Question

X and Y have joint density function

$f(x,y)=\frac{1}{x^2{y}^2}x\ge 1,y\ge 1$

(a) Compute the joint density function of U = XY, V = X/Y.

(b) What are the marginal densities?

Short answer

(a) $f_{U,V}(u,v)=\frac{1}{2u^{2}v}$

(b)Marginal density of U, ${\int }_0^{\infty }\frac{1}{2u^2}du=1$$u\ge 1,v\ge 1$

Marginal density of V,${\int }_0^{\infty }\frac{1}{v}dv=1$$u\ge 1,v\ge 1$

Step-by-step solution

(a) Joint probability density function :

The joint probability density function (joint pdf) is a function that is used to characterize a continuous random vector's probability distribution.

(a) Explanation :

Joint density function of X and Y,

$f\left(x,y\right)=\frac{1}{x^2{y}^2}x\ge 1,y\ge 1$

For X and Y,

$U=XY,V=X/Y$

We have,

localid="1647268773098" $f_{X,Y}(x,y)=\frac{1}{x^2{y}^2}$

Apply the transformation,

$g:{(1,\infty )}^2\to (1,\infty )\times (0,\infty )$

Such that,

$g(x,y)=(u,v)=\left(xy,\frac{x}{y}\right)$

By using theorem, the density function of random vector $(U,V)=g(X,Y)$as

localid="1647268791021" $f_{U,V}(u,v)=f_{X,Y}(x,y)·{\left|det(\nabla g(x,y\left)\right)\right|}^{-1}$

Calculate,

localid="1647268809890" $$\begin{gathered} \nabla g(x,y)=\left[\begin{array}{cc}y& x\\ \frac{1}{y}& -\frac{x}{y^2}\end{array}\right] \\ \left|det(\nabla g(x,y\left)\right)\right|=-\frac{x}{y}-\frac{x}{y}=-\frac{2x}{y} \\ {f}_{U,V}(u,v)=f_{X,Y}(x,y)·\frac{y}{2x}=\frac{1}{2x^{3}y} \end{gathered}$$

Obtain the range of $u$and $v$as follows:

Since $x\ge 1\Rightarrow x^2\ge 1$

Therefore,

$$\begin{gathered} uv\ge 1(because{x}^2=uvasx=\sqrt{uv}) \\ \Rightarrow v\ge \frac{1}{u} \end{gathered}$$

Also, $y\ge 1\Rightarrow y^2\ge 1$

Therefore,

$\frac{u}{v}\ge 1\Rightarrow \frac{1}{v}\ge \frac{1}{u}\Rightarrow v\le u(onrever\sin gtheinequality)$

Thus, from $v\ge \frac{1}{u}andv\le u$, the range of v is $\frac{1}{u}\le v\le u$.

Now, as $x\ge 1andy\ge 1$, therefore, $u\ge 1(x\ge 1,y\ge 1andu=xy)$

Hence, the range of u and v is$u\ge 1,\frac{1}{u}\le v\le u$, respectively.

Now, write x and y in terms of u and v and substitute,

But we have,

$x=\sqrt{uv}$

and

$y=\sqrt{\frac{u}{v}}$

That finally implies,

localid="1647268829405" $f_{U,V}(u,v)=\frac{1}{2u^{2}v}$

(b) Marginal density :

The marginal probability of a continuous variable may be calculated using a marginal density function.

(b) Explanation :

From part (a) result,

We have Joint density function of U and V,

$f_{U,V}(u,v)=\frac{1}{2u^{2}v}$

This probability distribution function can be factorized as

$\frac{1}{2u^{2}v}=\frac{1}{2u^2}·\frac{1}{v}$

Hence,

Marginal density of U,

${\int }_0^{\infty }\frac{1}{2u^2}du=1$$u\ge 1,v\ge 1$

Marginal density of V,

${\int }_0^{\infty }\frac{1}{v}dv=1$$u\ge 1,v\le 1$