Why is $\operatorname{argmax}_{\pi} P(\pi \mid x) = \operatorname{argmax}_{\pi} P(x, \pi)$?

Question

Why is $\operatorname{argmax}_{\pi} P(\pi \mid x) = \operatorname{argmax}_{\pi} P(x, \pi)$?

Could it be because of Bayes Theorem

$$P(\pi \mid x) = \frac{P(x, \pi)}{P(\pi)}$$

and since $\operatorname{argmax}_{\pi}P(\pi \mid x)$ gives possible values of $\pi$ for which $P(\pi \mid x)$ is maximum and since $P(\pi \mid x) = \frac{P(x, \pi)}{P(\pi)}$ so when $P(\pi \mid x)$ is maximum $P(x, \pi)$ will be maximum too. So $\operatorname{argmax}$ will be the same for both.

Am I right here?

Answer 1

Your reasoning is right but your formula is wrong. Bayes rule says that $P(\pi | x) = \frac{P(x, \pi)}{P(x)}$, where the denominator is clearly independent of $\pi$ so the $\mbox{argmax}_\pi$ of either $P(\pi | x)$ or $P(x,\pi)$ is the same.