# Find expected value of random variable, given sum with another random variable

I have two independent random variables $\tilde{u} \sim \mathcal{N}(m, v)$ and $\tilde{\epsilon} \sim \mathcal{N}(0, \eta)$, and their sum is $\tilde{x} = \tilde{u} + \tilde{\epsilon}$. I know that $$\mathbb{E} [\tilde{u} | \tilde{x} = x ] = m + \frac{v}{v + \eta} (x - m),$$ but I don't know why. The intuition makes sense, but what is this called? What is the search phrase to find a proof?

The bigger problem that I want to solve is what is $\mathbb{E} [\tilde{u} | \tilde{x} = x]$ if $\tilde{\epsilon}$ comes from one of two distributions; either $\tilde{\epsilon} \sim \mathcal{N} (0, l)$ with probability $p$, or $\tilde{\epsilon} \sim \mathcal{N} (0, h)$ otherwise. For this case can I say the following? $$\mathbb{E} [\tilde{u} | \tilde{x} = x] = m + (x-m) \left[ p \frac{v}{v + l} + (1-p) \frac{v}{v + h} \right]$$

• This is the conditional expectation. See en.wikipedia.org/wiki/…. I haven't been able to find a proof yet. – Clarinetist Mar 4 '17 at 19:42
• The formula follows from the fact that $(\tilde{x},\tilde{u})$ are bi-variate normal. You may find this post and this post helpful. – Francis Mar 4 '17 at 19:55
• @Francis Those make sense. Thanks. Can I say that $\mathbb{E} [\tilde{u} | \tilde{x} = x] = m + (x-m) \left[ p \frac{v}{v + l} + (1-p) \frac{v}{v + h} \right]$? – Richard Herron Mar 4 '17 at 20:12
• @RichardHerron: not really, in this case $E[\tilde{u}\mid\tilde{x}=x]$ is itself a random variable, as a function of random variable $\eta$ (the variance of $\tilde{\epsilon}$), so it cannot be deterministic as your second formula suggests. Maybe you want $E_{\eta}[E[\tilde{u}\mid\tilde{x}=x]]$? – Francis Mar 4 '17 at 20:28
• @Francis -- What is $\mathbb{E}_{\eta}$? I just want to show how my expectation of $\tilde{u}$ changes with the probability that $\tilde{\epsilon}$ is drawn from the low precision (high variance) distribution. I am OK with nested expectations. Is there a good standard reference on problems like this? It has been a long time since I took this class. :) – Richard Herron Mar 4 '17 at 20:37