Range of a marginal density function? Given the joint density function: $$p(x,y) = \frac{4x^3}{y^3} \text{ when } 0 < x < 1\text{ and } y > x; \text{ otherwise } p(x,y) = 0$$
How does one calculate the marginal density for $Y$? I do this by simply integration over the the interval in which $X$ lies (so from 0 to 1), which gives $y^{-3}$. But according to my textbook, the answer is $y^{-3}$ for $y > 1$, and $y$ for $0 < y \le 1$? How does that show?
 A: When you integrate with respect to $x$, you must also take into account that $x$ cannot be allowed to exceed $y$.  Sketch the region of integration and you will see what I mean.  Therefore, the correct calculation is $$f_Y(y) = \int_{x=0}^{\min(1,y)} \frac{4x^3}{y^3} \, dx = \frac{\min(1,y^4)}{y^3} = \min(y^{-3},y) = \begin{cases} y^{-3}, & 0 < y \le 1 \\ y, & y > 1. \end{cases}$$
A: The best way to understand this answer is to draw out a graph showing the ranges of $x$ and $y$. Draw a vertical line at $x=1$ to show that boundary value and draw the line $y=x$. Now shade in the region that corresponds to $y>x$. You should be able to see that the range of $y$ can be divided into two parts: $y>1$ and $0<y\le1$. Hopefully the limits of integration for each part should be clear by this point.
A: Once again an illustration that one should consider PDFs as functions defined on the whole product space. Here $p:\mathbb R^2\to\mathbb R$ is defined, for every $(x,y)$ in $\mathbb R^2$, as
$$p(x,y)=4x^3y^{-3}\,\mathbf 1_{0\lt x\lt1}\,\mathbf 1_{y\gt x},$$ hence, by definition, the density $p_Y$ of $Y$ is defined, for every $y$ in $\mathbb R$, by $$p_Y(y)=\int_{-\infty}^\infty p(x,y)\,\mathrm dx.$$
Now we can take into account the specific form of $p$, leading to $$p_Y(y)=4y^{-3}\int_0^1x^3\,\mathbf 1_{y\gt x}\,\mathrm dx=4y^{-3}\int_0^{\min(1,y)}x^3\mathrm dx=y^{-3}\,\left.x^4\right|_0^{\min(1,y)}=y^{-3}\,\min(1,y)^4,$$ which is equivalent to the two-parts formula you were given.
