Transformation of Random Variables

In this section we shall consider the problem of change of variables in the two dimensional random variable(X,Y).

Let the random variables U and V be defined by the transformation u = u(x,y) v=v(x,y).

The Jacobian of the transformation is

J =`(del (x,y))/(del(u,v))`

`[[(delx)/(delu),(dely)/(delu)],[(delx)/(delv),(dely)/(delv)]]`

The joint p.d.f `g_uv` (u,v) of the transformed variables U and V is given by `g_uv(u,v)=f_xy(x,y)|J| ` where |J| is the modulus value of the jacbian of transformation and f(x,y) is expressed in terms of u and v. Now we can see topic transformation of random variable given some examples problem are given below content.
Example Problem for Transformation of Random Variables

Example 1:-using transformation of random variables

If X and Y are independent continuous random variables; then the probability density functions of U = X + Y is given by h(u) = `int_-oo^oof_(x)(v)f_(v)(u-v)dv`

Solution:

Let f(x,y)be the joint p.d.f of X and Y. consider the transformation u = x +y, v =x

X=v,y=u-v

Then

J =`(del (x,y))/(del(u,v))`

`[[0,1],[1,-1]]`

= - 1

Thus the joint p.d.f of random variablems U and  V is given by

`g_(uv)`

=f_(xy)(x,y) |J|

g(u,v)=`f_x(v)f_y(u-v)`

The marginal density of U is given by

h(u)=`int_-oo^oog(u,v)dv`

h(u)=`int_-oo^oof_(x)f_(y)(u-v)dv`

which is the convolution of f_(x)(.) and f_(y)(.)

I am planning to write more post on Define Polynomial Function and Define Probability Distribution. Keep checking my blog.

Example 2:-using Transformation of Random Variables


Given the joint density function of X and Y as

f(x,y)=`{((1/2)xe^-y,00),(0 , elsewhere):}`

Find the distribution of X + Y.

Solution :

Transformation : u = x=y and v=y

x=u - v and y = v

`[[(delx)/(delu),(dely)/(delu)],[(delx)/(delv),(dely)/(delv)]]`

=1

The region 00 transforms into 00, is

The joint density function of U and V is given by

g(u,v)=(1/2)(u-v)e`^-v` , 0 0

The density function of U = X +Y is obtained by splitting the range of U into two parts

(i)0
(ii)u>2(Region II)

for o
h(u)=`int_0^u (u-v)e^-v(dv)`

=(1/2){(u-v)(-e`^-v` )-(-1)(e`^-v` )`]^u_0`

and for 2
h(u) = `int_0^u (u-v)e^-v(dv)`

=(1/2)[(u-v)(-e`^-v` )-(-1)(e`^-v` )`]^u_(u-2)`

Hence (u) = `{((1/2) (e^-(-u) + u -1),0