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Let $W = (1 - Z)X + ZY$ where $X \sim N(0, \sigma^2)$, $Y \sim N(\mu,\sigma^2)$ and $Z$ is Bernoulli with $P(Z = 0) = P(Z = 1) = 1/2$. $X,Y$ and $Z$ are independent. I try to find the distribution $p(Z|W)$. It seems $Z$ and $W$ are not independent. $p(Z\mid W) = \frac{p(Z,W)}{p(W)}$ does not really help. Any ideas to proceed?

Edit:

$$\begin{align} p_{Z|W}(z|w) &= \frac{p_{W|Z}(w|z)p_Z(z)}{p_W(w)} \\ p_W(w) &= p_{W|Z}(w|z=0)p_Z(z=0) + p_{W|Z}(w|z=1)p(z=1) \\ &= 0.5N(0, \sigma^2) + 0.5N(\mu, \sigma^2) \end{align}$$

Then: If $z = 1$, $$p_{Z|W}(z=1\mid w) = \frac{0.5N(\mu,\sigma^2)}{0.5N(0, \sigma^2) + 0.5N(\mu, \sigma^2)}$$ and if $z = 0$, $$\begin{align} p_{Z|W}(z = 0\mid w) &= \frac{0.5N(0,\sigma^2)}{0.5N(0, \sigma^2) + 0.5N(\mu, \sigma^2)} \\ &= \left.\frac{\frac{1}{\sqrt{2\pi\sigma^2}}\exp(\frac{-1}{2\sigma^2}(z-0)^2)}{0.5\frac{1}{\sqrt{2\pi\sigma^2}}\exp(\frac{-1}{2\sigma^2}(z-0)^2) + 0.5\frac{1}{\sqrt{2\pi\sigma^2}}\exp(\frac{-1}{2\sigma^2}(z-\mu)^2)} \right\vert_{z = 0} \end{align}$$

Lee David Chung Lin
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    Switch the conditioning, such as in Baye's rule. You want to find the conditional mass function $P[Z=z|W=w]$ but it is easier to find the conditional density $f_{W|Z}(w|z)$. – Michael Jun 15 '21 at 19:09
  • If you prefer, you can fix $\delta>0$ as a small value and find $P[Z=0|W \in [w, w+\delta]]$, which by Baye's rule is... – Michael Jun 15 '21 at 19:17
  • @Michael I edit the question. Is this approach true? – eet Jun 15 '21 at 19:22
  • I don't like the notation, but your edits are roughly doing manipulations that get a desired form of answer. Specific critiques: (i) $P[W]$ does not make sense because $W$ is a random variable, and you can only take probabilities of events; (ii) I do not know what $p(Z|W)$ means or what its variables are; (iii) Strictly speaking, $p(Z|W)$ cannot be related in an equality to $\frac{0.5N(0,\sigma^2)}{0.5N(0, \sigma^2) + 0.5N(\mu, \sigma^2)}$ because they are different types of things with different variables ($Z, W$ versus $\mu, \sigma$). – Michael Jun 15 '21 at 19:27
  • @Michael I changed the notation a little bit. Can you explain why $p_{W|Z}(w|z)$ cannot be related that? – eet Jun 15 '21 at 19:35
  • Looks kind of good as long as $p_{W|Z}(w|z)$ is a density and $p_Z(z)$ a mass function. There seems to be a factor of $0.5$ in the final numerator that is missing, and that final expression should be in terms of $w$, not $z$ (we already know $z=0$ in that expression). I would loose the imprecise $N(\mu,\sigma^2)$ symbolic representation of a PDF, as it is not helping you notice those issues. You can relate everything to basic formulas of conditional probabilities of events $P[A|B]$ by defining events $A = {Z=0}$ and $B= {W \in [w, w+\delta]}$. – Michael Jun 16 '21 at 04:03

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