- #1

- 376

- 0

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)

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Bachelier
- Start date

- #1

- 376

- 0

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

- #2

chiro

Science Advisor

- 4,790

- 132

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?

- #3

- 376

- 0

and M

hence I get M

- #4

chiro

Science Advisor

- 4,790

- 132

_{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 * ∫_{Y}E[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)].

- #5

- 376

- 0

M

and M

hence since M

this implies M

which implies Z~gamma(x+y, λ)

- #6

chiro

Science Advisor

- 4,790

- 132

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

- #7

- 376

- 0

thanks

Share: