Under what circumstance does $P(X|H) = P(H|X)$? I came across a Machine Learning exam question regarding the difference between the Frequentist and Bayesian approach to classification; it specifically asked what condition must be met for the two to be equivalent. 
Formulating a mathematical answer isn't too tricky. Our hypothesis is that an observation $X$ from our dataset belongs to a particular class $H$, and we want the conditional probabilities to be equal such that:
$$P(X|H) = P(H|X)$$
Apply Bayes theorem:
$$ P(X|H) = \frac{P(X|H)P(H)}{P(X)}\ $$
Which simplifies to: 
  $$ P(X) = P(H) $$
That's as far as I can get with respect to solving the problem. I have no idea how to put this into words, or give an example where this condition would be satisfied. Can someone please provide an intuitive explanation?
 A: Since $$P(H|X)=\frac{P(X|H)P(H)}{P(X)}$$ and $$P(X)=\int_{H\in\Theta}P(X|H)P(H)\mathrm{d}H,$$ where $\Theta$ is the parameter space and also assuming everything is continuous, it follows that since $$\frac{P(H)}{P(X)}=1,$$ that $$P(X)=\int_{H\in\Theta}P(X|H)P(H)\mathrm{d}H=P(H).$$
Since $$P(X)=\int_{H\in\Theta}P(X|H)P(H)\mathrm{d}H$$ is a constant, then $P(H)$ must be a constant.  It follows then that $P(H)\propto{k},k>0$ which is only satisfied by the uniform distribution.
$$\therefore{P(H|X)}=P(X|H)\iff{P(H)\propto{1}}. \square$$ 
The fundamental assumption of Frequentist methods is that no prior information exists.  No information outside the sample is usable in Frequentist methods.  This maps to Bayesian methods with uninformative prior densities.  It is rather important to note that not all Frequentist methods map to Bayesian methods with a uniform prior density.  A good example of this is the estimation of parameters for the binomial density.  It maps to a Bayesian solution with a Haldane prior density.  The uniform density actually provides a more precise solution, in general, than the Haldane prior density does.
Additionally, some Frequentist methods, such as those covered by Stein's Lemma, do not map to any Bayesian density precisely because they are inadmissible statistics and all Bayesian methods are admissible.  So, for example, $z=f(a,b,c)$ where all the assumptions are fully met has no Bayesian equivalent if OLS is used because OLS is inadmissible.  If you were to attempt to use a uniform prior density over the parameter space the posterior density would diverge.
A: For frequentist vs baysian maybe you are looking for
$$\frac{P(X\vert H_1)}{P(X\vert H_2)} = \frac{P(H_1\vert X)}{P(H_2\vert X)}$$
?
This maximum likelihood equals maximum posterior probability happens if the prior is flat $P(H_1)=P(H_2)$
