Let $f(x,y) = \frac{1}{2x}\mathbb{1}_D(x,y)$ with $D = \{(x,y) : 1\leq x \leq 3, 0 \leq y \leq x \}$

Question

Let $f(x,y) = \frac{1}{2x}\mathbb{1}_D(x,y)$ with $D = \{(x,y) : 1\leq x \leq 3, 0 \leq y \leq x \}$.

I want to calculate the probability that $2y > 3 - x$, I set up my integral like this: $$ \int_{1}^{3}\int_{-\frac{1}{2}x + \frac{3}{2}}^{x} \frac{1}{2x} dy dx$$ Is that right? My question is: how do I find $f_X(x)$ and $f_Y(y)$? I tried writing: $$ f_X(x) = \int_{0}^{x}\frac{1}{2x}dy$$ But that doesn't look too good to me, plus I don't know to set up the other one. Any help?

Answer 1

Your computation of the probability looks fine.

$$f_X(x) = \int_0^x \frac1{2x}\, dy = \frac{x}{2x}=\frac12$$

We have

$$f_Y(y) = \int_{\max(1, y)}^3 \frac1{2x} \, dx$$