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I'm not looking for rationale why log transformation doesn't change position of maximum. I came across this when reading about maximum likelihood and don't really get it about this log transformation. Until now I was taking it as given, but when I wanted to articulate it I couldn't. I was drawing few log graphs and thinking about it, but came to no conclusion.

So, in my case I'm looking for parameters in maximum likelihood method:

$$\theta^* = arg max_{\theta} \prod_{n=1}^{m} p_{data}(x; \theta)= arg max_{\theta} log(\prod_{n=1}^{m} p_{data}(x; \theta)) = arg max_{\theta}\sum_{n=1}^{m} log(p_{data}(x; \theta))$$

$p_{data}(x; \theta)$ is meant as probability distribution parametrized by parameters $\theta$ and we have m training points(or size of sample is m).

I understand log increases anywhere, but how does this prove this application of transformation?(This is rationale given in my book)

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  • $\begingroup$ A maximizer of the product is a maximizer of the log of that product and vice versa because log is strictly increasing. That's it. $\endgroup$ – Ian Nov 8 '17 at 11:37
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Suppose you have $x,y > 0$. Then $x > y \Leftrightarrow \log x > \log y$. Thus if we have a function $f: \mathbb R _{> 0} \to \mathbb R$, then $x > 0 $ is a maximizer of $f$ iff $f(x) > f(y)$ for all $y >0$ iff $\log f(x) > \log f(y)$ for all $y >0$ iff $x$ is a maximizer of $\log f$.

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