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Let X1, X2, and X3 be mutually independent random variables with Poisson distributions having means 2, 1, 4, respectively. Find the moment generating function of the sum Y = X1+X2+X3
 2 years ago
 2 years ago
S Group Title
Let X1, X2, and X3 be mutually independent random variables with Poisson distributions having means 2, 1, 4, respectively. Find the moment generating function of the sum Y = X1+X2+X3
 2 years ago
 2 years ago

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Zarkon Group TitleBest ResponseYou've already chosen the best response.1
where are you stuck?
 2 years ago

S Group TitleBest ResponseYou've already chosen the best response.0
I don't know how to start this. I missed class and I can't find much in the textbook. The poisson mgf is M(t) = e^(lambda((e^t)1)).. lambda is the mean, but i don't get how to put this all together.
 2 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
mgf of a sum of independent rv is the product of the MGF's
 2 years ago

S Group TitleBest ResponseYou've already chosen the best response.0
O, so does this mean I should plug in each mean into the formula and multiply them all together?
 2 years ago

S Group TitleBest ResponseYou've already chosen the best response.0
Oh ok, thank you!
 2 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
here is the property I was referring too. http://en.wikipedia.org/wiki/Momentgenerating_function#Sum_of_independent_random_variables
 2 years ago

S Group TitleBest ResponseYou've already chosen the best response.0
I kinda get it but i get confused with so many letters. X stands for my M(t) and what does a stand for then?
 2 years ago

S Group TitleBest ResponseYou've already chosen the best response.0
dw:1332814695156:dw I started with this but I'm not sure if this is the way.. do I also plug in 1 2 and 3 for t? Do I also multiply it by a since it's in the property? =\
 2 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
\[M_{Y}(t)=\mathbb{E}[e^{Yt}]=\mathbb{E}[e^{(X_1+X_2+X_3)t}]\] \[=\mathbb{E}[e^{X_1t+X_2t+X_3t}]=\mathbb{E}[e^{X_1t}e^{X_2t}e^{X_3t}]\] \[=\mathbb{E}[e^{X_1t}]\mathbb{E}[e^{X_2t}]\mathbb{E}[e^{X_3t}]\] \[=M_{X_1}(t)M_{X_2}(t)M_{X_3}(t)\]
 2 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
you can simplify that
 2 years ago

S Group TitleBest ResponseYou've already chosen the best response.0
Oooooo ok, that breakdown explains it, ok, i see now, thank you very very much!
 2 years ago

S Group TitleBest ResponseYou've already chosen the best response.0
Yea, I'll go on with the simplification, i just wasnt sure if this is where i start.
 2 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
The sum of independent poissons is poisson
 2 years ago

S Group TitleBest ResponseYou've already chosen the best response.0
Thanks a lot! I wish i could give you another medal =D
 2 years ago
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