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\title[Stat 609 Lecture 13]{Stat 609: Mathematical Statistics \\ Lecture 13}
\author[UW-Madison]{Jun Shao}
\institute[Statistics]{%
%  Department of Statistics\\
%  University of Wisconsin\\
%  Madison, WI 53706, USA\\
%~~~\\
  }
\date[2015]{}

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

\begin{document}

%\frame{\titlepage}

\frame {\vspace{-2mm}
\frametitle{Lecture 13: Noncentral $\chi^2$-, t-, and F-distributions}
\vspace{-1mm}
\begin{block}{}
The results on transformation lead to many useful results 
based on transformations of normal random variables. 	
\end{block}\vspace{-3mm}
\begin{alertblock}{Ratio of two normal random variables}
	If $X_1$ and $X_2$ are independent and both have the
	normal distribution $N(0,1)$, then,
	the pdf of $V_1=X_1/X_2$ is \vspace{-3mm}
	\beqrs
	f_{V_1}(v_1)  & = & \int_{-\infty}^\infty |v_2| \frac{e^{-(v_1v_2)^2/2}}{\sqrt{2\pi} }
	\frac{e^{-v_2^2/2}}{\sqrt{2\pi} } dv_2 \\
	& = &
	\frac{1}{2\pi} \int_{-\infty}^\infty |v_2| e^{-(1+v_1^2)v_2^2/2} dv_2 \\
	&=& \frac{1}{\pi} \int_{0}^\infty v_2 e^{-(1+v_1^2)v_2^2/2}  dv_2  \\
	& = &  \frac{1}{\pi} \int_0^\infty e^{-(1+v_1^{2})s}ds  \\
	& = &  \frac{1}{\pi(1+v_1^2)}
	\eeqrs
	
	\vspace{-3mm}
	which is the pdf of  $Cauchy(0,1)$.
\end{alertblock}
}
\frame{
\vspace{-2mm}
\begin{alertblock}{}
	The next result
	concerns a ratio of independent chi-squares random variables, or sums of squared independent normal random variables.  
\end{alertblock}\vspace{-3mm}
	
	\begin{block}{Ratio of chi-square random variables and F-distribution}
		Let $X_1$ and $X_2$ be independent random variables having the chi-square
		distributions with degrees of freedom $n_1$ and $n_2$, respectively.
		
		By the transformation theorem,  the p.d.f.\ of $Z=X_1/X_2$ is \vspace{-2mm}
		\beqrs
		f_Z(z)  & = &  \int_0^\infty  x_2 \frac{(zx_2)^{n_1/2-1} e^{-zx_2/2}}{2^{n_1/2}\Gamma(n_1/2)} \
		\frac{x_2^{n_2/2-1}e^{-x_2/2}}{2^{n_2/2}\Gamma(n_2/2)} dx_2 \\
		& = & \frac{z^{n_1/2-1}}{
			2^{(n_1+n_2)/2} \Gamma(n_1/2)\Gamma(n_2/2)}
		\int_0^\infty x_2^{(n_1+n_2)/2-1}e^{-(1+z)x_2/2}dx_2\\
		& = & \frac{\Gamma[(n_1+n_2)/2]}{
			\Gamma(n_1/2)\Gamma(n_2/2)}
		\frac{z^{n_1/2-1}}{(1+z)^{(n_1+n_2)/2}},  \quad \quad \quad z>0
		\eeqrs
		
