I have the following problem
Consider a random variable $s$ with p.d.f. $f_s()$ and c.d.f. $F_s()$ with support on $\mathbb{R}$ and expected value $\mu_s$, consider also a discrete random variable $\tau$ such that $P(\tau=1)=\gamma$ and $P(\tau=0)=1-\gamma$.
$\tau$ and $s$ are independent.
I want to calculate the following expectation for an arbitrary real number $z$
$\mathbb{E}[s | \{\tau=0\} \cup \{s \geq z\}]$
I have been trying to do the following, but probably is the wrong path:
$\mathbb{E}[s | \{\tau=0\} \cup \{s \geq z\}] = \int_{-\infty}^\infty s f_s(s |\{\tau=0\} \cup \{s \geq z\})ds $
Using Bayes Rule (This is the part where Im not sure) I replace the conditional pdf of $s$ with:
$f(\tau=0 \text{ }\lor \text{ } s\geq z \text{ } | \text{ }s) \frac{f_s(s)}{f(\tau=0 \text{ }\lor \text{ } s\geq z)}$
Then I use the fact that $\tau$ and $s$ are independent and rewrite the following probabilities as:
$f(\tau=0 \lor s \geq z)= P(\tau=0)+(1-F_s(z))-P(\tau=0)(1-F_s(z))$
Which is equal to
$(1-\gamma)+(1-F_s(z))\gamma$
Now focusing on the conditional
$f(\tau=0 \text{ }\lor \text{ } s\geq z \text{ } | \text{ }s)$
I use the independence of $s$ and $\tau$
$P(\tau=0 | s)+P(s \geq z | s) -P(\tau=0 \land s\geq z | s)$
Which I rewrite as
$P(\tau=0)+1_{s\geq z} -P(\tau=0)1_{s\geq z}=(1-\gamma)+1_{s\geq z}-(1-\gamma)1_{s\geq z}$
Which is equal to
$(1-\gamma)+1_{s\geq z}\gamma$
Here my intuition is that the probability of $s\geq z$ given $s$ will be $1$ if in fact the $s \geq z$ and 0 otherwise, that's why I have the indicator function.
Wrapping up things, the conditional p.d.f. becomes
$f_s(s |\{\tau=0\} \cup \{s \geq z\})=\frac{(1-\gamma)+1_{s\geq z}\gamma}{(1-\gamma)+(1-F_s(z))\gamma}f_s(s)$
Replacing in the integral
$\int_{-\infty}^\infty s \frac{(1-\gamma)+1_{s\geq z}\gamma}{(1-\gamma)+(1-F_s(z))\gamma}f_s(s) ds=\frac{1}{(1-\gamma)+(1-F_s(z))\gamma} \Big((1-\gamma)\int_{-\infty}^\infty s f_s(s)ds +\gamma \int_{-\infty}^\infty 1_{s\geq z} s f_s(s)ds\Big)$
which is equal to
$\frac{1}{(1-\gamma)+(1-F_s(z))\gamma} \Big((1-\gamma)\mu_s+\gamma \int_{z}^\infty s f_s(s)ds\Big)$
I am not sure about a couple of the steps in the middle, does it make sense?
Thank you
If $X$ is a random variable on probability space $(\Omega,\mathcal A,P)$ and $A\in\mathcal A$ with $P(A)>0$ then: $$\mathbb E[X\mid A]P(A)=\mathbb EX\mathbf1_A$$
Applying that here leads to $$\mathbb E[s\mid\{\tau=0\}\cup\{s\geq z\}]P(\tau=0\vee s\geq z)=\mathbb Es[\mathbf1_{\{\tau=0\}\cup\{s\geq z\}}]$$
Making use of independence and the fact that $s$ has continuous distribution we find: $$P\left(\tau=0\vee s\geq z\right)=1-P(\tau=1\wedge s<z)=1-\gamma F_{s}\left(z\right)$$
Further we have: $$s[\mathbf{1}_{\{\tau=0\}\cup\{s\geq z\}}]=s\left[1-\mathbf{1}_{\{\tau=1\}\cap\{s<z\}}\right]=s\left[1-\tau\mathbf{1}_{\left(-\infty,z\right)}\left(s\right)\right]=s-\tau s\mathbf{1}_{\left(-\infty,z\right)}\left(s\right)$$
Again using independence we find: $$\mathbb{E}s[\mathbf{1}_{\{\tau=0\}\cup\{s\geq z\}}]=\mathbb{E}s-\gamma\mathbb{E}s\mathbf{1}_{\left(-\infty,z\right)}\left(s\right)=\mu_s-\gamma\int_{-\infty}^{z}xf_s(x)dx$$
So our final result is:$$\mathbb E[s\mid\{\tau=0\}\cup\{s\geq z\}]=\frac{\mu_s-\gamma\int_{-\infty}^{z}xf_s(x)dx}{1-\gamma F_{s}\left(z\right)}=\frac{(1-\gamma)\mu_s+\gamma\int_{z}^{\infty}xf_s(x)dx}{1-\gamma F_{s}\left(z\right)}$$
Sanity check: if $z\downarrow-\infty$ then the numerator approaches $\mu_s$ and the denominator approaches $1$ so that the RHS converges to $\mu_s$ as it should.