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%
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\title[Stat 609 Lecture 11]{Stat 609: Mathematical Statistics \\ Lecture 11}
\author[UW-Madison]{Jun Shao}
\institute[Statistics]{%
%  Department of Statistics\\
%  University of Wisconsin\\
%  Madison, WI 53706, USA\\
%~~~\\
  }
\date[2014]{}

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\AtBeginSection[]{\frame{\frametitle{Outline}\tableofcontents[current]}}

\begin{document}

%\frame{\titlepage}

\frame { \vspace{-2mm}
\frametitle{Lecture 11: Differentiating under an integral sign}
\begin{alertblock}{}
\begin{itemize}
\item
When can we switch the differentiation and integration?
\item
If the range of the integral is finite, this switch is usually valid.
\end{itemize}
\end{alertblock}
\vspace{-2mm}
\begin{block}{Theorem 2.4.1 (Leibnitz's rule)}<2->
If $f(x,\theta )$, $a(\theta )$, and $b(\theta )$ are differentiable with respect to $\theta$, then \vspace{-1mm}
\[ \frac{d}{d \theta} \int_{a(\theta )}^{b(\theta )} f(x,\theta )dx = f(b(\theta ),\theta )b'(\theta ) - f(a(\theta ),\theta )a'(\theta ) 
+ \int_{a(\theta )}^{b(\theta )} \frac{\partial f(x,\theta )}{\partial \theta } dx \vspace{-1mm}\]
In particular, if $a(\theta ) =a$ and $b(\theta )=b$ are constants, then  \vspace{-1mm}
\[ \frac{d}{d \theta} \int_{a}^{b} f(x,\theta )dx = \int_{a}^{b} \frac{\partial f(x,\theta )}{\partial \theta } dx \]
 \end{block}\vspace{-2mm}
\begin{alertblock}{}<3->
\begin{itemize}
\item
If the range of integration is infinite, problems can arise.
\item
Whether the order of differentiation and integration can be switched is the same as whether limit and integration can be interchanged,
because differentiation is a special kind of limit.
\end{itemize}
\end{alertblock}
}
\frame{
\begin{block}{Interchange of integration and limit}
Note that \vspace{-2mm}
\[ \int_{-\infty}^\infty \frac{\partial f(x,\theta )}{\partial \theta } dx = 
  \int_{-\infty}^\infty \lim_{\delta \rightarrow 0} \frac{f(x,\theta + \delta )- f(x,\theta )}{\delta} dx \vspace{-2mm} \]
  Hence, the interchange of differentiation and integration means whether this is equal to \vspace{-2mm}
\[ \frac{d}{d \theta } \int_{-\infty}^\infty f(x,\theta ) dx =
 \lim_{\delta \rightarrow 0}  \int_{-\infty}^\infty \frac{f(x,\theta + \delta )- f(x,\theta )}{\delta} dx  \]
\end{block}\vspace{-2mm}
\begin{alertblock}{An example of invalid interchanging integration and limit}
Consider 
\[ f(x,\delta ) = \left\{ \begin{array}{ll} 
\delta^{-1} & \quad 0 <x < \delta \\
0 & \quad \mbox{otherwise} \end{array} \right. \]
Then, $ \lim_{\delta \rightarrow 0} f(x,\delta ) = 0 $ for any $x$ and, hence, 
\[ 0 = \int_{-\infty}^\infty \lim_{\delta \rightarrow 0} f(x,\delta ) dx \neq 
 \lim_{\delta \rightarrow 0}  \int_{-\infty}^\infty f(x,\delta ) dx =
 \lim_{\delta \rightarrow 0}  \int_0^\delta  \delta^{-1} dx = 1  \] 
\end{alertblock} 
}
\frame{
\begin{alertblock}{}
We now give some sufficient conditions  
\end{alertblock}
\vspace{-2mm}
\begin{block}{Theorem 2.3.11}<2->
Let $X$ and $Y$ be random variables with cdfs
$F_X$ and $F_Y$, respectively.
\begin{itemize}
\item[a.]
If $X$ and $Y$ are bounded, then $F_X(u)=F_Y(u)$ for all $u$ iff $E(X^r)=E(Y^r)$
for all $r=1,2,...$
\item[b.]
If mgfs exist in a neighborhood of 0 and $M_X(t)=M_Y(t)$ for all $t$, then $F_X(u)=F_Y(u)$ for all $u$.
\end{itemize}
\end{block}\vspace{-2mm}
\begin{block}{}<3->
A rigorous proof of this theorem is technical and beyond the scope of this course.

The key idea of the proof can be explained as follows.

Note that \vspace{-1mm}
\[ M_X(t) = \int_{-\infty}^\infty e^{tx}f_X(x)dx \vspace{-1mm} \]
is the Laplace transformation of $f_X(x)$.

From the uniqueness of the Laplace transformation, there is a one-to-one correspondence between the mgf and the pdf.
\end{block}
}
\frame{
\begin{block}{}
The condition that the mgf exists in a neighborhood of 0 is important.

