Thresholding rule for posterior distribution

Question

I have a binomial distribution $$P(Y=1)=q$$ and a a conditional distribution $P(x|y=0)=Normal(0,1)$, $P(x|y=1)=Normal(0,\sigma^2)$

I'm trying to determine the posterior mode $$arg\ max_{y}P(y|x)$$ as a thresholding rule so I think I need to set the two piecewise parts of $P(x|y)P(y)$ equal to each other and solve for $x$.

I'm getting something like $$2x^2\sigma^2+ln(\sigma^2)-2ln(q) <> x^2-2ln(1-q)$$

So am I thinking about this correctly that I should just solve for $x$ here or am I misunderstanding?

Answer 1

Exactly. Only check and correct your final inequality.

You need to find $\arg\max_{y}P(y|x)$, that is to find $y$ maximizing $P(y|x)$. There are 2 possible values of $y$: $0$ and $1$. So, you need to compare the values of $$P(x|0)P(0)=(1-q)\dfrac{1}{\sqrt{2\pi}}e^{-x^2/2} \text{ and } P(x|1)P(1)=q\dfrac{1}{\sqrt{2\pi\sigma^2}}e^{-x^2/(2\sigma^2)}.$$

When the l.h.s. is greater, $\arg\max_{y}P(y|x)=0$. When the r.h.s. is greater, $\arg\max_{y}P(y|x)=1$. Which of these options is actually happens will depend on $x$. You need only to know whether $\sigma>1$ or not. If so, $$(1-q)\dfrac{1}{\sqrt{2\pi}}e^{-x^2/2} < q\dfrac{1}{\sqrt{2\pi\sigma^2}}e^{-x^2/(2\sigma^2)}$$ is equivalent to $$x^2> \dfrac{2\sigma^2}{\sigma^2-1}\cdot\ln\left(\sigma\frac{1-q}{q}\right)$$ For this $x$ we get $\arg\max_{y}P(y|x)=0$.