Mixed Joint Probability Functions

Question

For P(X = n; Y = m) = a^2b^(n+m) with n, m greater than 1 and a + b = 1

Find

1. P(X = n) and P(Y = m)

2. E(X^2), E(Y^2)

3. E(XY) and Cov(XY)

Answer 1

One must have ∑_{m, n=1 to infinity) P(x=n, Y=m) = 1. That is: one must have

Sum_(m, n = 1 to ∞) a²b(m+n) = 1

But (since a and b are less than 1)

Sum_(m, n = 1 to ∞) a²b(m+n) = a² Sum_{m=1 to ∞}(Sum_{n=1 to ∞) b^(n+m)) =

a² Sum_{m=1 to ∞} (b^(1+m)/(1 - b)) = a²(b²/(1 - b)²) = a²b²/(1 - b)² = a²b²/a²=b²

^{Which is not equal to 1. So, for m and n starting at 1 P(X=n, Y=m) must be}

(a²/b²)b^(m+n)

But if we assume that m and n start at zero then

Sum_(m, n = 0 to ∞) a²b(m+n) = a²(1/(1 - b)²) = a²/a² = 1.

So let us assume that m and n start at 0. With this assumption we have

P(X = n) = Sum_{m=0 to ∞) a²b(m+n) = a²(bⁿ/(1-b)) = abⁿ

Similarly

P(Y = m) = Sum_{n = 0 to ∞) a²b(m+n) = ab^m

This shows that

P(X = n, Y = m) = P(X=n)P(Y=m)

Which implies that X and Y are independent. We need E(X) and E(Y).

E(X) = Sum_{n=0 to ∞} n abⁿ = ab/(1 - b)² = ab/a² = b/a

E(Y) = Sum_{n=0 to ∞) m ab^m = ab/(1 - b)² = b/a

Therefore

E(XY) = E(X) E(Y) = (b/a)². (Because X and Y are independent)

Also

E(X^2) = Sum_{n=0 to ∞} n² abⁿ = ab(1 + b)/(1 - b)² = (b/a)(1 + b)

Similarly

E(Y^2) = (b/a)(1 + b)

And

Cov(X, Y) = E(XY) - E(X)E(Y) = E(X)E(Y) - E(X)E(Y) = 0