Let X1, X2, and X3 be independent discrete random variables with common pmf p(x) = 1/3 if x = 0, 2/3 if x = 1, and 0 otherwise. Find the mgf of Y=X1X2X3. I made a table and came up with the same answer in the book, which is 19/27 + 8/27e^t. However, I would also like to do it without the table. I am trying to carry out the triple sum like this:
[sum(all x1) sum(all x2) sum(all x3)] p(x)e^(tX1X2X3)
=[sum(all x1) sum(all x2)](1/3 + 2/3 e^(tX1X2))
=[sum(all x1)] ???
I am stuck now, because I think this next step should eliminate X2, but I don't see how it will, but still give me the right answer.
[sum(all x1) sum(all x2) sum(all x3)] p(x)e^(tX1X2X3)
=[sum(all x1) sum(all x2)](1/3 + 2/3 e^(tX1X2))
=[sum(all x1)] ???
I am stuck now, because I think this next step should eliminate X2, but I don't see how it will, but still give me the right answer.
Finan 49.13
0 commentaires:
Enregistrer un commentaire