		\vspace{-2mm}
		where the last equality follows from the fact that\vspace{-2mm}
		\[\frac{1}{2^{(n_1+n_2)/2}\Gamma[(n_1+n_2)/2]}
		x_2^{(n_1+n_2)/2-1}e^{-x_2/2} \quad \quad \quad x_2>0 \vspace{-2mm} \]
		is the pdf of the chi-square
		distribution with degrees of freedom $n_1+n_2$.
	\end{block}}
	\frame{\vspace{-2mm}
		\begin{block}{}
			Making another transformation $Y=(n_2/n_1)Z = (X_1/n_1)/(X_2/n_2)$ and applying the (univariate) transformation theorem,
			we obtain that the pdf of $Y$ is\vspace{-2mm}
			\[ \frac{n_1^{n_1/2} n_2^{n_2/2} \Gamma[(n_1+n_2)/2]}{
				\Gamma(n_1/2)\Gamma(n_2/2)}
			\frac{y^{n_1/2-1}}{(n_1y+n_2)^{(n_1+n_2)/2}},  \quad \quad \quad y>0 \vspace{-2mm} \]
			which is the pdf of the well-known F-distribution with degrees of freedom $n_1$ and $n_2$.
		\end{block}\vspace{-2mm}
		\begin{alertblock}{t-distribution}
			Let $U_1$ and $U_2$ be independent random variables, $U_1 \sim N(0,1)$ and $U_2$ has the chi-square distribution with degrees of freedom
			$n$.
			\vspace{1mm}
			
			What is the distribution of $T=U_1/\sqrt{U_2/n}$?
			\vspace{1mm}
			
			Let $X_1=U_1^2$ and $X_2=U_2$.
			
			Then $X_1$ and $X_2$ are independent, because $U_1$ and $U_2$ are independent.
			
			By a result obtained previously,  $X_1$ has the chi-square distribution with degree of freedom 1.
			
			From the previous proof, $Y=X_1/(X_2/n)$ has the F-distribution with degrees of freedom 1 and $n$.
		\end{alertblock}}
		\frame{\vspace{-2mm}
			\begin{alertblock}{}
				The pdf of $Y$ is \vspace{-2mm}
				\[ \frac{n^{n/2}
					\Gamma[(n+1)/2]y^{-1/2}}{\sqrt{\pi}\Gamma(n/2)
					(n+y)^{(n+1)/2}} \quad \quad \quad y> 0 \vspace{-2mm} \]
				Applying the univariate transformation theorem, we obtain that the pdf of $W= \sqrt{Y}$ is ($d y = 2wdw $ when $y=w^2$) \vspace{-2mm}
				\[ \frac{2 n^{n/2}
					\Gamma[(n+1)/2]}{\sqrt{\pi}\Gamma(n/2)
					(n+w^2)^{(n+1)/2}} \quad \quad \quad w> 0 \vspace{-2mm} \]
				Note that\vspace{-2mm}
				\[ T = \left\{ \br{ll}
				W & \quad U_1 \geq 0 \\
				-W & \quad U_1 < 0 . \er \right.\vspace{-2mm}  \]
				and
				\[ P( T < -t ) = P( T > t) . \quad \quad \quad t > 0 .\vspace{-2mm}\]
				Hence the pdf of $T$ is \vspace{-2mm}
				\[ \frac{ n^{n/2}
					\Gamma[(n+1)/2]}{\sqrt{\pi}\Gamma(n/2)
					(n+t^2)^{(n+1)/2}} \quad \quad \quad t \in {\cal R} \vspace{-2mm} \]
				This  is the pdf of the well-known t-distribution with degrees of freedom $n$.
			\end{alertblock}
		}
		\frame{
			\begin{alertblock}{}
				When $n=1$, the pdf of the t-distribution is \vspace{-2mm} 
				\[ \frac{ \Gamma(1)}{\sqrt{\pi}\Gamma(1/2)
					(1+t^2)} = \frac{1}{\pi (1+t^2)} \quad \quad \quad t \in {\cal R} \vspace{-2mm} \]
				which is the pdf of $Cauchy(0,1)$.  
				\vspace{1mm} 
				
				Hence, if $X_1 \sim N(0,1)$ and $X_2 \sim N(0,1)$ are independent, then 
				we have just shown that $X_1 / |X_2|  \sim Cauchy(0,1) $.  
				\vspace{1mm}
				
