I'm getting a wrong answer to the following exercise:
Let $(X,Y)$ be a random vector with density $$\frac{2}{y}\centerdot 1_{D}$$ where $D=\{ (x,y)\in (0,1)\times (0,\infty ) : 0<x<y, 0<xy<1\}$. Determine whether $X$ has a probability density function and if it does, find it.
My attempt:
Letting $x\in (0,1)$ with $x<y,y^{-1}$ we have
$$P(X\leq x)=\int_{\mathbb{R}}\int_{-\infty}^{x}2y^{-1}\centerdot1_Ddt\ dy=\int_x^{1/x}\frac{x}{2y}dy=-x\ln x$$
We wish to find a function $f_X$ so that -for any $x$ with the retraints above-
$$-x\ln x=\int_{-\infty}^{x}f_X(t)dt$$
which gives $$f_X(x)=\frac{d}{dx}(-x\ln x)=-1-\ln x$$
yet this is impossible since the function takes negative values for some $x\in (0,1)$. According to my teacher's notes the correct answer should be $$f_X(x)=-\ln x$$
Where did I go wrong?
Best Answer
The image is poor, but good enough to understand the idea. The red and blue combined is the region to integral over to calculation $P(X\leq x)$. Specifically, the mistake that you made is that if you integrate with respect to $x$ first (through dummy variable $t$), you would have to split this over two iterated integrals. The answer should be in terms of $x$. Then differentiate to get $f_X(x)$.
A far easier method is to simply integrate the joint density from $y=x$ to $y=1/x$.