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\centerline{Mathematics 375 -- Probability and Statistics I}
\centerline{Solutions for Problem Set 7}
\centerline{October 28, 2005}
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Text Problems:
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4.73/ From the given information, then pdf for $Y$ is
$f(y) = \cases{{1\over 44} e^{-y/44} & if $y > 0$\cr 0 & otherwise\cr}$.
a)  $P(Y < 31) = \int_0^{31} {1\over 44} e^{-y/44}\ dy = 1 - e^{-31/44} \doteq .506$.
b)  From the general formula for exponentially-distributed $Y$,
$V(Y) = \beta^2 = 44^2 = 1936$.

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4.75/ We have by the definition of the discrete random variable $X$:
$$\eqalign{P(X=k) &= \int_{k-1}^k {1\over \beta} e^{-y/\beta}\ dy\cr
&= -e^{-y\over \beta}|_{k-1}^k\cr
&= -e^{-k\over \beta} + e^{-(k-1)\over \beta}\cr
&= (e^{-(k-1)\over \beta})(1 - e^{-1\over \beta})\cr
&= (e^{-1\over \beta})^{k-1} (1 - e^{-1\over \beta})\cr}$$
This is the form ($q^{k-1}p$) for the probability  function 
for a {\it geometric} random variable with $p = 1 - e^{-1\over \beta}$.

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4.77/ Exponential distribution with mean $\beta = 4$, 
so $\int_4^\infty e^{-y/4}/4 dy = e^{-1} \doteq .3679$. 

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4.78/ We want to determine $y_0$ so that $P(Y > y_0) = .05$.
This says
$$.05 = P(Y > y_0) = \int_{y_0}^\infty {1\over 4}e^{-y/4}\ dy =
e^{-y_0/4}.$$
We solve $.05 = e^{-y_0/4}$ for $y_0$ by taking natural log
of both sides:  $y_0 = -4\ln(.05) \doteq 11.98$.

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4.86/ a) Using the general formulas for a $\Gamma$-distribution,
$\mu = E(Y) = \alpha\beta = 1000\cdot 20 = 20000$ (units \$ ) and 
$V(Y) = \alpha\beta^2 =400000$ (units $\$^2$).
b)  From part a, $\sigma = \sqrt{V(Y)} \doteq 632.5$.
This means that $40000 \doteq \mu + 31\sigma$ so 
it is very unlikely that any income is that large.  
(Another way to answer this would be to compute the integral
$\int_{40000}^\infty f(y)\ dy$ for the gamma pdf.  This
would come out extremely small!)

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4.87/  First, let's derive the general formula for $E(Y^k)$ when 
$Y$ has a Gamma-distribution with parameters $\alpha, \beta$.  We
apply the substitution $u = {y\over \beta}$, so $dy = \beta\ du$,
then apply the functional equation $\Gamma(x+1) = x\Gamma(x)$ repeatedly:
$$\eqalign{
E(Y^k) &= {1\over \Gamma(\alpha) \beta^\alpha}\int_0^\infty y^k \cdot y^{\alpha-1} e^{-y\over \beta}\ dy\cr
&={1\over \Gamma(\alpha) \beta^\alpha}\int_0^\infty y^{\alpha+k-1} e^{-y\over \beta}\ dy\cr
&={1\over \Gamma(\alpha) \beta^\alpha}\cdot \beta^{\alpha+k} \int_0^\infty u^{\alpha+k-1} e^{-u}\ du\cr
&={1\over \Gamma(\alpha) \beta^\alpha}\cdot \beta^{\alpha+k} \cdot \Gamma(\alpha+k)\cr
&=\beta^k (\alpha+k-1)(\alpha+k-2)\cdots (\alpha+1)(\alpha)\cr}$$
In our problem, with $\alpha=3, \beta =2$, we get
$$\eqalign{E(Y) &= \alpha\beta = 6\cr
         E(Y^2) &= (\alpha+1)(\alpha)\beta^2 = 48\cr
         E(Y^3) &= (\alpha+2)(\alpha+1)\alpha\beta^3 = 480\cr
         E(Y^4) &= (\alpha+3)(\alpha+2)(\alpha+1)\alpha\beta^4=5760\cr}$$
Then
$$E(L) = 30 E(Y) + 2E(Y^2) = 30\cdot 6 + 2\cdot 48 = 276$$
and
$$E(L^2) = 900 E(Y^2) + 120E(Y^3) + 4E(Y^4) = 900\cdot 48 + 120\cdot 480 + 4\cdot
5760 = 123840.$$
Hence
$$V(C) = E(L^2) - (E(L))^2 = 123840 - 276^2 = 47664.$$

