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\title{\sc Math 20 -- Homework 7 \framebox{Solutions!}}
\author{due Wednesday, August 16}
\date{}
\begin{document}
\maketitle
\pagestyle{main}
\textbf{Instructions:} This assignment is due at the \emph{beginning} of class. Staple your work together (do not just fold over the corner). Please write the questions in the correct order. If I cannot read your handwriting, you won't receive full credit.\\
%\emph{You may use Wolfram Alpha to compute any necessary sums or integrals. If you have trouble with this, let me know.}\\
\textbf{If you're using facts about distributions to answer the questions, be very clear about which distribution you're using to model that problem and why that distribution is appropriate.}
\begin{enumerate}
\item Suppose that the height, in inches, of a 25-year old man is a normal random variable with parameters $\mu = 71$ and $\sigma^2 = 6.25$. What percentage of $25$-year old men are over $6$ feet $2$ inches tall? What percentage of men over $6$ feet tall are over $6$ feet $5$ inches?
\textbf{Solution:} First we should convert all height to inches. 6 feet is 72 inches, 6 feet 2 inches is 74 inches, and 6 feet 5 inches is 77 inches.
How many standard deviations above the mean is 74 inches? The answer is
\[
\frac{74 - \mu}{\sigma} = \frac{74 - 71}{\sqrt{6.25}} = \frac{3}{2.5} = 1.2.
\]
Let $f(x)$ be the density function for the standard normal distribution. Looking at a normal table tells us that
\[
f(1.2) \approx 0.8849.
\]
Therefore, the percentage of $25$-year old men over 6 feet 2 inches is
\[
1 - f(1.2) \approx 0.1151 = \framebox{\;11.51\%\;}\;.
\]
The second part of the question involves a conditional probability. It asks for
\begin{align*}
P(\text{a man is over 6 feet 5 inches}\;|\;\text{the man is over 6 feet}) &= \frac{P(\text{a man is over 6 feet 5 inches AND over 6 feet})}{P(\text{a man is over 6 feet})}\\
&= \frac{P(\text{a man is over 6 feet 5 inches})}{P(\text{a man is over 6 feet})}\\
&= \frac{P(\text{a man is over 77 inches})}{P(\text{a man is over 72 inches})}
\end{align*}
77 inches is $\frac{77 - 71}{2.5} = 2.4$ standard deviations above the mean, while 72 inches is $\frac{72-71}{2.5} = 0.4$ standard deviations above the mean. The conditional probability is thus
\[
\frac{P(\text{a man is over 77 inches})}{P(\text{a man is over 72 inches})} = \frac{1 - f(2.4)}{1 - f(0.4)} = \frac{1 - 0.9918}{1 - 0.6554} = \frac{0.0082}{0.3446} \approx 0.0238 = \framebox{\;2.38\%\;}\;.
\]
\hrulefill
\item Suppose that $X$ and $Y$ are two independent random variables with density functions
\[
f_X(x) = \left\{\begin{array}{ll}
\frac{3}{2}(x+1)^2,&\text{ if }-1 \leq x \leq 0\\[5pt]
\frac{3}{2}(x-1)^2,&\text{ if }0 \leq x \leq 1\\[5pt]
0,&\text{otherwise}
\end{array}\right.
\]
and
\[
f_Y(x) = \left\{\begin{array}{ll}
\frac{1}{2},&\text{ if } 0 \leq x \leq 2\\[5pt]
0,&\text{otherwise}
\end{array}\right..
\]
Find the density function for $Z = X + Y$.
\textbf{Solution:} By our formula from class, the density of $Z = X+ Y$ is
\[
f_Z(z) = \int_{-\infty}^\infty f_X(z-y)f_Y(y)dy.
\]
The second function in the integrand, $f_Y$ is only nonzero when $y$ is between $0$ and $2$. Thus,
\[
f_Z(z) = \int_{-\infty}^\infty f_X(z-y)f_Y(y)dy = \int_0^2 f_X(z-y)f_Y(y)dy.
\]
In fact, in this range, it is equal to the constant $1/2$, which makes our lives a little easier.
\[
f_Z(z) =\int_0^2 f_X(z-y)f_Y(y)dy = \frac{1}{2} \int_0^2 f_X(z-y)dy.
