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While I am studying maximum a posteriori, somehow it just came across my mind that if I have a uniform prior probability distribution, MAP looks like MLE, am I right?

Tim
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user122358
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1 Answers1

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In short yes, a one-liner goes like this

$$h_{MAP} = argmax_{h}\ log(p(D|h) + log(p(h)) = argmax_h\ log(p(D|h)) = h_{MLE} $$

Since $log(p(h))$ under uniform prior is constant. What is more, given enough data, it is true that regardles of prior

$$h_{MAP} \underset{|D| \to \infty}\longrightarrow h_{MLE}$$

A general argument is that $p(D|H)$ depends exponentially on $|D|$ and prior probability $p(h)$ stays constant

Łukasz Grad
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  • You are saying that MAP becomes MLE when the data size becomes huge, almost infinity? I don't think the size has to be infinite, I think you are saying when we have a reasonably large size, MAP approaches MLE. Am I right? – user122358 Mar 09 '17 at 12:14
  • @user122358 I simply say that MAP converges to MLE – Łukasz Grad Mar 09 '17 at 12:17
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    @user122358 for extended discussion of this see http://stats.stackexchange.com/questions/200982/do-bayesian-priors-become-irrelevant-with-large-sample-size/201059#201059 – Tim Mar 09 '17 at 13:04