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\centerline{Mathematics 375 -- Probability and Statistics 1}
\centerline{Problem Set 8, {\it due:}  Thursday, 11/12}
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\noindent
From Wackerly, Mendenhall, and Scheaffer: 4.136, 4.139, 4.140, 4.147, 
4.149, 4.150, 5.4, 5.6, 5.7, 5.8, 5.11, 5.14, 5.15.
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{\it Additional Problem}
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\noindent
A) 
As we proceed into Chapter 5, we will need to make increasingly
heavy use of material on multiple integrals from MATH 241 (Multivariable
Calculus).  As a ``warm-up'' for this, and to fill in a proof
for a fact we just stated in class, in this problem, we will
develop a proof of the formula
$$B(p,q) = {\Gamma(p)\Gamma(q)\over \Gamma(p+q)}$$
that we used in working with Beta-distributed random variables.
(Thanks to Professor Anderson for suggesting this argument to 
derive the formula!!)

The idea is similar to what we did in class to derive the formula
$$\int_{-\infty}^\infty e^{-y^2/2} dy = \sqrt{2\pi}.$$
We will take two one-variable integrals, multiply them, 
treat as a double integral, do coordinate transformations, and 
derive the desired result.
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\noindent
The starting point will be the product: 
$$\Gamma(p)\Gamma(q) = \int_0^\infty s^{p-1} e^{-s}\ ds \cdot
                       \int_0^\infty t^{q-1} e^{-t}\ dt\leqno(1)$$
\item{1)}  Show that the product on the right side in (1) is the same
as the double integral
$$\int_0^\infty\int_0^\infty s^{p-1} t^{q-1} e^{-(s+t)}\ dt\ ds
 = \int\int_R s^{p-1} t^{q-1} e^{-(s+t)}\ dA$$
(where the region of integration $R$ is the first quadrant in the $(s,t)$-plane).
\item{2)}  To work with this, note that we can transform the integral
by introducing a new coordinate system defined by $u = s+t; s = s$.
This is a linear mapping from the $(s,t)$ plane to the $(u,s)$
plane defined by the matrix equation
$$\pmatrix{u\cr s\cr} = \pmatrix{1&1\cr 1&0\cr} \pmatrix{s \cr t\cr}$$
The matrix has determinant of absolute value 1, hence defines an 
area-preserving mapping.
The area element in the new coordinate system in this case
is just $dA = ds\ du$.  Show that this change of coordinates leads to 
a new equation
$$\Gamma(p)\Gamma(q) = \int\int_S s^{p-1} (u-s)^{q-1} e^{-u}\ dA\leqno(2)$$
where $S$ is the image of $R$ under the change of coordinate mapping.
\item{3)}  Show that (2) can also be written as the following iterated
integral:
$$\Gamma(p)\Gamma(q) = \int_0^\infty\int_0^u s^{p-1} (u-s)^{q-1} e^{-u}\ ds\ du\leqno(3)$$
\item{4)}  Now, we will do a second change of coordinates, letting
$\lambda = s/u; u = u$.  This coordinate change is {\it not} area-preserving.  
The area element transforms to 
$$ds\ du = u\ d\lambda\ du.$$
Assuming this (we will discuss the general change of variables theorem
for multiple integrals in class if you have not seen it before),
show that (3) becomes
$$\Gamma(p)\Gamma(q) = 
\int_0^\infty\int_0^1 \lambda^{p-1}(1-\lambda)^{q-1} u^{p+q-1} e^{-u}\ d\lambda\ du\leqno(4)$$
(Pay careful attention to the limits of integration on the ``inside'' integral.
Where do they come from?)
\item{5)} Deduce the desired formula 
$$B(p,q) = {\Gamma(p)\Gamma(q)\over \Gamma(p+q)}$$
from (4).  Very Cool!!!!
\bye
