# Fixing the sum of normal distributed random variables

Let $(x_1, x_2, x_3) \in \mathbb{R}^3$ be jointly normally distributed with mean $\mu = (\mu_1,\mu_2, \mu_3)$ and covariance matrix $\Sigma \in \mathbb{R}^{3 \times 3}$.

It is known that $y := x_1 + x_2+ x_3$ is also normally distributed with mean $\mu_y = \mu_1+\mu_2+\mu_3$ and variance $\sigma^2_y = \sum_{i=1}^3 \sum_{j=1}^3 \Sigma_{i,j}$.

My question is concerned with the conditional expectation and variance of $(x_1, x_2, x_3) \in \mathbb{R}^3$ given that the sum $y=x_1 + x_2+x_3 = a \in \mathbb{R}$ is fixed, i.e. $$\mathbb{E}[(x_1, x_2, x_3) | x_1 + x_2+x_3 = a] .$$ What can be said about this setting? Is the conditional distribution still normal? How can the density be computed? Thanks!

• Do you know how to compute the conditional expectation of x_1| x_1+x_2? If so, denote it by $E[x_1 | x_1+x_2]$. Then $E[x_2| x_1+x_2]=E[x_1+x_2 |x_1+x_2]- E[x_1|x_1+x_2]$, i.e. the joint distribution of $(x_1,x_2)$ given their sum is trivial. The same carries over to $x_1+x_2+x_2$. Is this clear? Do you have any problems finding $E[x_1|x_1+x_2]$? – Ecthelion Feb 2 '18 at 21:28
• As mentioned above the conditional distribution is 2-dimensional only due to the linear dependency. Say you write $X_3 = a - X_1 - X_2$ and "discard" it. Now consider the joint distribution $(X_1, X_2, X_1 + X_2 + X_3)$ which is again multivariate normal as it is an affine transformation from $(X_1, X_2, X_3)$. Afterward you can compute the distribution of $(X_1, X_2)|X_1 + X_2 + X_3 = a$ with ease, by the standard formula. – BGM Feb 3 '18 at 3:00