The sum of two gamma distributions

AI Thread Summary
The discussion centers on proving that the sum of two independent gamma-distributed random variables, X and Y, results in another gamma distribution, Z, with parameters (x+y, λ). The moment generating functions (MGFs) are highlighted as a key method for this proof, with the relationship MZ(t) = MX(t) * MY(t) established. The participants confirm that the MGFs for X and Y can be expressed as MX(t) = (λ / (λ - t))^x and MY(t) = (λ / (λ - t))^y. By multiplying these MGFs, it is shown that MZ(t) corresponds to the MGF of a gamma distribution with parameters (x+y, λ). The conclusion affirms that the result holds true under the assumption of independence between X and Y.
Bachelier
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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)
 
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Bachelier said:
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?
 
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
 
Bachelier said:
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)].
 
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)
 
Bachelier said:
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 that's it. As long as you have the assumption that X and Y are independent, you have your result which is correct.
 
thanks
 
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