				But previously we showed that $X_1 / X_2  \sim Cauchy(0,1) $.
				\vspace{1mm}
				
				If they are both true, then $X_1/X_2 \sim X_1/|X_2|$. 
				\vspace{1mm}
				
				This is in fact true, because for $t \in {\cal R}$,  \vspace{-2mm} 
				\[  \begin{aligned}
				P( X_1/X_2 < t ) & = P(X_1 < t X_2 , X_2 >0 ) + P( X_1 > t X_2  , X_2 < 0 ) \\
				& = P(X_1 < t |X_2|, X_2 >0 ) + P( X_1 > -t |X_2 | , X_2 < 0 ) \\
				& = P(X_1 < t |X_2|, X_2 >0 ) + P( - X_1 < t |X_2 | , X_2 < 0 ) \\
				& = P(X_1 < t |X_2|, X_2 >0 ) + P(  X_1 < t |X_2 | , X_2 < 0 )  \\
				& = P(X_1 < t |X_2|  ) =  P(X_1 / |X_2|  < t)
				\end{aligned}\vspace{-2mm} \]
				and the 4th equality holds since $(X_1,X_2) \sim (-X_1,X_2)$. 
			\end{alertblock}}
\frame{

\begin{alertblock}{}
The previously defined chi-square ($\chi^2$-), t-, and F-distributions previously defined are special cases of the
non-central $\chi^2$-, t-, and F-distributions, respectively, which are useful in statistics. 
\end{alertblock}
\vspace{-2mm}
\begin{block}{Definition (the noncentral chi-square distribution)}
Let $X_1,...,X_n$ be independent random variables and
$X_i \sim N(\mu_i,\sigma^2)$, $i=1,...,n$. The distribution of the random
variable $Y = (X_1^2 +\cdots + X_n^2)/\sigma^2$ is called the
\cblue{noncentral chi-square} distribution with degrees of freedom $n$ and the noncentrality parameter
$\delta = (\mu_1^2+ \cdots + \mu_n^2)/\sigma^2$.
\end{block}\vspace{-2mm}
\begin{alertblock}{}
The chi-square distribution defined earlier is
a special case of the noncentral chi-square distribution
with $\delta
=0$ and, therefore, is sometimes called a central chi-square distribution.
\vspace{1.5mm}

It follows from the definition of noncentral
chi-square distributions that if $Y_1,...,Y_k$ are independent
random variables and $Y_i$ has the noncentral chi-square distribution with degrees of freedom
$n_i $ and the
\end{alertblock}
}
\frame{
\begin{alertblock}{}
noncentrality parameter $\delta_i$, $i=1,...,k$, then
$Y=Y_1+\cdots + Y_k$ has the noncentral chi-square distribution with degrees of freedom
$n_1+\cdots + n_k$ and the noncentrality parameter $\delta_1+ \cdots + \delta_k$.
\end{alertblock}\vspace{-2mm}
\begin{block}{Theorem (properties of the noncentral chi-square distribution)}
Let $Y$ be a random variable having the noncentral chi-square
distribution with degrees of freedom $k$ and noncentrality parameter $\delta$.
\begin{itemize}
\item[(i)]
The pdf of $Y$ is \vspace{-2mm}
\[ g_{\delta , k} (x) =
e^{-\delta /2} \sum_{j=0}^\infty \frac{(\delta /2)^j}{j!} f_{2j+k}(x), \vspace{-3mm} \]
where $f_v(x)$
is the pdf of the central chi-square distribution with degrees of freedom $v$, $v=1,2,...$;
\item[(ii)] The mgf of  $Y$ is \vspace{-1mm}
\[ \frac{ e^{\delta t/(1-2t)}}{(1-2 t)^{ k/2}}, \quad\quad t < 1/2;\vspace{-2mm} \]
\item[(iii)]
$E(Y) = k+\delta$;
\item[(iv)]
 Var$(Y) = 2k +4\delta$.
\end{itemize}
\end{block}}
\frame{
\begin{block}{Proof.}
We first prove result (ii).
\vspace{1mm}

