Given the following:
Let $r = ab + n$, where $a, b,$ and $n$ are independent zero-mean Gaussian random variables with variances $\sigma_a$, $\sigma_b$, and $\sigma_n$, respectively. Find the MAP estimate of $a$, $\hat{a}_{map}(R)$.
My thought is to find the conditional density $p_{r|a}(R|A)$ and substitute it into the MAP equation $$\frac{\partial l_B(A;R)}{\partial A}|_{\hat{a}_{map}(R)} = \frac{\partial ln\;p_{r|a}(R|A)}{\partial A}|_{\hat{a}_{map}(R)} + \frac{\partial ln\;p_a(A)}{\partial A}|_{\hat{a}_{map}(R)} = 0$$ I am stuck trying to find the conditional pdf $p_{r|a}(R|A)$. I do not understand how to obtain this density from the given information. It seems that $ab$ has a product-normal density, and adding $n$ would result in convolution of that density with a Gaussian. However, I was told that $$p_{r|a}(R|A) = \frac{1}{\sqrt{2\pi}\sigma_r}exp(-\frac{R^2}{2\sigma_r^2}), \sigma_r = A^2\sigma_b^2 + \sigma_n^2$$ without any explanation, as though it should be obvious. I do not see where this comes from. Is this because the conditional density treats $A$ as a constant, so that $R$ becomes a linear combination of Normal random variables?