Assume that $(X_1, X_2, X_3)$ are jointly normal random variables with the mean vector $(a,b,c)$ and the covariance matrix: $$\left( \begin{array}{ccc} \sigma_1^2 & \alpha & \beta \\ \alpha & \sigma_2^2 & \gamma \\ \beta & \gamma & \sigma_3^2 \end{array} \right)$$ What is the distribution of $X_1+X_2+X_3$?

  • $\begingroup$ Is it simply $\mathcal{N}(a+b+c, \sigma_1^2+\sigma_2^2+\sigma_3^2$)? $\endgroup$ – nwtqswvq Feb 8 '14 at 23:27
  • $\begingroup$ No. You can see this by considering ${\rm Var}[X_1 + X_2 + X_3] = \sigma_1^2 + \sigma_2^2 + \sigma_3^2 + 2(\alpha+\beta+\gamma).$ $\endgroup$ – heropup Feb 8 '14 at 23:30
  • $\begingroup$ Then how about $\mathcal{N}(a+b+c, \sigma_1^2+\sigma_2^2+\sigma_3^2+2\alpha+2\beta+2\gamma)$? :) $\endgroup$ – nwtqswvq Feb 8 '14 at 23:32

For every deterministic vector $U=(u,v,w)^T$ and every gaussian vector $X$ with mean $M$ and covariance matrix $C$, the distribution of $U^TX=uX_1+vX_2+wX_3$ is gaussian with mean $m=U^TM$ and variance $\sigma^2=U^TCU$.

In your case, $m=a+b+c$ and $$ \sigma^2=\begin{pmatrix}1&1&1\end{pmatrix}\cdot C\cdot\begin{pmatrix}1\\ 1\\ 1\end{pmatrix}=\sigma_1^2+\sigma_2^2+\sigma_3^2+2\alpha+2\beta+2\gamma. $$


Use the variance formula:

$Var(X_1+X_2+X_3) = \sigma_1^2+\sigma_2^2+\sigma_3^2+2\alpha+2\beta+2\gamma)$ its just the sum of all elements in the covariance matrix. The resulting sum will have mean a+b+c and variance as given above and be normally distributed.


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