# Moment Generating Functions

• October 17th 2013, 11:07 PM
RedXIII
Moment Generating Functions
Let Y be a random variable that takes on positive integers.

$p(k) = \frac{1}{2}p(k-1), k = 1, 2, ... ,$

Where
$p(k) = P(Y = k) for\ k = 0, 1, 2, ... ,$

a) Find the probability that $Y = 0$
b) Find the moment generating function of $Y$.
c) Find the first and second moments

Attempt:
(A)
I don't really know what to do with it. I understand that it is an infinite geometric series that will sum to 1. But I'm not sure how to find Y = 0.
Using the formula for geometric series, I can find various p(k) in terms of p(0), but I don't know how to find p(0) itself.

$a_n=a_0r^n$

$p(1)=\frac{1}{2}p(0)$

(B)
$m(t)=E(e^t^y)=\sum_{y=0}^\infty e^t^y\frac{1}{2}p(y-1)=\frac{1}{2}\sum_{y=1}^\infty e^t^yp(y-1)=...?$

I'm I on the right track? I'm not sure how to proceed after that.

(C)
That just involves finding the first and second derivative of the result I find in B, correct?
• October 17th 2013, 11:15 PM
SlipEternal
Re: Moment Generating Functions
You declare $Y$ to be a variable that takes on positive integers. $0$ is not a positive integer. So, the probability that $Y=0$ must be 0, right? But, that implies that the probability that $Y=k$ must be 0 for all positive integers $k$. Did you mean $Y$ is a variable that can be any nonnegative integer?

Now, $\sum_{k=0}^\infty p(k) = 1$.

$p(1) = \dfrac{1}{2}p(0)$
$p(2) = \dfrac{1}{4}p(0)$
And in general, $p(k) = \dfrac{1}{2^k}p(0)$

Plugging that in, we have
$\sum_{k\ge 0} 2^{-k}p(0) = p(0)\sum_{k\ge 0}2^{-k} = 1$.

Now, you know $\sum_{k\ge 0}\left(\dfrac{1}{2}\right)^k = 2$, so $2p(0) = 1$ implies $p(0)=\dfrac{1}{2}$.

Does that help at all?
• October 18th 2013, 05:48 PM
RedXIII
Re: Moment Generating Functions
Sorry, my mistake. It is indeed non-negative integers.

That definitively makes a lot more sense now.
Just the clarify my understanding, in the last step, the summation equals 2 because:
$\sum_{k\ge 0}\left(\dfrac{1}{2}\right)^k =\dfrac{1}{1-\frac{1}{2}}=\dfrac{1}{\frac{1}{2}}=2$?

Then for part B, would this be the solution?
$m(t)=E(e^t^k)=\sum_{k=0}^\infty e^t^k\left(\dfrac{1}{2}\right)^k=e^t \sum_{k=0}^\infty \left(\dfrac{e}{2}\right)^k=e^t\left(\frac{1}{1-\dfrac{e}{2}}\right)$

Thanks!
• October 18th 2013, 06:02 PM
SlipEternal
Re: Moment Generating Functions
Quote:

Originally Posted by RedXIII
Sorry, my mistake. It is indeed non-negative integers.

That definitively makes a lot more sense now.
Just the clarify my understanding, in the last step, the summation equals 2 because:
$\sum_{k\ge 0}\left(\dfrac{1}{2}\right)^k =\dfrac{1}{1-\frac{1}{2}}=\dfrac{1}{\frac{1}{2}}=2$?

Then for part B, would this be the solution?
$m(t)=E(e^t^k)=\sum_{k=0}^\infty e^t^k\left(\dfrac{1}{2}\right)^k=e^t \sum_{k=0}^\infty \left(\dfrac{e}{2}\right)^k=e^t\left(\frac{1}{1-\dfrac{e}{2}}\right)$

Thanks!

Yes for the reason why the summation equals 2. No for part B.

$e^{tk} \neq e^te^k = e^{t+k}$

So,
\begin{align*}m(t) & = E(e^{tk}) \\ & = \sum_{k=0}^\infty e^{tk}\left(\dfrac{1}{2}\right)^kp(0) \\ & = \sum_{k=0}^\infty \left(\dfrac{e^t}{2}\right)^kp(0) \\ & = \dfrac{\tfrac{1}{2}}{1-\tfrac{e^t}{2}} \\ & = \dfrac{1}{2-e^t}\end{align*}

Then $m(0) = \dfrac{1}{2-e^0} = 1$

$m'(t) = \dfrac{e^t}{(2-e^t)^2}, m'(0) = \dfrac{e^0}{(2-e^0)^2} = 1$

$m''(t) = \dfrac{e^t(2-e^t)^2 + 2e^{2t}(2-e^t)}{(2-e^t)^3} = \dfrac{e^t(e^t+2)}{(2-e^t)^3}, m''(0) = \dfrac{e^0(e^0+2)}{(2-e^0)^3} = 3$