what is the marginal density of U where fUVuvfXYv1u1uV2V1u2

what is the marginal density of U where fU,V(u,v)=fX,Y[(v(1+u)/(1-u),V](2V/(1-u)^2)

assumed that X and Y are independent and follows a X,Y~Exp(1) distribution.

Solution

The joint pdf of X and Y is given by

fX,Y(x,y) = e-(x+y) if x>0,y>0

              = 0    otherwise

U and V are such defined that

x = v(1+u)/(1-u)    and y=v

=> x = y(1+u)/(1-u)

=> (1+u)/(1-u) = x/y

=> u = (x-y)/(x+y)

=> -1<u<1

We have 0<y<

=> 0<v<

Now the joint pdf of U and V is

fU,V (u,v) = fX,Y(v(1+u)/(1-u) , v) * 2v/(1-u)2     if -1<u<1 and 0<v<

                = e-(v(1+u)/(1-u) + v) * 2v/(1-u)2     if -1<u<1 and 0<v<     

                = e-v(2/(1-u)) 2v/(1-u)2     if -1<u<1 and 0<v<

The marginal pdf of U is then given by

g(u) = integration(from=0,to=,fU,V(u,v) dv )

          = int(0, , e-v(2/(1-u)) 2v/(1-u)2dv)

          = 2/(1-u)2 int(0,, e-v(2/(1-u)) v dv)

          = 2/(1-u)2 * (2) (2/(1-u))-2    using gamma integral

          = 1/2          since (2) = 1

Hence the marginal pdf of U is

g(u) = ½   if -1<u<1

          = 0    otherwise

what is the marginal density of U where fU,V(u,v)=fX,Y[(v(1+u)/(1-u),V](2V/(1-u)^2) assumed that X and Y are independent and follows a X,Y~Exp(1) distribution.S

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