Question

Let W1,W2 be i.i.d (W) (i.e., independent and identically distributed as W) where W = z^2. Find the moment generating function for W2+W2. Also generalize the result to n.i.i.d. (W) random variables

Answer 1

\[f(w _{1},w _{2}) = fw _{1}(fw _{1}^{2})*(fw _{2}^{2})\]

Answer 2

is that how you start this

Answer 3

with w1 and w2 being independent

Answer 4

?

Answer 5

Correct, you have
\[\mathrm{M}_W(t)=\mathrm{M}_{W_1}(t)\times\mathrm{M}_{W_2}(t)\]
From the last question, you know that
\[\mathrm{M}_{W_1}(t)=\frac{1}{\sqrt{1-2t}}\]
so what would your answer be here?

Answer 6

To check your answer, note the following generalization: http://en.wikipedia.org/wiki/Chi-squared_distribution#Definition

Answer 7

okay I'm trying to figure it out

Answer 8

i would have to use integration to break up Mw1 and Mw2 ? @SithsAndGiggles

Answer 9

Nope, it's much, much simpler than that. The work is already done in the previous problem. We know that the mgf for both \(W_1\) and \(W_2\) are the same: \(\dfrac{1}{\sqrt{1-2t}}\).

Since \(W_1\) and \(W_2\) are i.i.d. random variables, you have that the mgf of the sum of these random variables is equivalent to the product of the individual mgf's:
\[\mathrm M_{W_1+W_2}=\mathrm M_{W_1}\times \mathrm M_{W_2}\]
So,
\[\mathrm M_{W_1+W_2}=\frac{1}{\sqrt{1-2t}}\times \frac{1}{\sqrt{1-2t}}=\frac{1}{1-2t}\]
In the link above, it says that the sum of \(k\) squared random variables has the mgf \(\dfrac{1}{(1-2t)^{k/2}}\). Notice that in the previous problem, \(k=1\), and we found that indeed, \(\mathrm M_{Z^2}=\dfrac{1}{(1-2t)^{1/2}}\). Here, \(k=2\), and we get \(\mathrm M_{W_1+W_2}=\dfrac{1}{(1-2t)^{2/2}}=\dfrac{1}{1-2t}\).

Answer 10

okay thank you that made it easier to follow being most of the work was done in the first part thanks! great help !!!

Answer 11

yw