# Confusion in MAP estimation

Consider the situation $$r = a+n$$, where $$n \sim \mathcal{N}(0,\sigma_n^2)$$. I am having confusion with respect the computation of $$p_{a|r}(A)$$ for the above scenario.
Option 1, $$r = a+n \implies a = r - n \implies p_{a|r}(A) = \mathcal{N}(R, \sigma_n^2)$$
Option 2, from Bayes theorem gives, $$p_{a|r}(A) = \dfrac{p_{r|a}(R)p_a(A)}{p_r(R)}$$. Assuming $$p_a(A) = \mathcal{N}(0, \sigma_a^2)$$, we get $$p_{a|r}(A) = \mathcal{N}\left(\dfrac{R}{1 + \frac{\sigma_n^2}{\sigma_a^2}}, \dfrac{\sigma_n^2}{1 + \frac{\sigma_n^2}{\sigma_a^2}}\right)$$.
1. My first question is why do these two approches give different anwers? I understand option 2 depends on the prior for $$a$$, but even in that scenario we compute the likelihood $$p_{r|a}(R)$$ using the approach described in option 1.
2. Option 2 converges to option 1 when $$\sigma_a^2 \to \infty$$. This can be seen with the gaussian prior, however is it true for other prio distributions on $$a$$ as well?

The dependence structure of $$a,n$$ and $$r$$ is important! Typically in such "inverse problem" scenarios you assume that $$n$$ is independent of $$a$$ and therefore $$p_{r|a} = \mathcal{N}(A, \sigma_n^2),$$ but since $$r$$ and $$n$$ are not independent (knowing $$r$$ gives some information on $$n$$), you can not conclude $$p_{a|r} = \mathcal{N}(R, \sigma_n^2).$$ (I tried to adopt your notation, which is not very intuitive to me). So, under the above assumption, Option 1 is not correct, while Option 2 is (for that prior).
• Thanks for the response, but whenever we have a random variable say $z = x + y$, to find $z|x=x_0$ we simply use the distribution of $x_0+y$ , rarely do we have to check for independence between $x$ and $y$. Please help me figure out where am I making a mistake? – sh10 Aug 6 at 4:02
• Independence of $x$ and $y$ is crucial for this step. Imagine the extreme case $x=y$ ("maximal dependence"), then $z|x=x_0$ would equal $2x_0$ almost surely. Since $y$ is often some sort of noise, the independence is sometimes not mentioned explicitly, which is negligent, but it is definitely necessary for writing down the distribution of $z|x=x_0$ in that form. – iljusch Aug 6 at 7:26