• S

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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

  • S

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

where are you stuck?
  • S
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.
mgf of a sum of independent rv is the product of the MGF's

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

  • S
O, so does this mean I should plug in each mean into the formula and multiply them all together?
yes
  • S
Oh ok, thank you!
yw
here is the property I was referring too. http://en.wikipedia.org/wiki/Moment-generating_function#Sum_of_independent_random_variables
  • S
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?
  • S
|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? =\
\[M_{Y}(t)=\mathbb{E}[e^{Yt}]=\mathbb{E}[e^{(X_1+X_2+X_3)t}]\] \[=\mathbb{E}[e^{X_1t+X_2t+X_3t}]=\mathbb{E}[e^{X_1t}e^{X_2t}e^{X_3t}]\] \[=\mathbb{E}[e^{X_1t}]\mathbb{E}[e^{X_2t}]\mathbb{E}[e^{X_3t}]\] \[=M_{X_1}(t)M_{X_2}(t)M_{X_3}(t)\]
you have it
you can simplify that
  • S
Oooooo ok, that breakdown explains it, ok, i see now, thank you very very much!
  • S
Yea, I'll go on with the simplification, i just wasnt sure if this is where i start.
The sum of independent poissons is poisson
  • S
Thanks a lot! I wish i could give you another medal =D
;)...yw

Not the answer you are looking for?

Search for more explanations.

Ask your own question