Why in finding PDF of e^X in terms of the PDF of X,which is uniformly distributed between 0 and 1, the answer is 1/y, rather than 1/(2y)?

Question

The question here is: "Find the PDF of e^x in terms of PDF of X. Specialize the answer to the case where X is uniformly distributed between 0 and 1 (Problem 4.1 of the book "Introduction to Probability, 2nd edition"). Using the fact that px(x) is 1/2 (1/(1-0+1) = 1/2), I come out with 1/2y for fy(y). However, the solution is only 1/y. What have I missed or done wrong?

Answer 1

Since $X$ takes values in $[0,1]$, $Y=e^X$ takes values in $[1,e]$. For $1\leqslant t\leqslant e$ we have $$\mathbb P(e^{X}\leqslant t) = \mathbb P(X\leqslant \log t), $$ so the density of $Y$ is obtained by differentiating: $f_Y(y) = \frac 1y\cdot\mathsf 1_{[1,e]}(t)$.

Answer 2

Pdf of uniform distribution on $(0,\,1)$ is ($a=0$, $b=1$), $0\leqslant x \leqslant 1$ $$ p_X(x) = \frac{1}{b-a}=\frac{1}{1-0}= 1. $$