Let $X_k$ be a random variable having the standard normal distribution
and $\mu = \sqrt{\delta} $ be a positive number.
\vspace{1mm}

For $t<1/2$, the mgf of $(X_k+\mu )^2$ is \vspace{-2mm}
\beqrs
\psi_\delta (t) \back &=& \back
\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty  e^{-x^2/2}e^{t(x+\mu )^2} dx\\
&=& \back \frac{e^{\mu^2 t/(1-2t)}}{\sqrt{2\pi}}
\int_{-\infty}^\infty  e^{ -(1-2t)[x-2\mu t/(1-2t)]^2/2} dx\\
&=& \back \frac{e^{\delta t/(1-2t)}}{ \sqrt{1-2t}} .
\end{eqnarray*}
$~~$ \vspace{-5mm}

By definition, $Y \sim X_1^2+\cdots +
X_{k-1}^2 + (X_k + \sqrt{\delta})^2$,
where $X_i$'s are independent and have the standard normal distribution.
\vspace{1mm}

From the obtained result, the mgf of $Y$ is \vspace{-2mm}
\[ E\left\{ e^{t[X_1^2+ \cdots + X_{k-1}^2+(X_k+\sqrt{\delta})^2]}\right\}
= [\psi_0(t)]^{k-1} \psi_{\sqrt{\delta}}(t)
=   \frac{e^{\delta  t/(1-2t)}}{ (1-2t)^{k/2}} . \]
\end{block}}
\frame{\vspace{-2mm}
\begin{block}{}
We now use the proved result (ii) to show result (i).
\vspace{1mm}

By the uniqueness of the mfg, it suffices to show that the pdf $g_{\delta,k}(x)$ in (i) has mgf exactly the same as the one in (ii).
\vspace{1mm}

Note that the mgf of the central chi-square
distribution with degrees of freedom $v$ is, \vspace{-1mm}
\[ \frac{1}{(1-2t)^{v/2}}  , \quad\quad t < 1/2 , \vspace{-1mm} \]
The mgf of the pdf $g_{\delta,k}(x)$ in (i) is, for $t<1/2$, \vspace{-2mm}
\begin{eqnarray*}
\int_{-\infty}^\infty e^{tx} g_{\delta , k} (x) dx
\back &=& \back e^{-\delta/2} \int_{-\infty}^\infty \sum_{j=0}^\infty \frac{(\delta/2)^j}{j!}
 e^{tx} f_{2j+k} (x) dx  \\
\back &=& \back e^{-\delta/2} \sum_{j=0}^\infty \frac{(\delta/2)^j}{j!}
\int_{-\infty}^\infty  e^{tx} f_{2j+k} (x) dx  \\
&=& \back e^{-\delta/2} \sum_{j=0}^\infty \frac{(\delta/2)^j}{j!(1-2t)^{(j+k/2)}} \\
&=& \back \frac{e^{-\delta/2}}{(1-2t)^{k/2}}\sum_{j=0}^\infty
\frac{\{\delta/[2(1-2t)]\}^j}{j!}
\eeqrs
 \end{block}
}
\frame{
\begin{block}{}
\[ = \frac{e^{-\delta/2+\delta/[2(1-2t)]}}{(1-2t)^{k/2}}
= \frac{e^{\delta t/(1-2t)}}{(1-2t)^{k/2}}\]
To establish (iii), let $X_i$'s be as defined in the proof of (i).
\vspace{1mm}

 Then, \vspace{-2mm}
\beqrs
 E(Y) \back & = & \back E(X_1^2) + \cdots + E(X_{k-1}^2) + E(X_k+\sqrt{\delta})^2 \\
& = & \back k-1 + E(X_k^2) + \delta + E(2 \sqrt{\delta} X_k) \\
& = & \back k + \delta
\eeqrs
$~~$ \vspace{-7mm}

because $E(X_i^2) =1$ and $E(X_i) =0$.
\vspace{1mm}

To show (iv), note that Var$(X_i^2)=2$ and
Cov$(X_k^2,X_k) = E(X_k^3)  =0$.
\vspace{1mm}

