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\centerline{Mathematics 375 -- Probability and Statistics I}
\centerline{Problem Set 5 Solutions}
\centerline{October 20, 2005}
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3.84/  Hypergeometric with $r = 5, N = 10, n = 3$ (think green and
``not green''), so 
$P(Y = 3) = {{5\choose 3}{5\choose 0}\over {10\choose 3}} = {1\over 12}.$

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3.88/  $Y$ is the number of nonlocal firms.
a) $P(Y = 1) = {{2\choose 1}{4\choose 2}\over {6\choose 3}} = {3\over 5}.$
b) $P(Y \ge 1) = {3\over 5} + P(Y = 2) = {3\over 5} + {{2\choose 2}{4\choose 1}\over {6\choose 3}}
={4\over 5}$. c) $P(Y \le 1) = P(Y = 0) + P(Y = 1) = {4\over 5}$.

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3.89/ For a) hypergeometric with $N = 10, n = 3, r = 2$ so table
of values of the pdf is
$$\matrix{ y &   0   &  1    & 2 \cr
       P(Y=y)& 14/30 & 14/30 & 2/30\cr}$$
For b), $N = 10, n = 3, r = 4$, so
 $$\matrix{ y &   0   &  1    & 2   & 3 \cr
       P(Y=y)& 5/30 & 15/30 & 9/30\ & 1/30 \cr}$$

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3.97/ $Y$ is Poisson with $\lambda = 2$.
\item{a)} $P(Y = 4) = {2^4\over 4!} e^{-2} \doteq .090$.
\item{b)} Using the table of Poisson probabilities in text:
$P(Y \ge 4) = 1 - P(Y \le 3) = 1 - .857 = .143$.
\item{c)}  $P(Y < 4) = P(Y \le 3) = .857$.
\item{d)}  $P(Y\ge 4 | Y \ge 2) = .143/.594 = .241$ (using definition of
conditional probability).

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3.101/ Poisson with $\lambda = 4$.  $P(Y \le 4) = .629$ from 
the text's table of Poisson probabilities.

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3.102/  {\it Method 1 -- ``direct approach'':} Let $I$ be the number of cars arriving at entrance I and
$II$ similarly for entrance II.  Let $T = I + II$ be the total number.  We can 
get $T = 3$ in four different ways ($I = 3$ and $II = 0$, or
$I = 2$ and $II = 1$, etc.)  Then 
using independence:
$$\eqalign{
P(T = 3) &= \sum_{k=0}^3 P(I = k) P(II = 3-k)\cr
&= \sum_{k=0}^3 {3^k e^{-3}\over k!}{4^{3-k}e^{-4}\over (3-k)!}\cr
&= e^{-7}\left({9\over 2} + 18 + 24 + {32\over 3}\right)\cr
&\doteq .0521\cr}$$

\noindent
{\it Method 2 -- ``clever'':}  The fact that there are two entrances
is really irrelevant for the total number $T$ of cars entering if the 
numbers from the two entrances are independent.  It should be the same 
as a situation where you have a Poisson distribution 
with $\lambda = 3 + 4 = 7$ cars entering per hour.
Then $P(T = 3) = {7^3 e^{-7}\over 3!} \doteq .0521$.  (Same as above!)

