Question

Let X be the number of customers arriving in a given minute at the drive-up window of a local bank, and let Y be the number who make withdrawals. Assume that X is Poisson distributed with expected value E(X) = 3, and that the conditional expectation and variance of Y given X = x are E(Y|x) = x/2 and Var(Y |x)= (x + 1)/3. (a) Find E(Y) (b) Var(Y) (c) Find E(XY). 20.

Answer 1

Answer:

Given that:

Assume that X is Poisson distributed with expected value E(X) = 3, and that the conditional expectations and variance of Y given X = x are E(Y | x) = x/2 and Var(Y | x) = (x + 1)/3.

Since X~Poisson(3) ,E(X)=3 and V(X)=3

(a) Find E(Y).

E(Y)=E(E(Y|X)) =E(X/2)

E(Y)=3/2

b) Find Var(Y).

Var(Y) = E[V(Y|X)]+V[E(Y|X)]

Var(Y) = E[(X+1)/3]+V[X/2]

Var(Y) = 4/3 +3/4

Var(Y) = 25/12

Var(Y) = 2.0833

c) Find E(XY).

E(XY) = E[E(XY|X)]=E[XE(Y|X)]

E(XY) = E[X2/2]

E(XY) = 1/2 E[X2]

E(XY) = 1/2 (V(X)+E2(X))=3+32 /2=6