Let X be a discrete r.v. and Y be a continuous r.v. such that the conditional distribution of X given Y = y is a (discrete) geometric distribution with probability for success p, and such that Y has pdf f_Y(y) = 3y for 0 < y < 1 (and zero otherwise).
a) Compute the pmf of X.
b) Compute E[X].
c) Does the r.v. Var(X | Y) have a finite expectation?
The PDF given in question f_Y(y) = 3y for 0 < y < 1 is wrong.
I take it as fY(y) = 3y2; 0 < y < 1
Given X is a discrete RV and Y be a continuous RV such that the conditional distribution of X given Y =y is geometric distribution with probability for success P.
So, the conditional PMF is
P (X = x | Y = y) = (1 – p)x p; x = 0,1,2,3…
So, X,Y are independent RVs.
Also, the PDF of Y is fY(y) n= 3y2; 0 < y < 1
The conditional expectation is E (X | Y) = (1 - p) / p
The joint mixture PDF is,
P (X = x, y) = P (X = x | Y = y) fY(y)
P (X = x, y) = (1 – p)xp(3y2)
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