# How do Normals combine with Bayes? [duplicate]

If $X = N_x$ and $Y = 4*X + N_y$, where $N_x$, $N_y \sim N(0, 1)$, then we have that $P(Y|X=x) = N(4x, 1)$. If we want to then find $P(X|Y=y)$, I would think that Bayes would work:

$P(X|Y=y) = P(Y|X=x)P(x)/P(y) = N(4x, 1)P(x)/P(y)$.

For x=2, the solution is N(8/17, 1/17). I don't see how this results. Can someone clarify please?

EDIT: Another user cited this as being a duplicate, but I don't see why.

EDIT: Originally, I asked "what is $P(x)$, $P(y)$? Are they the pdf of $N(0, 1)$ at their respective values?"

• Could you please use $\TeX$ formatting? In current form the formulas are not very readable. Check also: en.wikipedia.org/wiki/Bayes%27_theorem#Simple_form_2 since it seems to answer your question.
– Tim
Mar 21 '18 at 12:43
• Edited with latex. Thanks. Regarding your link, does that mean that they are just the evaluation of the normal at their respective values? Mar 21 '18 at 13:33

$P(x) = N_x \sim N(0,1)$ by hypothesis.
You can find $P(y)$ using Bayes rule:
$P(y) = \int_\mathbb{R}{P(y,x)dx} = \int_\mathbb{R}{P(y|x)P(x)dx}$.