 Then, \vspace{-2mm}
\beqrs
\mbox{Var}(Y) \back & = & \back \mbox{Var}(X_1^2) + \cdots +
\mbox{Var}(X_{k-1}^2) + \mbox{Var}\bigl((X_k+\sqrt{\delta})^2\bigr) \\
& = & \back 2(k-1) + \mbox{Var}(X_k^2+ 2\sqrt{\delta}X_k)\\
& = & \back 2(k-1) + \mbox{Var}(X_k^2) + 4 \delta \mbox{Var}(X_k)
+ 4\sqrt{\delta} \mbox{Cov}(X_k^2,X_k) \\
& = & \back 2k + 4 \delta
\eeqrs

\end{block}}
\frame{
\begin{alertblock}{Definition (the noncentral t-distribution)}
Let $X \sim N(\delta , 1 )$, $\delta \in {\cal R}$, $U \sim $ the central chi-square with degrees of freedom $n$, and $X$ and $U$ be independent.
The distribution of $T= X/\sqrt{U/n}$ is the \cred{noncentral t-distribution} with degrees of freedom $n$ and noncentrality parameter $\delta $.
\end{alertblock}
\vspace{-2mm}
\begin{block}{}
The t-distribution previously defined can be
called a central t-distribution, since it is
a special case of the noncentral t-distribution with $\delta =0$.

 Using the formula for the ratio of two independent random variables, we can show that $T$ has the following pdf: \vspace{-2mm}
\[ \frac{1}{2^{(n+1)/2} \Gamma (n/2) \sqrt{\pi n}}
\int_0^\infty y^{(n-1)/2} e^{-[(x\sqrt{y/n}-\delta)^2+y]/2} dy \vspace{-2mm} \]
Since $U \sim $ the central chi-square with degrees of freedom $n$, $W = \sqrt{U/n}$ has pdf \vspace{-2mm}
\[ \frac{2n^{n/2}}{\Gamma (n/2) 2^{n/2}} w^{n-1} e^{-nw^2/2}, \quad\quad w> 0  \]
\end{block}
}
\frame{
\begin{block}{}
Then, $T$ has pdf \vspace{-2mm}
\beqrs
& & \back
 \int_0^\infty w  \left( \frac{n^{n/2}}{\Gamma (n/2) 2^{n/2-1}} w^{n-1} e^{-nw^2/2} \right)
 \left( \frac{1}{\sqrt{2\pi} } e^{-(x w-\delta )^2/2 } \right) dw \\
 & = & \back \frac{n^{n/2}}{\Gamma (n/2) 2^{(n-1)/2}\sqrt{\pi}} \int_0^\infty w^n e^{[(xw-\delta)^2 + nw^2]/2 } dw
 \eeqrs
 $~~$ \vspace{-5mm}

 Letting $y = nw^2$, we obtain the desired result.
 \vspace{1.5mm}

 By the independence of $X$ and $U$ and the fact that $X \sim N(\delta , 1)$ and $U \sim $ the central chi-square, \vspace{-2mm}
 \beqrs
  E(T) \back & = & \back E\left( \frac{X}{\sqrt{U/n}}\right) = \sqrt{n}  E( X ) E \left(\frac{1}{\sqrt{U}}\right) \\
 \back & = & \back
  \sqrt{n} \delta \int_0^\infty
 \frac{1}{\sqrt{u}}  \left( \frac{1}{\Gamma (n/2) 2^{n/2} } u^{n/2 -1} e^{-u/2}\right) du \\
 \back & = & \back
 \frac{ \sqrt{n} \delta}{\Gamma (n/2) 2^{n/2} } \int_0^\infty
  u^{(n-1)/2 -1} e^{-u/2} du =
 \frac{ \sqrt{n} \delta \Gamma((n-1)/2)}{\sqrt{2} \Gamma (n/2) }
 \eeqrs
$~~$ \vspace{-5mm}

