# The sum of two gamma distributions

Bachelier
let X~gamma(x,λ), Y~gamma(y,λ)
then Z = X+Y is gamma (x+y, λ)

I'm trying to prove this. Is using the moment generating functions the only way to do it.

and in such case, can I assume that MZ(t)= MX(t)*MY(t)

let X~gamma(x,λ), Y~gamma(y,λ)
then Z = X+Y is gamma (x+y, λ)

I'm trying to prove this. Is using the moment generating functions the only way to do it.

and in such case, can I assume that MZ(t)= MX(t)*MY(t)

Hint: Do you know what Moment Generating Functions are? Do you know the consequence of equating a MGF to a particular distribution?

Bachelier
I know MX(t) = ∫X E[et*x] dx

and MY(t) = ∫X E[et*y] dy

hence I get MZ(t) = ∫X E[et*x] dx * ∫YE[et*y] dy

I know MX(t) = ∫X E[et*x] dx

and MY(t) = ∫X E[et*y] dy

hence I get MZ(t) = ∫X E[et*x] dx * ∫YE[et*y] dy

Let Z = X + Y and calculate E[e^(tZ)]. Also you can use the fact that since X and Y are independent then E[e^(tZ)] = E[e^(t[X+Y])] = E[e^(tX + tY)] = E[e^(tX)] x E[e^(tY)].

Bachelier
I know how to get to

MX(t) = (λ/ λ-t)x

and MY(t) = (λ/ λ-t)y

hence since MZ(t) = MX(t)*MY(t)

this implies MZ(t) = (λ/ λ-t)x+y

which implies Z~gamma(x+y, λ)

I know how to get to

MX(t) = (λ/ λ-t)x

and MY(t) = (λ/ λ-t)y

hence since MZ(t) = MX(t)*MY(t)

this implies MZ(t) = (λ/ λ-t)x+y

which implies Z~gamma(x+y, λ)

Yep thats it. As long as you have the assumption that X and Y are independent, you have your result which is correct.

Bachelier
thanks