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Assume $X$ and $Y$ are independent Gaussian random variables and $Z = X +Y$. Then we know that $X$ and $Z$ are jointly Gaussian and $P(X=x \vert Z=z)$ is also Gaussian. Is there any simple way to show this? I know one way that we use $P(X=x \vert Z=z) = \dfrac{P(X=x, Z=z)}{Z=z}$ and write the explicit pdfs. Then we show the result has the form of Gaussian. Can we show this somehow using properties of jointly Gaussian variables?

Now, consider a Bernoulli random variable $W$ which is independent of $X$ and $Y$. Now we want to compute the distribution of $P(X=x \vert Z=z, W=w, WZ = q)$. Is this one also Gaussian?

My approach: If $wz =q$, then \begin{align} P(X=x \vert Z=z, W=w, WZ = q) = P(X=x \vert Z=z, W=w) = P(X=x \vert Z=z) \end{align} where the last equality is true because $W$ is independent of $X,Y$. Hence, the distribution is Gaussian.

But, what if $wz \neq q$?

m0_as
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1 Answers1

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If $wz\ne q$, then you are conditioning on the impossible event, and this was aleady addressed here or here

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Momo
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  • I updated the question, can you take a look at the first question? – m0_as Dec 22 '16 at 00:30
  • Other than plugging in the formula, I don't know of any way to prove that the conditional of a multivariate Gaussian is Gaussian. The only thing I can think about is to simplify the computation using matrices, as shown here – Momo Dec 22 '16 at 00:42
  • Okay, Is the fact that "the conditional of a multivariate Gaussian is Gaussian" a theorem? any reference? Assume we know this, how can we use this to conclude that $P(X=x | Z=z)$ is Gaussian? – m0_as Dec 22 '16 at 00:48
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    In the section "Conditional distributions" of the link I provided, wikipedia not only state this, but also gives you a formula for the mean and variance (both in matrix form and explicit form for bivariate case) – Momo Dec 22 '16 at 00:51
  • Yes, I found that there. Using that we can say that $P(X=x | Y=y, X=x)$ is Gaussian. How to get the result we want form there? Using the fact that linear function of a multivariate Gaussian is Gaussian? – m0_as Dec 22 '16 at 00:57
  • Then your multivariate should be $(X,Y,X)$, which is a degenerate multivariate normal, so the formula in the link does not apply. – Momo Dec 22 '16 at 01:09