Derived PDF calculation - where did it go wrong?

Question

Modify from this problem, if I have:

$X$ is has a uniform distribution with PDF $f(x) = 1/2$, $x \in [0, 2]$, zero elsewhere.

Find the PDF of $Y = \ln X$.

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Using substitution, I have:

$$F_Y(y)=\mathbf{P}(Y\leq y)=\mathbf{P}(\ln(X)\leq y)=\mathbf{P}(X\leq e^y)=F_ X(e^y).$$

Then:

$$f_{Y}(y) = \frac{dF_Y}{dy}(y) = \frac{dF_X}{dy} (h(y)) = \frac{d}{dy}\left(\frac{1}{2}\right) = 0.$$

This is problematic, where did I go wrong? Thanks.

Answer 1

Found my error:

$$f_{Y}(y) = \frac{dF_{Y}(y)}{dy} = \frac{dF_{X}(e^y)}{dy} = f_{X}\bigl(e^y\bigr)\cdot\frac{d}{dy}(e^y) = \frac{1}{2}\cdot e^y = \frac{e^y}{2}.$$

Answer 2

$F_X$ is not constant; rather its derivative, $f_X,$ is constant.