The following question comes from Sheldon Ross's Example 6b in Properties of Expectation chapter.
Suppose that if a signal value s is sent from location A, then the signal value received at location B is normally distributed with parameters (s, 1). If S is the value of the signal sent at A, is normally distributed with parameters (μ, $σ^2$), what is the vest estimate of the signal sent if R, the value received at B, is equal to r?
Solution: Let us start by first computing the conditional density of S given R. We have $$ f_{(S|R)}(s|r) = \frac{f_{(S,R)}(s,r)}{f_R(r)} $$
$$ = \frac{f_{(S)}(s)f_{(R|S)}(r|s)}{f_R(r)} $$
$$ Ke^{-(s-μ)^2/2σ^2}e^{-(r-s)^2/2} $$
where K does not depend on s.
I mostly follow what happens after this statement, but I'm confused as to how this original equation is derived, specifically $e^{-(r-s)^2/2}$. I'm not understanding where this value comes from, and where $f_R(r)$ comes into play.
Thank you!
We are given that the conditional distribution of $R$ given $S=s$ is normal with mean $s$ and variance $1$. This means the conditional density of $R$ given $S=s$ is the following function of $r$: $$ f_{(R|S)}(r|s) = \frac1{\sqrt{2\pi}} e^{(r-s)^2/2}. $$ Now substitute this into $$\frac{f_{(S)}(s)f_{(R|S)}(r|s)}{f_R(r)}. $$ You can collect everything that doesn't depend on $s$ into a multiplicative constant $K$. This $K$ includes the density $f_R(r)$ -- which explains why it seems to have disappeared.