Why doesn't this work for finding the PDF of $X_1 X_2$ where $X_1,X_2$ iid Unif(0,1)

Question

Let $X_1,X_2$ be iid Unif(0,1) random variables.
Find the PDF of $Y = X_1 X_2$.

This is my method and I'm unsure why it's wrong:

$$\mathbb{P}(Y\leq y) = \mathbb{P}(X_1 X_2 \leq y) \\ = \int_{\text{all} x} \mathbb{P}(X_2 \leq y/x | X_1 = x)f_{X_1}(x) \ dx \\ = \int_{\text{all} x} F_{X_2}(y/x) f_{X_1}(x) \ dx \\ = \int_{\text{all} x} \int_{0}^{y/x} I(0<y/x < 1)I(0<x<1) \ dx \\ = \int_{\text{all} \ x} y/x I(0<y/x < 1)I(0<x<1) \ dx \\ = \int_{y}^1 \frac{y}{x} \ dx \\ = -y\log y $$ Hence, the PDF is the derivative which is $-(1+\log y)$
but the solution says that it's actually just $-\log y$. Where did I go wrong?

Answer 1

The problem is that$$\mathbb{P}(X_2 \leq y/x | X_1 = x)\not=\frac yx.$$

This is because if $y>x$ then $y/x>1$ and then $$F_{X_2}(y/x)=1.$$

So

$$\int_0^1\mathbb{P}(X_2 \leq y/x | X_1 = x)dx=\int_0^y1dx+y\int_y^1\frac 1x\ dx=y-y\ln(y)$$

the derivative of which is, indeed $-\ln(y)$.