# Conditional expectation of (degenerate) trivariate normal?

Say $$A$$ and $$B$$ are both independent, normally distributed. $$A\sim\mathcal{N}(\mu_{A},\sigma^{2}_{A})$$ and $$B\sim\mathcal{N}(0,\sigma^{2}_{B})$$ define $$C=A+B$$. Then $$(A,B,C)'$$ are degenerate, joint trivariate normal because $$\rho_{A,C}=\rho_{B,C}=1$$ and the correlation matrix of $$(A,B,C)'$$ is singular. What would be the conditional expectation $$\mathbb{E}[B|C]$$?

Naive approach: Plug in values in the formula for conditional expectation of bivariate normal: $$\mathbb{E}[X|Y]=\mathbb{E}[X]+\rho\frac{\sigma_X}{\sigma_Y}(Y-\mathbb{E}[Y])$$

So: $$\mathbb{E}[B|C]=\mu_{B}+(\rho_{B,C})\frac{\sigma^{2}_{B}}{\sigma^{2}_{C}}(C-\mu_{C})=0+(1)\frac{\sigma^{2}_{B}}{\sigma^{2}_{A}+\sigma^{2}_{B}}(C-\mu_C)$$

But then I'm stuck for 2 reasons: (a) Since $$C=A+B$$ should I write $$(A+B-\mu_{A})$$? (b) Does it even make sense to use the formula for the conditional expectation of the bivariate normal by just using $$1$$ for $$\rho=1$$?

• those correlations are not 1 Oct 25, 2019 at 18:01
• It isn't 1? Then what is it? I thought since C was a linear combination of A and B, then the correlation with A or B would be 1.
• just compute it from variances and covariances: $Cov(A, C) = Var(A)$, $Cov(B, C) = Var(B)$, there is always some variability in C that can't be explained by A or B alone, you need both. Oct 25, 2019 at 19:15