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\begin{document} 
\title{Written Assignment 5}
\section{Written Assignment 5}  
\subsection{Problem 1}
\par\vspace{0.3 cm}
A class of functions known as \textit{linear operators} is defined by the following properties:  If $L$ is a linear operator, then for any $X,Y$ in the domain of $L$ and any real number $\alpha$,
\[
L(\alpha X) = \alpha[L(X)]\quad\mbox{and}\quad L(X+Y) = L(X)+L(Y)
\]
Assuming that the "expectation operator" $E()$ has these properties, show that if $X$ is a random variable and $c$ is a constant,
\[
E[c]=c
\]
(hint: Define a function $f(X)$ that maps all values of $X$ into $c$, and use the properties of linear operators)
\subsection{Problem 2}
Assuming that the "expectation operator" $E()$ is a linear operator, show that if $X$ is a discrete random variable and $E(X)=\mu$, 
\par\vspace{0.3 cm}\noindent
a) 
\[
V(X)=E[(X-\mu)^2] = E(X^2)-2\mu E(X)+[E(X)]^2
\]
provided all of the sums converge.
\par\vspace{0.3 cm}\noindent
b) Show that the above expression simplifies to:
\[
V(X)=E[(X-\mu)^2] = E(X^2)-\mu^2 = E(X^2)-[E(X)]^2
\]
\subsection{Problem 3}
The moment generating function of a random variable $X$ is defined as:
\[
m(t) = E(\exp(tX))
\]
provided the expected value exists in some neighborhood of $t=0$.  For our purposes, we can treat $t$ as a constant.
\par\vspace{0.3 cm}
Use the binomial expansion formula
\[
(a+b)^n = {n\choose n}a^nb^0+{n \choose n-1}a^{n-1}b^1+\cdots+{n \choose 0}a^0b^n = \sum_{k=0}^n{n\choose k}a^kb^{n-k}
\]
to show that the moment generating function of the binomial distribution with $n$ trials and probability of success $p$ on each trial is:
\[
m(t) = (pe^t+1-p)^n
\]
\subsection{Problem 4}
\par\vspace{0.3 cm}
We can think of a probability distribution as a combination of a probability space $(\Omega,\mathcal{F},\rho)$ and a random variable $X$ defined on that probability space.  For example, in the case of the binomial experiment we define a random variable $X$ whose value is the number of successes in the outcome of the experiment on the underlying probability space corresponding to $n$ independent Bernoulli trials with probability of success $p$.  We say that the random variable $X$ has a binomial distribution.  
\par\vspace{0.3 cm}
There is a 1-1 correspondence between probability distributions and moment generating functions, that is, if you know the moment generating function, you can identify the distribution and vice versa.
\par\vspace{0.3 cm}
An important theorem states that if $X_1,X_2,\ldots,X_n$ are independent random variables with moment generating functions $m_1(t),\ldots,m_n(t)$, then the moment generating function of the random variable $X$ defined as their sum, 
\[
X=X_1+X_2+\cdots+X_n = \sum_{i=1}^nX_i
\] 
is the product of the individual moment generating functions:
\[
m_X(t) = E(e^{tX})=E(\exp[t(X_1+X_2+\cdots+X_n)]) = \prod_{i=1}^ne^{tX_i} = \prod_{i=1}^nm_i(t)
\]
\par\vspace{0.3 cm}
a) Show that the moment generating function of the Bernoulli distribution with probability of success $p$ is:
\[
m(t) = pe^t+1-p
\]
\par\vspace{0.3 cm}
b) Use the fact that there is a 1-1 correspondence between moment generating functions and probability distributions to show that a random variable defined as sum of $n$ independent Bernoulli trials each with probability of success $p$ should have a binomial distribution.
\subsection{Optional Extra Credit Problem 5}
\par\vspace{0.3 cm}
Suppose $X$ is a discrete random variable and $A$ is a subset of the possible values of $X$, that is,
\[
A\subseteq X[\Omega]
\]
Define the \textbf{indicator function} of $A$ by:
\[
I_A:X[\Omega]\rightarrow\{0,1\}\quad\mbox{by}\quad I_A(x)=\left\{\begin{array}{lcl}
1 &\mbox{ if }& x\in A\\
0 &\mbox{ if }& x\notin A
\end{array}\right.
\] 
\par\vspace{0.3 cm}
Suppose $X$ is a discrete random variable and $c$ is a constant, and let
\[
A = \{x: x\in X[\Omega]\mbox{ and } x>c\}
\]
Show that
\[
P(X>c) = E(I_A(X))
\]
\end{document}



