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

  • $\begingroup$ 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. $\endgroup$ – Greenparker Apr 10 '16 at 15:20

I will explain a symmetric PDF by example. Suppose the proposal distribution in the Metropolis-Hastings algorithm is a Normal distribution with variance 1. Then

$$f(x|y) = \text{density of } N(y,1). $$

This proposal distribution is symmetric since $f(x|y) = f(y|x)$. I will demonstrate this by writing the pdf of both.

\begin{align*} f(x|y) = \dfrac{1}{\sqrt{2\pi}}\exp \left\{-\dfrac{(x - y)^2}{2} \right\}\\ f(y|x) = \dfrac{1}{\sqrt{2\pi}}\exp \left\{-\dfrac{(y - x)^2}{2} \right\}\\ \end{align*}

Thus, you see that the pdfs are the same, and hence we say that the proposal distribution is symmetric.


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