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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 $\rho=1$?

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    $\begingroup$ those correlations are not 1 $\endgroup$
    – carlo
    Commented Oct 25, 2019 at 18:01
  • $\begingroup$ 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. $\endgroup$
    – Adam
    Commented Oct 25, 2019 at 18:43
  • $\begingroup$ 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. $\endgroup$
    – carlo
    Commented Oct 25, 2019 at 19:15
  • $\begingroup$ I see. Thank you. And what about using the formula for the conditional expectation? Is that OK even though C is a function of A and B? $\endgroup$
    – Adam
    Commented Oct 25, 2019 at 21:46
  • $\begingroup$ I don't know it, but... why not? $\endgroup$
    – carlo
    Commented Oct 25, 2019 at 21:56

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