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

**f(x,y)=λ**^{2}**e(-λ(x+y)) zero otherwise x>0 y>0**

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**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**

“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}n dx/du & dx/dv \\n dy/du & dy/dvn\end{vmatrix}=\begin{vmatrix}n V & U \\n -V & 1-Un\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.

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