Answer to Question #252321 in Statistics and Probability for pas

Question

Suppose the joint p.d.f of the continuous random variable X and Y is given by

f(x,y)=λ²e(-λ(x+y)) zero otherwise x>0 y>0

i) Obtain the joint p.d.f of the new random variables V=X+Y and U=X/(X+Y)

ii)Find the marginal probability distribution functions of U and V

iii) Determine whether or not U and V are independent

Answer 1

$f(x,y)=\lambda^2e^{-\lambda(x+y)}, \space x,y\gt0$

The new random variables are defined by,

$V=X+Y$ and $U=X/(X+Y)$

$i)$

To find the joint probability distribution functions for these random variables, we shall apply the transformation of continuous random variable using $Jacobian$ transformation as described below.

Let us first make the random variables $X$ and $Y$ subject of the formula then,

$X=UV$ and $Y=V(1-U)$. We define $Jacobian(J)$ as a $2\times2$ matrix given as,

We now apply the Jacobian transformation as follows,

$J=\begin{vmatrix} dx/du & dx/dv \\ dy/du & dy/dv \end{vmatrix}=\begin{vmatrix} V & U \\ -V & 1-U \end{vmatrix}=(V*(1-U))+UV=V$

The joint probability distribution function of $U$ and $V$ is given as,

$g(u,v)=\lambda^2e^{-\lambda(uv+v-uv)}*|J|=\lambda^2e^{-\lambda(v)}*v$

Therefore, the joint pdf for $U$ and V is given as,

$g(u,v)=\lambda^2ve^{-\lambda v},\space 0\lt u\lt1,\space v\gt 0$

$0, \space elsewhere$

$ii)$ The marginal distributions of the random variables are obtained as described below.

Given that the joint pdf is given by,

$g(u,v)=\lambda^2ve^{-\lambda v}, \space 0\lt u\lt 1, v\gt0$ ,

Then the marginal distribution of the random variable $V$ is given as,

$h(v)=\int^1_0g(u,v)du$

$h(v)=\int^1_0(\lambda^2ve^{-\lambda v})du$

$h(v)=(\lambda^2ve^{-\lambda v})*u|^1_0=\lambda^2ve^{-\lambda v}$

Therefore,

$h(v)=\lambda^2ve^{-\lambda v},\space v\gt0$

$0,\space elsewhere$

and the marginal distribution of the random variable $U$is given as,

$f(u)=\int^\infin_0g(u,v)dv$

$f(u)=\int^\infin_0\lambda^2ve^{-\lambda v}dv$

Let us evaluate the integral first,

$\int^\infin_0\lambda^2ve^{-\lambda v}dv=\lambda^2\int^\infin_0ve^{-\lambda v}dv$

This integral can be solved using integration by parts method as below,

Let

$x=v$ and $dy/dv=e^{-\lambda v}$

we need to find $dx/dv$ and $y$

$dx/dv=1$ and $y=\int e^{-\lambda v }dv=-(1/\lambda)e^{-\lambda v}$

Thus the integral can be written as,

$\lambda^2\int^\infin_0ve^{-\lambda v}dv=\lambda^2((-v/\lambda)e^{-\lambda v}|^\infin_0+(1/\lambda)\int^\infin_0 e^{-\lambda v}dv)$

$=\lambda^2(-(v/\lambda)e^{-\lambda v}-(1/\lambda^2)e^{-\lambda v})|^\infin_0$

$=\lambda^2(1/\lambda^2)=1$

Therefore the marginal distribution for the random variable $U$ given as,

$f(u)=1,\space 0\lt u\lt 1$

$0, \space elsewhere$

$iii)$

If the random variables $U$and $V$ are independent then we expect that their joint probability distribution function$g(u,v)$ must equal the product of their marginal distribution functions as written below.

If $U$ and $V$are independent then,

$g(u,v)=h(v)*f(u)$

so,

$\lambda^2ve^{-\lambda v}=1*\lambda^2ve^{-\lambda v}$

Since the product of the marginal distributions is equal to the joint distribution then, $U$ and $V$ are independent.