The sum of two gamma distributions

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  • #1
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Main Question or Discussion Point

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)
 

Answers and Replies

  • #2
chiro
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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)
Hint: Do you know what Moment Generating Functions are? Do you know the consequence of equating a MGF to a particular distribution?
 
  • #3
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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
 
  • #4
chiro
Science Advisor
4,790
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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)].
 
  • #5
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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)
 
  • #6
chiro
Science Advisor
4,790
131
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.
 
  • #7
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thanks
 

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