Marginal pdf for $X$ given the simultaneous pdf for $(X,Y)$ (hard)

Question

PDF for $(X,Y)$ is given by

$$ f(x, y)=\left\{\begin{array}{cl} \frac{4 y^{3}}{x^{3}} & \text { for } 0<y<1, y<x \\ 0 & \text { else } \end{array}\right.$$

By drawing the support of $(X,Y)$ I can see that I have to construct $g(x)$ piecewise as follows:

For $x \in (0,1]$ I have $g(x)=\int_0^x4x^{-3}y^3dy=x$.

For $x \in [1,\infty)$ I have $g(x)=\int_0^1 4x^{-3}y^3dy=x^{-3}$.

For $x \not \in (0,\infty)$ I have $f(x,y)=0$, and therefore $g(x)=0$.

So $g(x)$ written as a piecewise function

$$ g(x)=\left\{\begin{array}{ll} x & , x \in (0,1] \\ x^{-3} & , x \in [1, \infty) \\ 0 & , x \not \in (0, \infty ) \end{array}\right.$$

Can someone confirm that the method and result are correct? It will be nice to know :-)

Answer 1

It is fine. The marginal pdf is simply

\begin{align} f_X(x)&=\int_0^{\min(x,1)}\frac{4y^3}{x^3}\,dy\,1_{(0,\infty)}(x) \\&=\frac{(\min(x,1))^4}{x^3}1_{(0,\infty)}(x) \\&=x1_{(0,1)}(x)+\frac1{x^3}1_{(1,\infty)}(x) \end{align}

Here $1_A(x)$ equals for $1$ if $x\in A$ and equals $0$ if $x\notin A$.