2
$\begingroup$

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? $$\mathbb{E} [\tilde{u} | \tilde{x} = x] = m + (x-m) \left[ p \frac{v}{v + l} + (1-p) \frac{v}{v + h} \right]$$

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ This is the conditional expectation. See en.wikipedia.org/wiki/…. I haven't been able to find a proof yet. $\endgroup$
    – Clarinetist
    Mar 4, 2017 at 19:42
  • 1
    $\begingroup$ The formula follows from the fact that $(\tilde{x},\tilde{u})$ are bi-variate normal. You may find this post and this post helpful. $\endgroup$
    – Francis
    Mar 4, 2017 at 19:55
  • $\begingroup$ @Francis Those make sense. Thanks. Can I say that $\mathbb{E} [\tilde{u} | \tilde{x} = x] = m + (x-m) \left[ p \frac{v}{v + l} + (1-p) \frac{v}{v + h} \right]$? $\endgroup$
    – Richard Herron
    Mar 4, 2017 at 20:12
  • 1
    $\begingroup$ @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 $E_{\eta}[E[\tilde{u}\mid\tilde{x}=x]]$? $\endgroup$
    – Francis
    Mar 4, 2017 at 20:28
  • $\begingroup$ @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. :) $\endgroup$
    – Richard Herron
    Mar 4, 2017 at 20:37

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.