Vrftjkry

I have a continuous random variable $X$ whose distribution is conditioned on another discrete random variable $S$. The conditional density function of $X$ is

• $f_{X|S}(x|1) = alpha e^{-alpha x}, x geq 0$


• $f_{X|S}(x|-1) = beta e^{beta x}, x leq 0$

I can see that when $S=1$, $X sim Exp(alpha)$ with $alpha >0$ and when $S=-1$, $-X sim Exp(beta)$ with $beta >0$. That is, $S$ is the sign of $X$.

I then define $Z = |X|$ and I have to calculate $mathbb{P}(S=1|Z=z)$. I know that this is the posterior probability of $S$ and I can use the Bayes rule for the case of a discrete unknown and continuous observation. The chosen formula is:
$$p_{S|Z}(1|z)= frac{p_S(1)f_{Z|S}(z|1)}{f_Z(z)}$$
and I know from the prior of $S$ that $p_S(1) = p$.

My approach to calculate $f_{Z|S}(z|1)$ would be to consider that, if I know that $S=1$ therefore $Z = X$ and $f_{Z|S}(z|1) = f_{X|S}(x|1)$. What I don't understand is how to come out with the marginal of $Z$. The marginal of $X$ is:

• 
  $p cdot alpha e^{-alpha x}$ when $x geq 0$
  


• 
  $(1-p) cdot beta e^{beta x}$ when $x leq 0$
  

and my idea here is that the marginal of $Z$ is $p cdot alpha e^{-alpha z} + (1-p) cdot beta e^{beta z}$ since the $|X|$ mirrors $-X$ for the negative part, but I get a bit lost at this point.

EDIT: I also thought that, since $f_{X|S}(x|-1) = beta e^{beta x}$ then $f_{Z|S}(z|-1) = beta e^{-beta x}$, therefore
$$f_Z(z) = p cdot alpha e^{-alpha z} + (1-p) cdot beta e^{-beta z}$$
but I don't know whether this is correct.

For $z ge 0,$
$$begin{align*}
Pr[Z le z] &= Pr[-z le X le z] \
&= Pr[-z le X < 0 mid S = -1]Pr[S = -1] + Pr[0 le X le z mid S = 1]Pr[S = 1] \
&= (1 - e^{beta (-z)})(1-p) + (1 - e^{-alpha z})p.
end{align*}$$

Therefore,
$$f_Z(z) = p alpha e^{-alpha z} + (1-p) beta e^{-beta z}.$$

Yes your thoughts are correct.

Assuming $P(S=1)=p=1-P(S=-1)$,

we have for all $zge 0$,

begin{align}
P(Zle z)=P(|X|le z)&=P(|X|le zmid S=1)P(S=1)+P(|X|le zmid S=-1)P(S=-1)
\&=P(Xle zmid S=1)p+P(-Xle zmid S=-1)(1-p)
\&=pF_{Xmid S=1}(z)+(1-p)F_{-Xmid S=-1}(z)
end{align}

, where $F$ is the distribution function.

So the density of $Z$ is

begin{align}
f_Z(z)&=pf_{Xmid S=1}(z)+(1-p)f_{-Xmid S=-1}(z)
\&=palpha e^{-alpha z}mathbf1_{zge0}+(1-p)beta e^{-beta z}mathbf1_{zge0}
\&=left[palpha e^{-alpha z}+(1-p)beta e^{-beta z}right]mathbf1_{zge0}
end{align}