Question

Two random variable X; Y has the following joint moment generating function

M_X;Y (s; t) = 0:5 + 0:1e^t + 0:1e^s + 0:1e^s+t + 0:15e^2s + 0:05e^2s+t

Find the probability of Y = 1 given X < 2

Hint:

> Find the PGF

> Deduce the joint probability mass function PMF f_X,Y(x,y)

>Plot the PMF on a 3-D plane x, y and z = f_X,Y(x,y) or alternatively use Cartesian (x-y) plane with f_X,Y(x,y) plots

>Use conditional Probability rues to obtain the answer!

Answer 1

P(Y = 10x<2) = P(Y = 1, X = 0) + P(Y = 1, X = 1) = 0.1 + 0.1 = 0.2 P(X = 0) = P(X = 0,9 = y) y=0,1 = P(x = 0, y = 0) + P(X = 0,7 = 1) = 0.5 + 0.1 = 0.6 P(X = 1) = P(X = 1,9 = y) y=0,1 = P(x = 1, Y = 0) + P(x = 1, Y = 1) = 0.1 + 0.1 = 0.2 :: P(X<2) = P(X = 0) + P(X = 1) = 0.6 + 0.2 = 0.8 Hence, P(Y = 10X<2) P(Y = 1|X<2) = P( X2) = 0.2 0.8 (OR 0.25)