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 M_Z(t)= M_X(t)*M_Y(t)

 

[chiro]

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 M_Z(t)= M_X(t)*M_Y(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 M_X(t) = ∫_X E[e^t*x] dx

and M_Y(t) = ∫_X E[e^t*y] dy

hence I get M_Z(t) = ∫_X E[e^t*x] dx * ∫_YE[e^t*y] dy

 

[chiro]

I know M_X(t) = ∫_X E[e^t*x] dx

and M_Y(t) = ∫_X E[e^t*y] dy

hence I get M_Z(t) = ∫_X E[e^t*x] dx * ∫_YE[e^t*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

M_X(t) = (λ/ λ-t)^x

and M_Y(t) = (λ/ λ-t)^y

hence since M_Z(t) = M_X(t)*M_Y(t)

this implies M_Z(t) = (λ/ λ-t)^x+y

which implies Z~gamma(x+y, λ)

 

[chiro]

I know how to get to

M_X(t) = (λ/ λ-t)^x

and M_Y(t) = (λ/ λ-t)^y

hence since M_Z(t) = M_X(t)*M_Y(t)

this implies M_Z(t) = (λ/ λ-t)^x+y

which implies Z~gamma(x+y, λ)

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

 

[Bachelier]

thanks