Springer Texts in Business and Economics

The Gamma Distribution

a. Using theMGF for the Gamma distribution derived in problem 2.14f, we get MxO) = (1 — pt)-“.

Differentiating with respect to t yields

MX(t) = —a(1 — "t)-a-1(—") = a"(1 — "t)-a-1.

Therefore

Mx(0) = a" = E(X).

Differentiating MX0 (t) with respect to t yields

Mx0(t) = —a"(a + 1/(1 — "t)-a-2 (—") = a"2(a + 1/(1 — "t)-a-2.

Therefore

Mx0 (0) = a2"2 + a"2 = E (X2).

Hence

var(X) = E (X2) — (E (X))2 = a"2.

b. The method of moments equates E(X) = X = a"

and

n

E(X2) = J2 X2/n = a2"2 + a"2.

i=1

These are two non-linear equations in two unknowns. Substitute a = X/" into the second equation, one gets

n

J^Xl/n = X2 + X"

i=1

Hence,

n

E(Xi - X )2

" = ——=-----------

H nX

and

nX2

a =-------------------- .

n2 E(Xi - XX)2

i=1

c. For a = 1 and " = 9,we get

f (X; a = 1," = 9) = щ)9X1-1e-X/9 for X >0 and 9 > 0,

= 1 e-X/9 9

which is the exponential p. d.f.

d. For a = r/2 and " = 2, the Gamma (a = r/2, " = 2) is x2. Hence, from part (a), we get

E(X) = a" = (r/2)(2) = r

and

where