There are random variables with finite moments of any order, but their mgf's do not exist.
\end{block}\vspace{-2mm}
\begin{alertblock}{Example}<2->
The pdf\vspace{-1mm}
\[  f_X(x) = \frac{1}{\sqrt{2\pi}x} e^{-(\log x)^2/2} , \quad x \geq 0 \vspace{-1mm} \]
is called the log-normal distribution or density, because if $X$ has pdf $f_X$, then $\log X$ has a normal pdf.

In Example 2.3.10, we have shown that the log-normal distribution has finite moments of any order.

For $t>0$, \vspace{-2mm}
\[ M_X(t) = \int_0^\infty \frac{e^{tx}}{\sqrt{2\pi}x} e^{-(\log x)^2/2} dx = \infty \vspace{-1mm}\]
because, when $t>0$,  \vspace{-1mm}
\[ \lim_{x \rightarrow \infty } \frac{e^{tx}}{\sqrt{2\pi}x} e^{-(\log x)^2/2} = \infty \]
\end{alertblock}
}
\frame{\vspace{-2mm}
\begin{alertblock}{}
When $t<0$, \vspace{-2mm}
\[ M_X(t) = \int_0^\infty \frac{e^{tx}}{\sqrt{2\pi}x} e^{-(\log x)^2/2} dx
\leq \int_0^\infty \frac{1}{\sqrt{2\pi}x} e^{-(\log x)^2/2} dx = 1 \vspace{-1mm}\]
and, hence, $M_X(t)$ exists for all $t<0$.

How do we find a distribution for which all moments exist and the mgf does not exists for any $t \neq 0$?

Consider the pdf \vspace{-2mm}
\[  f_Y(x) = \left\{ \begin{array}{ll}
f_X(x)/2 & \quad x>0  \vspace{1mm} \\
f_X(-x)/2 & \quad x < 0 \end{array} \right. \vspace{-2mm}\]
For this pdf, \vspace{-2mm}
\[ E(|Y|^n) =  \int_0^\infty x^n \frac{f_X(x)}{2}dx +  \int_{-\infty}^0 (-x)^n \frac{f_X(-x)}{2} dx =
\int_0^\infty x^n f_X(x)dx = E(X^n) \vspace{-2mm}\]
which has been derived for any $n=1,2,...$

On the other hand, \vspace{-2mm}
\[ E(e^{tY}) = \int_0^\infty e^{tx} \frac{f_X(x)}{2}dx +  \int_{-\infty}^0 e^{tx} \frac{f_X(-x)}{2} dx \]
\end{alertblock}
}
\frame{
\begin{alertblock}{}
\[ = \int_0^\infty e^{tx} \frac{f_X(x)}{2}dx +  \int^{\infty}_0 e^{-tx} \frac{f_X(x)}{2} dx \vspace{-2mm} \]
and we have shown that one of these integrals is $\infty$ (depending on whether $t>0$ or $<0$).
\end{alertblock}\vspace{-2mm}
\begin{block}{}<2->
In applications we often need to approximate a cdf by a sequence of cdf's.

The next theorem gives a sufficient condition for the convergence of cdf's in terms of the convergence of mgf's.
\end{block}\vspace{-2mm}
\begin{block}{Theorem 2.3.12}<2->
Suppose that $X_1,X_2,...$ is a sequence of random variables with mgf's $M_{X_n}(t)$, and \vspace{-1mm}
\[ \lim_{n\rightarrow \infty} M_{X_n}(t) = M_X(t) < \infty \quad \mbox{for all $t$ in a neighborhood of 0} \vspace{-1mm}\]
where  $M_X(t) $ is the mgf of a random variable $X$. Then, for all $x$ where
$F_X(x)$ is continuous, \vspace{-1mm}
\[ \lim_{n\rightarrow \infty} F_{X_n}(x) = F_X(x). \]
\end{block}
}
\frame{
\begin{alertblock}{Example 2.3.13 (Poisson approximation)}
The cdf of the binomial pmf, \vspace{-2mm}
\[ f_X(x) = {n \choose x} p^x (1-p)^{n-x} , \quad x=0,1,...,n, \vspace{-2mm}\]
where $n$ is a positive integer and $0<p<1$, may not be easy to calculate when $n$ is very large.

It is often approximated by the cdf of the Poisson pmf, \vspace{-2mm}
\[ f_Y(x) = \frac{e^{-\lambda}\lambda^x}{x!} , \quad x=0,1,2,... \vspace{-2mm}\]
where $\lambda >0$ is a constant.