when $n >1$ and is $\infty$ when $n=1$.
 \end{block}
}
\frame{\vspace{-2mm}
\begin{block}{} \vspace{-5mm}
\beqrs
  E(T^2) \back & = & \back E\left( \frac{X^2}{U/n}\right) = n  E( X^2 ) E \left(\frac{1}{U}\right) \\
 \back & = & \back
  n(1+ \delta^2)\frac{1}{\Gamma (n/2) 2^{n/2} } \int_0^\infty
  u^{n/2 -2} e^{-u/2}du \\
\back & = & \back
 \frac{ n(1+ \delta^2)}{\Gamma (n/2) 2^{n/2} } \Gamma(n/2-1)2^{n/2-1} = \frac{n(1+\delta^2)}{n-2}
 \eeqrs
 $~~$ \vspace{-5mm}

 when $n>2$ and is $\infty$ when $n \leq 2$.
 \vspace{1mm}

 Hence, when $n>2$,
 \vspace{-3mm}
 \[ {\rm Var}(T) =  \frac{n(1+\delta^2)}{n-2} -  \frac{n\delta^2 }{2} \left[ \frac{ \Gamma((n-1)/2)}{\Gamma (n/2) }\right]^2  \]
\end{block}\vspace{-2mm}
\begin{alertblock}{Definition (the noncentral F-distribution)}
Let $X_1 \sim $  the noncentral chi-square
distribution with degrees of freedom $n_1$ and noncertrality parameter $\delta \geq 0$, $X_2 \sim $ the central
chi-square distribution with degrees of freedom $n_2$, and $X_1$ and $X_2$ be
independent. The distribution of $F= (X_1/n_1)/(X_2/n_2)$ is called the
\cred{noncentral F-distribution} with degrees of freedom  $n_1$ and $n_2$ and noncentrality parameter $\delta $.
\end{alertblock}}
\frame{\vspace{-2mm}
\begin{alertblock}{}
The F-distribution introduced previously can be
called a central F-distribution, since
it is a special case of the noncentral F-distribution with $\delta =0$.
\vspace{1.5mm}

Using the formula for the ratio of two independent random variables and the pdf of $X_1$ we derived previously,
we can obtain the pdf for $F = (X_1/n_1)/(X_2/n_2)$.
\vspace{1.5mm}

Let $f_v$ denote the pdf of the central chi-square with degrees of freedom $v$ and
$f_{k_1,k_2}(x)$ be the pdf of the central
F-distribution with degrees of freedom $k_1$ and $k_2$.
\vspace{1.5mm}

Then the pdf of $F$ is \vspace{-2mm}
\beqrs
\back & & \back
\int_0^\infty y \left[ e^{-\delta /2} \sum_{j=0}^\infty \frac{(\delta /2)^j}{j! n_1n_2} f_{2j+k}\left(\frac{xy}{n_1} \right)\right] f_{n_2}\left(\frac{y}{n_2}\right) dy  \\
\back & = & \back
e^{-\delta /2} \sum_{j=0}^\infty   \frac{(\delta /2)^j}{j!} \int_0^\infty \frac{y}{n_1n_2}
f_{2j+k}\left(\frac{xy}{n_1} \right) f_{n_2}\left(\frac{y}{n_2}\right) \!\! dy  \\
\back & = & \back
 e^{-\delta /2}  \sum_{j=0}^\infty \frac{(\delta/2)^j}{ j!}\frac{n_1}{(2j+n_1)} f_{2j+n_1,n_2}\left( \frac{n_1x}{2j+n_1}\right)
\eeqrs
\end{alertblock}
}
\frame{
\begin{alertblock}{}
Here, we used the following result: if $g_j(x) \geq 0$ for all $j=0,1,2,...$ and $x \in {\cal R}$, then \vspace{-1mm}
\[ \int_{-\infty}^\infty \sum_{j=0}^\infty g_j(x) dx = \sum_{j=0}^\infty \int_{-\infty}^\infty g_j(x) dx \vspace{-1mm} \]
which holds even when one of side is $\infty$.
 \vspace{1.5mm}

