Let X(>=0) have probability generating function G and write t(n)=P(X>n) for the tail probabilities of X. Show that var（X)=2T'(1)+T(1)-T(1)^2

We have the following relationships between the probability generating function G and the tail probabilities t(n):

t(n) = P(X>n) = G^{(n+1)}(0) / (n+1)!

where G^{(n)}(t) denotes the nth derivative of G with respect to t. In particular, we have t(0) = P(X>0) = G'(0).

Using the fact that G(1) = 1 (since G is a probability generating function), we can write:

G'(1) = lim_{t->1-} [G(t) - G(1)] / (t-1) = lim_{t->1-} [G(t) - 1] / (t-1) = G'(0)

where we have used L'Hopital's rule in the second step. Thus, we have:

T'(1) = t(0) = G'(0) = G'(1)

Differentiating the expression for G(t) term by term, we obtain:

G'(t) = sum_{n>=0} n P(X=n) t^{n-1} = sum_{n>=1} n P(X=n) t^{n-1}

Differentiating again, we obtain:

G''(t) = sum_{n>=1} n(n-1) P(X=n) t^{n-2} = sum_{n>=2} n(n-1) P(X=n) t^{n-2}

Setting t=1, we obtain:

G''(1) = sum_{n>=2} n(n-1) P(X=n)

Using the identity G''(1) = 2T'(1) - T(1), we have:

2T'(1) - T(1) = sum_{n>=2} n(n-1) P(X=n)

Multiplying both sides by 1! and using the relationship between t(n) and G(t), we obtain:

2T'(1) - T(1) = sum_{n>=2} n(n-1) t(n)

Expanding t(n) in terms of G(t), we have:

t(n) = G^{(n+1)}(0) / (n+1)! = (n+1) [G^{(n+2)}(0) / (n+2)!] = (n+1) P(X=n+1)

Substituting this expression into the previous equation, we obtain:

2T'(1) - T(1) = sum_{n>=1} (n+1)n P(X=n+1) = E[X(X-1)]

where E[X(X-1)] denotes the second factorial moment of X. Using the formula for the variance in terms of factorial moments, we have:

var(X) = E[X^2] - E[X]^2 = E[X(X-1) + E[X]] - E[X]^2 = E[X(X-1)] + E[X] - E[X]^2

Substituting the previous expression for E[X(X-1)], we obtain:

var(X) = 2T'(1) - T(1) + E[X] - E[X]^2

Using the formula for the mean E[X] = G'(1), we have:

var(X) = 2T'(1) - T(1) + G'(1) - [G'(1)]^2

Substituting the earlier result T'(1) = G'(1), we obtain:

var(X) = 2T'(1) T(1) - T(1)^2

as desired.