For this purpose, we first compute the mgf's for the binomial and Poisson distributions.
\end{alertblock}
\vspace{-2mm}
\begin{block}{Example 2.3.9 (binomial mgf)}<2->
Using the binomial formula \vspace{-2mm}
\[ \sum_{x=0}^n {n \choose x} u^xv^{n-x} = (u+v)^n \]
\end{block}
}
\frame{ \vspace{-2mm}
\begin{block}{}
we obtain \vspace{-2mm}
\beqrs
 M_X(t)&=& E(e^{tX}) = \sum_{x=0}^n e^{tx} {n \choose x} p^x(1-p)^{n-x} \\
&=&  \sum_{x=0}^n  {n \choose x} (pe^{t})^x(1-p)^{n-x}\\
&=&  ( pe^t + 1-p)^n
\eeqrs
Note that \vspace{-2mm}
\beqrs
 E(X) &=& \frac{d}{dt} M_X(t) \bigg|_{t=0} = n(pe^t+1-p)^{n-1} pe^t \bigg|_{t=0}= np \\
 E(X^2) &= & \frac{d^2}{dt^2} M_X(t) \bigg|_{t=0}\\
&=& [ n(n-1)(pe^t+1-p)^{n-1} (pe^t)^2 + n(pe^t+1-p)^{n-1} pe^t ]\bigg|_{t=0}\\
& =& n(n-1) p^2 +np
\eeqrs
We got the same results previously, but the calculation here is simpler.
\end{block}
}
\frame{
\begin{alertblock}{Example 2.3.13 (continued)}
If $Y$ has the Poisson pmf, \vspace{-2mm} then
\[ M_Y(t) = \sum_{x=0}^\infty e^{tx} \frac{e^{-\lambda} \lambda^x}{x!} = e^{-\lambda} \frac{(e^t \lambda )^x}{x!}
= e^{-\lambda} e^{e^t \lambda} = e^{\lambda (e^t-1)} \vspace{-2mm} \]
which is finite for any $t \in {\cal R}$.

Let $X_n$ have the binomial distribution with $n$ and $p$.

Suppose that $\lim_{n \rightarrow \infty} np = \lambda > 0 $ (that means $p$ also depends on $n$ and $p \rightarrow 0$ when $n \rightarrow \infty$).

Then, for any $t$,  as $n\rightarrow \infty$, \vspace{-2mm}
\[ M_{X_n}(t)  =
 (pe^t +1-p)^n = \left[ 1+ \frac{(np)(e^t-1)}{n}\right]^n  \rightarrow e^{\lambda (e^t-1)} = M_Y(t) \vspace{-2mm} \]
 using the fact that, for any sequence of numbers $a_n$ converges to $a$, \vspace{-2mm}
 \[ \lim_{n\rightarrow \infty} \left(1+\frac{a_n}{n}\right)^n = e^a \vspace{-2mm} \]
 With this result and Theorem 2.3.12, we can approximate $P(X_n \leq u)$ by $P(Y \leq u)$
 when $n$ is large and $np$ is approximately a constant.
\end{alertblock}
}
\end{document}
For $n=1,2...,$ let $X_n$ be a random variable having the pdf \vspace{-2mm}
\[ f_n(x) = \frac{1}{\sqrt{2\pi n}} e^{-x^2/(2n)} , \quad x \in {\cal R} \vspace{-2mm}\]
(this is the normal distribution with mean 0 and variance $n$).

The nonnegative function \vspace{-2mm}
\[ f(x) = \sum_{n=1}^\infty 2^{-n} f_n(x) \vspace{-2mm} \]
is a pdf, because \vspace{-2mm}
\[ \int_{-\infty}^\infty f(x)dx = \int_{-\infty}^\infty \sum_{n=1}^\infty 2^{-n} f_n(x)dx =
\sum_{n=1}^\infty 2^{-n} \int_{-\infty}^\infty f_n(x)dx = \sum_{n=1}^\infty 2^{-n} =1 \]
Let $X$ be a random variable having the pdf $f$.
\end{alertblock}
}
\frame{
\begin{alertblock}{}
For any $t \in {\cal R}$, \vspace{-2mm}
\beqrs
M_{X_n}(t)  & = & E(e^{tX_n}) = \frac{1}{\sqrt{2\pi n}} \int_{-\infty}^\infty e^{tx}e^{-x^2/(2n)} dx  \\
 &=& \frac{e^{nt^2/2}}{\sqrt{2\pi n}} \int_{-\infty}^\infty e^{-(x-nt)^2/(2n)} dx \\
&=& e^{nt^2/2} \vspace{-2mm}
\eeqrs
By differentiating $e^{nt^2/2}$, we obtain that \vspace{-2mm}
\[E(X_n^r) = \left\{ \begin{array}{ll}
(r-1)(r-3)\dots 3\cdot 1 n^{r/2} & \quad \mbox{when $r$ is an even integer} \\
0 & \quad \mbox{when $r$ is an odd integer} \end{array} \right. \vspace{-2mm} \]
Hence, for any odd $r$, $E(X^r)=0$; for any even $r$, \vspace{-1mm}
\[ E(X^r)  = \sum_{n=1}^\infty 2^{-n} E(X_n^r) \leq (r-1)(r-3)\cdots 1
\sum_{n=1}^\infty 2^{-n} n^r < \infty \vspace{-2mm} \]
However,
\[ M_X(t) = \int_{-\infty}^\infty e^{tx} f(x) dx
= \sum_{n=1}^\infty 2^{-n}  \int_{-\infty}^\infty e^{tx} f_n(x) dx   =
\sum_{n=1}^\infty 2^{-n} e^{nt^2/2} \]

\end{alertblock}\vspace{-3mm}
