Sum of Gamma distributions?

Question

Let's say we have two gamma distributions X~G(a,b) Y~(c,d)

What's the distribution of X+Y?

Oops, I meant Y~G(c,d)

Answer 1

I don't think there is a general answer. If the gamma variates are equal b=d, then the shape parameters add: X+Y = (a+c,b=d) but you probably knew that already.

Answer 2

X ~ Γ(a, b)
Y ~ Γ(c, d)

Let Z = X + Y

Assuming that X and Y are independent then the moment generating function of Z is:

Mz(t) = (1 - bt)^a (1-dt)^c
given t < 1/b and t < 1/c

if b = d then
Mz(t) = (1 - bt) ^(a + c) and in this case

Z ~ Γ( a+ c, b)

if b ≠ d then you are at an impasse

you could look at Mz(t) as the moment generating function of the sum of a copies of Exp(b) and c copies of Exp(d).


you can also define the distribution by the convolution

the pdf of the gamma distribution is:

f_X(x) = 1 / (Γ(b) b^a) * x^(a-1) * Exp(-x/b)

the sum Z follows f_Z as defined as:

f_Z(z) = ∫ f(w) * f(z-w) dw

with the limits of integration being -∞ to ∞.