# Symmetric PDFs in Metropolis-Hastings

My textbook says that a symmetric PDF satisfies $$f(x|y)=f(y|x).$$ Can anyone explain this? Is it equivalent to $f(x+a)=f(x-a)$?

• I am not sure what you mean by $f(x+a) = f(x-a)$ and exactly what $a$ is here, but I have answered the question about symmetric PDFs. – Greenparker Apr 10 '16 at 15:20

$$f(x|y) = \text{density of } N(y,1).$$
This proposal distribution is symmetric since $f(x|y) = f(y|x)$. I will demonstrate this by writing the pdf of both.