\noindent
{\it Comment:}  In fact, this problem illustrates a general fact
about Poisson random variables.  If you have $X$ Poisson with 
parameter $\lambda$ and $Y$ Poisson with parameter $\mu$ and 
$X,Y$ are independent, then $Z = X+Y$ also has a Poisson distribution
with parameter $\lambda + \mu$.  One proof comes from an argument 
like the ``direct approach'' above:
$$\eqalign{
P(Z = z) &= \sum_{x=0}^z P(X = x)P(Y = z - x)\cr
&= \sum_{x=0}^z P(X = x)P(Y = z - x)\cr
&= \sum_{x=0}^z {\lambda^x e^{-\lambda}\over x!} {\lambda^{z-x} e^{-\lambda}\over (z-x)!}\cr
&= e^{-(\lambda+\mu)} \sum_{x=0}^z {1\over x! (z-x)!} \lambda^x \mu^{z-x}\cr
&= {e^{-(\lambda+\mu)}\over z!} \sum_{x=0}^z {z\choose x} \lambda^x \mu^{z-x}\cr
&={e^{-(\lambda+\mu)}\over z!} (\lambda + \mu)^z\cr }$$
(using the binomial theorem).  This is exactly the 
right probability mass function for a Poisson random variable
with parameter $\lambda + \mu$.  There's another proof via
the moment generating functions too(!)  If $X,Y$ are independent,
and $Z = X+Y$, then 
$$m_Z(t) = E(e^{tX + tY}) = E(e^{tX})E(e^{tY}) = 
e^{\lambda(e^t-1)} e^{\mu(e^t-1)} = e^{(\lambda + \mu)(e^t - 1)}$$
By the uniqueness theorem, $Z$ has a Poisson distribution with 
parameter $\lambda + \mu$.
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3.114/  
\item{a)}  We have 
$$p(y)/p(y-1) = {{\lambda^y\over y!} e^{-\lambda} \over 
{\lambda^{y-1}\over (y-1)!} e^{-\lambda}} = {\lambda \over y}$$
\item{b)}  
Hence $p$ is increasing for $y < \lambda$, and decreasing for $y > \lambda$.
\item{c)}  $p(y)$ is maximized with $y = \lfloor \lambda\rfloor$
(the greatest integer less than or equal to $\lambda$).

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3.115/  By the binomial theorem, 
$$\eqalign{ m(t) = E(e^{tY}) &=\sum_{y=0}^n e^{ty}p^y q^{n-y}\cr
&=\sum_{y=0}^n (e^tp)^y q^{n-y}\cr
&=(pe^t + q)^n.\cr}$$

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3.116/ The idea of this problem is to use the moment 
generating function for a binomial $Y$ to give a different
derivation of the
formulas we know for $E(Y)$ and $V(Y)$.
By the chain rule for derivatives,
$m'(t) = n(pe^t + q)^{n-1}pe^t$.  So $E(Y) = m'(0) = np$.
Then 
$$m''(t) = n(n-1)(pe^t + q)^{n-2}p^2e^{2t} + n(pe^t + q)^{n-1}pe^t$$
so 
$E(Y^2) = m''(0) = n(n-1)p^2 + np$.  Then 
$$V(Y) = E(Y^2) - E(Y)^2 = n(n-1)p^2 + np - n^2p^2 = -np^2 + np = 
np(1 - p) = npq.$$


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3.121/ By the Uniqueness Theorem for the moment generating function,
and the formulas from class and the text for moment generating
functions for standard types of random variables,
\item{a)} Binomial, $n = 5$, $p = 1/3$.
\item{b)} Geometric, $p = 1/2$.
\item{c)} Poisson, $\lambda = 2$.

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3.123/ a) $E(Y) = m'(0) = 1/6 + 2/3 + 3/2 = 7/3$.  b)  $E(Y^2) = m''(0)
= 1/6 + 4/3 + 9/2 = 6$. Then $V(Y) = E(Y^2) - E(Y)^2 = 6 - (7/3)^2 = 5/9$
c) By thinkning about the form of $m_Y(t)$, we see that 
$Y$ is a discrete random variable with $P(Y = 1) = 1/6, P(Y = 2) = 2/6$,
and $P(Y = 3) = 3/6$ (all other probability values zero).   

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3.124/ Since $e^{bt}$ is not random (it's a fixed function of $t$),
$$m_{aY+b}(t) = E(e^{(aY+b)t}) = E(e^{bt} e^{(at)Y}) = 
e^{bt}E(e^{(at)Y}) = e^{bt}m_Y(at).$$
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