Question

Consider the function f(x/y)=(y^xe^-y)/x!, x=0,1,... y>=0

show that for each fixed y, f(x/y)id a p.d.f, the conditional p.d.f of r.v. X, given another r.v. Y equals y

If the marginal p.d.f of Y is Negative Exponentioal with parameter lambds=1, what is the joint p.d.f of X,Y?

Show that the marginal p.d.f of X is given by f(x)=(1/2)^(x+1), x=0,1,2...

Answer 1

Given,

   for x=0,1,... y>=0

By exponential series, we know that

Therefore,

and hence f(x|y) is a p.d.f

Given, marginal p.d.f of Y is Negative Exponential with parameter lambda = 1, then

   for y > 0

The joint p.d.f of X,Y is given as,

   for x=0,1,... y>=0

The marginal p.d.f of X is given as,

Let z = 2y, then dz = 2dy and the limits of the integration is from z = 0 to z =

So,

Using Gamma function,

  

for x=0,1,2,...

     for x=0,1,2,...

answered by: anonymous