# How does the MAP estimate approach the MLE?

I had a question concerning the determination of the Maximum Likelihood Estimate (MLE) of a parameter (or vector of parameters) by taking the natural logarithm of the posterior distribution.

$p(D|h)$ is the likelihood of the data set.

$D$ is the data set.

$p(h)$ is the prior probability of the parameter $h$ (I would use $\theta$ instead of $h$, but for this example I'd rather use discrete probability instead of continuous).

So the posterior distribution is proportional to the likelihood times the prior:

$p(h|D) \propto p(D|h) \times p(h)$

My textbook first explains how to find the MAP estimate. MAP stands for "maximum a posteriori". It is basically the $h$ parameter vector in the posterior distribution where the probability mass (or density, depending on whether it is discrete or continuous) is the highest.

It says that to find the MAP of $h$, we do the following:

$h^{MAP} = \underset{h}{argmax}(p(D|h) p(h)) = \underset{h}{argmax}(log[p(D|h) p(h)]) = \underset{h}{argmax}(log[p(D|h)] + log[p(h)])$

This is possible of course because $log(x)$ is an increasing function for all $x \gt 0$, the log function has the property $log_{a}(p \times q) = log_{a}(p) + log_{a}(q)$, and finally, the $\underset{h}{argmax}()$ function simply wants to find the $h$ parameter that gives the largest value for whatever is in the parentheses.

My question is this: the book also says that as we get more and more data (as $D$ gets larger), the training data likelihood overwhelms the prior. In other words, the MAP estimate approaches the Maximum Likelihood Estimate (MLE) like so:

$\underset{h}{argmax}(log[p(D|h)] + log[p(h)]) \ \xrightarrow{} \ \underset{h}{argmax}(log[p(D|h)]) \ \$ as $D$ gets large.

It makes perfect sense why the likelihood would overwhelm the prior, but for the left side of the above equation to approach the right as the set $D$ gets larger does not make sense.