 To show this, note that $G_n(x) = \sum_{j=0}^n g_j(x)$ is increasing in $n$ for each $x$.
 \vspace{1.5mm}

 By the monotone convergence theorem, \vspace{-2mm}
 \beqrs
  \sum_{j=0}^\infty \int_{-\infty}^\infty g_j(x) dx \back & = & \back
  \lim_{n\rightarrow \infty}  \sum_{j=0}^n \int_{-\infty}^\infty g_j(x) dx
 = \lim_{n\rightarrow \infty}  \int_{-\infty}^\infty \sum_{j=0}^n  g_j(x) dx \\
 \back & = & \back
 \lim_{n\rightarrow \infty}  \int_{-\infty}^\infty G_n(x) dx
 =   \int_{-\infty}^\infty \lim_{n\rightarrow \infty} G_n(x) dx \\
 \back & = & \back
  \int_{-\infty}^\infty \lim_{n\rightarrow \infty} \sum_{j=0}^n g_j(x) dx
 = \int_{-\infty}^\infty  \sum_{j=0}^\infty g_j(x)  dx
 \eeqrs
 Finally, let's calculate the mean and variance of $F$.
 \end{alertblock}}
\frame{\vspace{-2mm}
\begin{alertblock}{}
From the previous calculation, \vspace{-3mm}
\[ E\left(\frac{1}{X_2}\right) = \left\{ \br{ll}
\frac{1}{n_2-2} & \quad n_2 >2 \\
\infty & \quad n_2 \leq 2 \er \right. \vspace{-3mm} \]
 Then, when  $n_2 > 2$, \vspace{-2mm}
 \[ E(F) = E \left(\frac{X_1/n_1}{X_2/n_2}\right)
 =  E \left(\frac{X_1}{n_1}\right) E\left( \frac{n_2}{X_2}\right)
 =  \frac{n_1+\delta }{n_1} \frac{n_2}{n_2-2} = \frac{n_2(n_1+\delta)}{n_1(n_2-2)} \vspace{-3mm} \]
 Also, \vspace{-3mm}
 \[  E\left(\frac{1}{X_2^2} \right)  =
\frac{1}{\Gamma (n_2/2) 2^{n_2/2} }
\int_0^\infty x^{n_2/2-3} e^{-x/2} dx
 = \left\{  \br{ll}
 \frac{1}{(n_2-2)(n_2-4) } & \quad n_2>4 \\
 \infty & \quad n_2 \leq 4 \er \right. \vspace{-2mm} \]
Thus, when $n_2 >4$, \vspace{-3mm}
\beqrs
 \mbox{Var}(F) \back & = & \back
E\left( \frac{X_1^2/n_1^2}{X_2^2/n_2^2}\right) - [E(F)]^2 =
\frac{n_2^2}{n_1^2} E(X_1^2)E\left(\frac{1}{X_2}\right)  -
\left( \frac{n_2(n_1+\delta )}{n_1(n_2-2)} \right)^2 \\
& = & \back
\frac{n_2^2}{n_1^2} \left(
\frac{2n_1+4 \delta + (n_1+\delta )^2}{(n_2-2)(n_2-4)}
- \frac{(n_1+\delta )^2}{(n_2-2)^2} \right) \\
& = & \back
\frac{ 2n_2^2 [ ( n_1+ \delta )^2 + (n_2 -2) (n_1 + 2\delta )]}
{ n_1^2 (n_2-2)^2 (n_2 -4)}
\eeqrs

 \end{alertblock}
}
\end{document}
