sigma_3^2 +2\rho_{12}\sigma_1\sigma_2 +2\rho_{13}\sigma_1\sigma_3 +2\rho_{23}\sigma_2\sigma_3 \end{align}$$ as you thought.

In short: in case of doubt, going back to the definition mught be a good idea.

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    $\begingroup$ (You can also go there, if you are comfortable with that, by recalling that $\operatorname{Cov}$ is bilinear, so $\operatorname{Cov}(X_1+X_2,X_3)=\operatorname{Cov}(X_1,X_3)+\operatorname{Cov}(X_2,X_3)$) $\endgroup$
    – Clement C.
    Apr 10, 2019 at 18:40

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