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\centerline{Mathematics 375 -- Probability and Statistics 1}
\centerline{Discussion 5 -- The ``Method of Moment-Generating Functions"}
\centerline{November 30, 2005}
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{\it Background}
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Because of the Uniqueness Theorem (Theorem 6.1 in our text), whenever we can compute
the moment-generating function for a random variable and recognize
it as one of our standard forms, then we know its distribution:
that is, its probability density function, mean, variance, and
hence ``everything about it''(!)  Today, we want to use this 
idea to work several examples and identify what we have.   
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{\it Discussion Questions}
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A) First, we will verify one point that
we deferred in discussing the use of the standard normal table.
Recall, we said that if $Y$ is normal with mean $\mu$ and 
standard deviation $\sigma$, then 
$$Z = {Y - \mu\over\sigma}$$
would have a standard normal distribution (i.e. normal with
mean 0 and standard deviation 1).  We never really justified this 
claim before (except in a review problem for Exam 2).  
But we {\it can do it now}!  Note that 
$$Z = {1\over\sigma} Y - {\mu\over\sigma}.$$
Find the moment generating function of $Z$ given the
moment generating function for the normal $Y$:
$$m_Y(t) = e^{{t^2\sigma^2\over 2} + \mu t}.$$
Deduce
that $Z$ must have a standard normal distribution.
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\noindent
C)  Let $Y$ be a standard normal, and let  $X = Y^2$.  
\item{1)}  Set up the integral to compute 
$$m_X(t) = E(e^{tX})$$
using the density for $Y$.  
\item{2)}  For the rest of the problem, assume $1 - 2t > 0$.  
Combine terms in your integral, make the substitution $u = \sqrt{1-2t}\cdot y$,
and show that 
$$m_X(t) = {1\over \sqrt{1 - 2t}}.$$
\item{3)}  What is the distribution of $X$?  (What ``type of random
variable'' is $X$, according to the Uniqueness Theorem?)
\item{4)}  Suppose $Y_1, \ldots, Y_k$ are independent random
variables, each with a standard normal distribution.  What
is the distribution of $X = Y_1^2 + \cdots + Y_k^2$?
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{\it Assignment}
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Group writeups due in class on Friday, December 2.
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