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...
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,...
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