MMSE of conditional normal distribution

Question

I have 2-step experiment: Flip a fair coin, $C = 0$ or $1$ wp $1/2$. Than generate normally distribution, $X|C$~$N(C, v)$, while $v$ is known. I trying to find MMSE estimator. First I calculate the conditional distribution P(C|X) with bayes rule and than I just take a E: \begin{align}P(C|X) = \frac{N(C,v)}{N(0, v) + N(1, v)} \end{align} Since $C$ can get only $0$,$1$, the $E$ is: \begin{align} C_{MMSE} = E[C|X] = \frac{N(1,v)}{N(0, v) + N(1, v)} \end{align}

Now I want to prove that this estimator is unbiased. It is easy to show that $E[C] = 1/2$, but I failed to calculate the $E$ of MMSE estimator of $C$. Any help? Thanks!

Answer 1

well, finally i handle it: \begin{align} E[C_{MMSE}] = E[E[C|X]] = \int_{-\infty}^{\infty}C_{MMSE}\cdot f(x)dx =...=\frac{1}{2} \end{align}