A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 years ago
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
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

This Question is Closed

S
 2 years ago
Best ResponseYou've already chosen the best response.0I 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.

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

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

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

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

S
 2 years ago
Best ResponseYou've already chosen the best response.0dw: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? =\

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

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

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

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

S
 2 years ago
Best ResponseYou've already chosen the best response.0Thanks a lot! I wish i could give you another medal =D
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.