# Problem obtaining a marginal from the joint distribution

Suppose $X_1$ and $X_2$ have the joint pdf

$$f_{X_1,X_2}\left(x_1,x_2 \right)=4x_1 x_2,\quad\text{for}\quad 0<x_1<1, 0<x_2<1$$

From that I found the joint pdf of $Y_1=\frac{X_1}{X_2}$ and $Y_2=X_1 X_2$ to be

$$f_{Y_1,Y_2} \left(y_1,y_2 \right)=2 \frac{y_2}{y_1}\quad \text{for}\quad \left(y_1,y_2 \right) \in \mathcal{T}$$

where $\mathcal{T}=\{ \left(y_1,y_2 \right): y_1,y_2>0,y_2<\frac{1}{y_1},y_2<y_1 \}$

That is, in the $\left(y_1, y_2 \right)$ plane, the area of the pdf is bounded by the 45 degree line and the hyperbola $\frac{1}{y_1}$. Based on that I can easily derive the marginal of $Y_1$ for the two cases

1. $0<y_1<1$

2. $1<y_1<\infty$

But what about the marginal of $Y_2$? $Y_1$ is not bounded above and is bounded below by zero, right? How can I integrate out $Y_1$ if the integral from 0 to infinity diverges? Which begs the question, have I done something wrong in the above?

Draw pictures of the regions of integration.

The region where $0 \le x_1 \le 1, 0 \le x_2 \le 1,$ and $x_1 x_2 \le y$ (for $0 \le y \le 1$) looks like the shaded part of

The colors denote the varying values of the density $f(x_1,x_2)$, ranging from blue (low) to red (high).

The integral of $f(x_1,x_2)dx_1 dx_2 = 4 x_1 x_2 dx_1 dx_2$ is readily found by integrating separately over the rectangle to the left of the dashed line and the region to its right, which is bounded above by the curve $x_1 x_2 = y$; it gives

$$\Pr(Y_2 \le y) = y^2-2 y^2 \log (y).$$

Here is a plot of this distribution: it is the marginal CDF for $Y_2$.

Differentiate this to obtain the PDF of $Y_2$.

The region where $0 \le x_1 \le 1, 0 \le x_2 \le 1,$ and $x_1 / x_2 \le y$ (for $0 \le y \lt \infty$) looks like the shaded part of

The lower curve is a portion of the line $x_2 = x_1 / y.$

When $y \gt 1$ the integral of $4 x_1 x_2 dx_1 dx_2$ can be broken into the two triangles shown; when $y \le 1$ only an upper triangle appears. The marginal CDF of $Y_1$ works out to

$$\Pr(Y_1 \le y) = y^2/2, \ 0 \lt y \le 1; \quad 1 - 1/(2y^2),\ y \ge 1.$$

A partial plot of this marginal CDF for $Y_1$ is

The full plot extends infinitely far to the right. Differentiate this to obtain the PDF of $Y_1$, the first marginal.

• Thank you very much, I see it now. This is perhaps one of the problems for which using the CDF technique instead of the change of variables is more helpful. Commented Oct 17, 2014 at 22:07
• What I do not get is how we got the squares in the cdf of $Y_1$, for example the area of the upper triangle is just $y/2$ isn't it? Commented Oct 17, 2014 at 22:26
• You are correct: that is the area. But areas are not relevant here; what does matter are the integrals of the probability densities.
– whuber
Commented Oct 17, 2014 at 22:27

Just to add visual input, it is easily found that

$$F_X(x_i) = x_i^2$$ and since for $U_i \sim U(0,1)$

$$F^{-1}(U_i) = X_i \Rightarrow X_i = \sqrt{U_i}$$

the $X's$ are the square roots of uniform RV's in $(0,1)$. They are also independent.So simulate two uniforms, and then take their product. The resulting empirical relative frequency curve is

Now take the CDF of $Y_2 = X_1X_2$

$$F_{Y_2}(y_2) = y^2-2 y^2 \log (y) \Rightarrow f_{Y_2}(y_2) =-4y_2\ln y_2$$

Graph this function in (0,1) to get

More mathematically, we have that $Y_2 = \sqrt {U_1U_2}$ i.e. it is the square root of the product of two independent standard uniform RV's. The density of the product of $n$ standard uniform independent RV's can be found here. . For $n=2$ it is simply $$f_{U_1U_2}(u_1u_2) = -\ln(u_1u_2)$$

For $Y_2 =\sqrt{ U_1U_2}$ we immediately obtain by the change of variable formula that $f_{Y_2}(y_2) =-4y_2\ln y_2$.