Uniformly distributed transformation

Question

We let X be uniformly distributed on the interval $(0,1)$. We must find the distribution of $Y=-log(X)$. I'm not sure how to do that but maybe I think maybe I can use that if X be uniformly distributed on the interval $(a,b)$ then will the distribution be: $$P(Y \leq y)=\frac{y-a}{b-a}$$, but how to use this and what if $P(Y > y)$?

Answer 1

$Y=-\log X=g(X)$ is a monotonic transformation thus $f_Y(y)$ can be derived immediately by derivating $g^{-1}(y)$

$$f_Y(y)=e^{-y}$$

$y>0$

that is a Negative Exponential distribution with mean 1

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This because you can apply the fundamental transformation theorem

$$f_Y(y)=\underbrace{f_X[g^{-1}(y)]}_{=1}\cdot|\frac{d}{dy}g^{-1}(y)|$$

Answer 2

Slower: first, derive the CDF of $X=h^{-1}(Y)$. Since the function $h(X)$ is monotone decreasing, the sign changes: $$ P(Y \leq y) = P(X \geq h^{-1}(y)) = 1 - e^{-y}=F_{Y}(y) $$ This is because CDF of $X$ is $\frac{e^{-y}-0}{1-0}$. Hence PDF of $Y$ is $$ f_{Y}(y) = e^{-y} $$