Prove that $P(E,X|\hat{X})=P(E|\hat{X})P(X|E,\hat{X})$

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I would like to prove that $P(E,X|\hat{X})=P(E|\hat{X})P(X|E,\hat{X})$ where $X$,$\hat{X}$ and $E$ are three (not necessarily independent) RVs.

I tried to use Bayes' rule and the chain rule in the following manner:

$$P(E,X|\hat{X})=\frac{P(E,X,\hat{X})}{P(\hat{X})}=\frac{\require{cancel}\cancel{P(\hat{X})}P(X|\hat{X})P(E|X,\hat{X})}{\require{cancel}\cancel{P(\hat{X})}}=P(X|\hat{X})P(E|X,\hat{X})$$

But I am not sure how to proceed from here.