Good evening, I'm trying to solve Exercise 12.3:
Could you please verify if my proof looks fine or contains logical gaps/errors? Thank you so much for your help!
My attempt:
$y \le 0$. Then $f_Y (y) = 0$.
$0 < y \le 1$. Then $1/y \ge 1$. It follows that $$f_Y (y) = \int_{\mathbb R} \frac{1}{2x^2y} \textbf{1}_{(1,\infty)}(x) \textbf{1}_{[1/x,x]}(y) \, \mathrm{d} x = \int_{1/y}^\infty \frac{1}{2x^2y} \, \mathrm{d} x = \frac{-1}{2xy} \Big|_{1/y}^\infty = \frac{1}{2}$$
$y > 1$. It follows that $$f_Y (y) = \int_{\mathbb R} \frac{1}{2x^2y} \textbf{1}_{(1,\infty)}(x) \textbf{1}_{[1/x,x]}(y) \, \mathrm{d} x = \int_{y}^\infty \frac{1}{2x^2y} \, \mathrm{d} x = \frac{-1}{2xy} \Big|_{y}^\infty = \frac{1}{2y^2}$$
Update: Thank you @Chinny84 for pointing out my misunderstanding of the requirement of the question. I fixed it here.
Let $A = \{(x,y) \in \mathbb R^2 \mid x >1 \, \text{and} \, y \in [1/x, x] \}$.
$(x,y) \notin A$. Then $f_{Y|X=x} (y) = 0$.
$(x,y) \in A$. It follows that $$\begin{aligned} f_{Y|X=x} (y) &= \frac{\frac{1}{2x^2y}}{\int_{ 1/x}^{x} \frac{1}{2x^2y} \, \mathrm{d} y} = \frac{\frac{1}{2x^2y}}{\frac{\ln y}{2x^2} \big|_{1/x}^{x}} = \frac{\frac{1}{2x^2y}}{\frac{\ln x}{x^2}} = \frac{1}{2y\ln x}\end{aligned}$$