3 Suppose XY Binomial Y p07 and Y Poisson 10 aWhat is the

3. Suppose X|Y ~ Binomial (Y, p=0.7) and Y ~ Poisson (=10)
(a)What is the expected value of X?

(b)What is the variance of X? (Hint: recall the iterated expectation and iterated variance formulae.)
Is the variance of X bigger or smaller than the variance of Z ~ Binomial (10, p=0.7)?

(c)What is the covariance of X and Y? Hint: you may use Cov(X,Y)=E(XY)-E(X)E(Y)

(d)What is the variance of X + Y? (careful: they are not independent!)

Solution

Given X|y ~ Bin(y, 0.7) where y is a realised value of Y ~ P(10)

(a)

E[x] = E[E(X|Y)]

          =E[0.7Y]               (Since if X~Bin(n,p), E[x]= np)

          =0.7E[Y]

          =0.7*10               (Since E[Y] = 10)

          =7

(b)

Var(X)

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

=Var[0.7Y] + E[Y*.7*(1-.7)]             (Since if X~Bin(n,p), Var(x)=np(1-p) )

=.7² Var[Y] + .21 E[Y]

=.49*10 + .21*10                  (Since Var(Y)=10 )

=4.9+2.1

=7

(c)

Z ~ Bin(10,0.7)

Var(Z) = 10*.7*.3 = 2.1

We find that Var(Z) < Var(X).

(d)

Cov (X,Y) = E[XY] – E[x]E[Y]

E[x]=7 (from part (a) )

E[Y]=10 (given)

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

          = E[YE(X|Y)]

          = E[Y*0.7Y]

          = .7 E[Y²]

          = .7 (Var(Y) + E²[Y])                (Since Var(Y) = E(Y²)-E²(Y))

          = .7(10+10²)

          = 77

Then cov(X,Y) = 77-7*10 = 7

(e)

Var(X+Y)

=Var(X) + Var(Y) +2cov(X,Y)

=7+10+7

=24