Find $P(X=x|Y=y)$ while giving P(X), P(Y) and $P(Y=y|X=x)$

Question

Assume that X is uniformly distributed on (0, 1) and that the conditional distribution of Y given $X = x$ is a binomial distribution with parameters $(n, x)$. Then we say that Y has a binomial distribution with fixed size n and random probability parameter.

I have to find the conditional distribution of X given Y: We have that: $$P(X=x|Y=y)\propto P(Y=y|X=x)P(X=x)$$ but the PDF for a uniform distribution on (0, 1) is just $\frac{1}{1+0}=1$ so we just have that $$P(X=x|Y=y)\propto P(Y=y|X=x)=\binom{n}{y} x^y(1-x)^{n-y}$$

But it seems to easy. Am I doing something wrong and have understand something wrong, this could not be so simple? (I have earlier found the marginal distribution of Y if this should be use)

Answer 1

\begin{align} P(X=x|Y=y) &=\frac{P(X=x)P(Y=y|X=x) }{P(Y=y)} \\ &=\frac{P(X=x)P(Y=y|X=x)}{\int_0^1 P(X=t) P(Y=y|X=t) \, dt} \end{align}

Since you have found the marginal distribution of $Y$ earlier, use it in the denominator.