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4.92/ $P(Y > .40) = \int_{.40}^1 12y^2(1-y)\ dy = \left(4y^3 - 3y^4\right|_{.40}^1 
\doteq .8208$.


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4.93/  Being ``creatively lazy,'' 
we do not need to grind through the ``from-scratch'' 
computation of the expected value and variance because
we recognize that $Y$ has a Beta distribution 
with $\alpha = 3$ and $\beta = 2$ from the formula
for the pdf.  This means we can use the general formulas
to obtain $E(Y) = \alpha/(\alpha + \beta) = 3/5$ and
$V(Y) = \alpha\beta/((\alpha + \beta)^2(\alpha + \beta + 1))
= 6/(25\cdot 6) = 1/25$.

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{\it Additional Problem}
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A)  1) We must show that when $\alpha$ is a positive integer, 
$${1\over \Gamma(\alpha)}\int_\lambda^\infty y^{\alpha-1} e^{-y}\ dy = 
\sum_{n=0}^{\alpha-1} {\lambda^n\over n!}e^{-\lambda}.\leqno(1)$$
We will use induction on $\alpha$.  The base case is $\alpha = 1$.  Then
the left of (1) is
$$1\cdot\int_\lambda^\infty e^{-y}\ dy = e^{-\lambda}.$$
With $\alpha = 1$, the sum on the right of (1) is:
$$\sum_{n=0}^0 {\lambda^n\over n!}e^{-\lambda} = e^{-\lambda}$$
So the base case is established.  

Now for the induction step, assume that (1) is true for $\alpha = k$.
In other words, our induction hypothesis is:
$${1\over \Gamma(k)}\int_\lambda^\infty y^{k-1} e^{-y}\ dy = 
\sum_{n=0}^{k-1} {\lambda^n\over n!}e^{-\lambda}.\leqno(2)$$
Consider the integral on the left of (1) for $\alpha = k+1$:
$${1\over \Gamma(k+1)} \int_\lambda^\infty y^k e^{-y}\ dy$$
Apply integration by parts with $u = y^k$, and $dv = e^{-y}\ dy$.  We
get:
$$\eqalign{
{1\over \Gamma(k+1)} \int_\lambda^\infty y^k e^{-y}\ dy
&= {1\over \Gamma(k+1)}\left(-y^k e^{-y}|_\lambda^\infty 
+ k\int_\lambda^\infty y^{k-1} e^{-y}\ dy\right) \cr
&= {\lambda^k\over k!} e^{-\lambda} 
+ {1\over \Gamma(k)} \int_\lambda^\infty y^{k-1} e^{-y}\ dy\cr}$$
Now we use the induction hypothesis (2) on the remaining integral.
We get that this is equal to 
$${\lambda^k\over k!} e^{-\lambda} + \sum_{n=0}^{k-1}{\lambda^n\over n!}e^{-\lambda}
=\sum_{n=0}^k {\lambda^n\over n!}e^{-\lambda}$$
Hence our formula also works with $\alpha = k+1$, and by induction
we conclude that it works for all positive integers $\alpha$.

2) Using the Poisson table in the text with $\lambda = 5$, $\alpha = 3$
$$\sum_{n=0}^2 {5^n\over n!}e^{-5} = .125$$

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