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\centerline{Mathematics 375 -- Probability and Statistics 1}
\centerline{Discussion 2 -- Moment Generating Functions}
\centerline{October 15, 2009}
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\noindent
{\it Background}
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For a discrete random variable $Y$, we have introduced the {\it expected
value}, 
$$E(Y) = \sum_{{\rm values\ } y} y P(Y = y).$$
We have seen that the variance can be computed in terms of $E(Y)$ and
$$E(Y^2) = \sum_{{\rm values\ } y} y^2 P(Y = y),$$
since $V(Y) = E(Y^2) - (E(Y))^2$.  The two quantities $E(Y)$ and $E(Y^2)$
are two early terms in the sequence of {\it moments of $Y$ about the origin}, which
are defined in general as follows:  For each $k \ge 1$, the $k$th moment
about the origin is 
$$E(Y^k) = \sum_{{\rm values\ } y} y^k P(Y = y).$$
We can also extend this to $k = 0$ by setting $E(Y^0) = E(1) = 1$.  

When dealing with a sequence of numbers like the $E(Y^k)$, $k \ge 0$, it is
common in mathematics to try to ``package'' the sequence into a single 
object we can deal with using techniques for {\it functions defined by 
infinite series}.  The way we will do this is to form the so called
{\it moment (exponential) generating function}:
$$
m_Y(t) = \sum_{k = 0}^\infty E(Y^k) {t^k\over k!}.\leqno(1)
$$
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\noindent
{\it Discussion Questions}
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\item{A)} For the purposes of this question, you can take it as given
that the expected value operator will behave in the same linear fashion
for infinite sums of functions of $Y$ that we have seen with finite sums.
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\itemitem{1)}  Assuming that, show that 
$$m(t) = E(e^{tY}),$$
where $e^{tY}$ is treated as a function of the random variable $Y$, depending
on the parameter $t$.  (Hint: What does the Taylor series of $e^u$ look like?)
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\itemitem{2)}  Show that if you know $m_Y(t)$ as a function of $t$, then
you can recover all the moments of $Y$ about the origin by differentiating
$m_Y(t)$ sufficiently often.  The $k$th moment of $Y$ is the value at $t = 0$
of the $k$th derivative of $m_Y(t)$:
$$
E(Y^k) = m_Y^{(k)}(0).
$$
(This is one reason why putting the factorials in the denominator in (1) is
a good thing to do!  Without them, we would have to keep track of extra 
factorials in the derivatives.)
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\item{B)}  Let $Y$ have a {\it binomial} distribution based on $n$ trials with success 
probability $p$.  By computing $E(e^{tY})$ from the definition of the 
expected value, show that 
$$m_Y(t) = (pe^t + q)^n.$$
Also, use this to compute $E(Y)$ and $E(Y^2)$ and check our formula for $V(Y)$
in this case.  
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\item{C)}  Let $Y$ have a {\it geometric} distribution with success probability $p$.
By computing $E(e^{tY})$ from the definition of the expected value, show that 
$$m_Y(t) = {pe^t\over 1 - qe^t}.$$
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\noindent
The main importance of moment generating functions comes from the 
following Uniqueness Theorem:

\item{} \proclaim Theorem. Let $X$ and $Y$ be two (discrete) random variables
and assume $m_X(t)$ and $m_Y(t)$ both exist and are equal.  Then $X$ and $Y$
have the same distribution (that is, $X$ and $Y$ have 
the same probability mass function).

\item{D)}  Suppose you have random variables $U$ and $W$ whose moment generating
functions look like this:
$$m_U(t) = {.45 e^t \over 1 - (.55)e^t} \qquad {\rm and} \qquad m_W(t) = (.7 + .3e^t)^{12}.$$
What can you say about the distributions of $U$ and $W$?
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\item{E)}  Say $m_Z(t) = {1\over 6}e^t + {1\over 3} e^{2t} + {1\over 2}e^{3t}$.  
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\itemitem{1)}  Compute $E(Z)$ and $V(Z)$ from the information in the moment generating
function.
\itemitem{2)}  What is the distribution of $Z$?

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\noindent
{\it Assignment}
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Group write-ups due in class on Tuesday, October 20.
\bye