\]
The next question is: where in the interval from $y=0$ to $y=2$ is $f_X(z-y)$ nonzero? In other words, where in the interval from $y=0$ to $y=2$ is $-1 \leq z-y \leq 1$? Of course, it will depend on the value of $z$.
Manipulating this inequality, we find
\begin{align*}
-1 \leq z-y \leq 1 &\Longrightarrow -1 - z \leq -y \leq 1 - z\\
&\Longrightarrow \phantom{-}1 + z \geq \phantom{-}y \geq z - 1\\
&\Longrightarrow \phantom{-}z -1 \leq \phantom{-}y \leq z + 1.
\end{align*}
(Don't forget that when you multiply or divide by a negative number, you have to flip the inequalities.)
Now we have the restrictions: for a fixed $z$ value, we need to integrate over whatever $y$ values satisfy both
\[
0 \leq y \leq 2 \qquad \text{and} \qquad z-1 \leq y \leq z + 1.
\]
Algebraically, this means the region of integration is
\[
\max(0, z-1) \leq y \leq \min(2, z+1),
\]
but we're far better off drawing a picture.
\begin{center}
\begin{tikzpicture}[scale=0.5]
\draw[help lines] (-2.5,-2.5) grid (3.5, 3.5);
\draw[<->, thick, gray] (-2.5,0) -- (3.5,0);
\draw[<->, thick, gray] (0,-2.5) -- (0,3.5);
\fill[gray, opacity=0.3] (-1,0) -- (1,0) -- (3,2) -- (1,2) --cycle;
\draw[dashed, thick] (-3,0) -- (4,0) node[right] {$y=0$};
\draw[dashed, thick] (-3,2) -- (4,2) node[right] {$y=2$};
\draw[dashed, thick] (-1.5, -2.5) -- (3.5, 2.5) node[above=6pt, right] {$y=z-1$};
\draw[dashed, thick] (-2.5, -1.5) -- (2.5, 3.5) node[above] {$y=z+1$};
\draw[dashed, gray, thick] (3.5, 3.5) -- (-2.5, -2.5) node[left] {$z-y=0$};
\end{tikzpicture}
\end{center}
This gives us the region of integration for a fixed $z$ value, but the situation become more complicated. For example, suppose we want to find the region of integration at $z = 1/2$. The picture shows us that the valid $y$ values range from $y=0$ to $y = 3/2$. However, $f_X(x)$ is piecewise, and $f_X(z-y)$ takes one value when $z-y$ is greater than $0$ (i.e., when $y$ is between $0$ and $1/2$) and another when $z-y$ is greater than $0$ (i.e., when $y$ is between $1/2$ and $3/2$). In the picture above the shaded region is divided into two parts by the line $z-y=0$; $z-y > 0$ to the right of the line and $z - y < 0$ to the left.
In the end, we need to split the region into \emph{four} separate $z$-intervals, so that the bounds of integration are the same on each region: $[-1,0]$, $[0,1]$, $[1,2]$, and $[2,3]$.
On the interval where $z \in [-1,0]$:
\begin{align*}
f_Z(z) &= \frac{1}{2} \int_0^{z+1} f_X(z-y)dy\\
&= \frac{1}{2} \int_0^{z+1} \frac{3}{2}(z-y+1)^2dy\\
&= \frac{3}{4} \int_0^{z+1}((z+1)-y)^2dy\\
&= \frac{3}{4} \left[-\frac{1}{3}((z+1)-y)^3\right]_0^{z+1}\\
&= \frac{1}{4} \left(0 + (z+1)^3\right)\\
&= \frac{(z+1)^3}{4}.
\end{align*}
On the interval where $z \in [0,1]$:
\begin{align*}
f_Z(z) &= \frac{1}{2} \int_0^{z+1} f_X(z-y)dy\\
&= \frac{1}{2} \int_0^{z} f_X(z-y)dy + \frac{1}{2} \int_z^{z+1} f_X(z-y)dy\\
&= \frac{1}{2} \int_0^{z} \frac{3}{2}(z-y-1)^2dy + \frac{1}{2} \int_z^{z+1} \frac{3}{2}(z-y+1)^2dy\\
&= \frac{3}{4} \left[ -\frac{1}{3}((z-1)-y)^3\right]_0^z + \frac{3}{4} \left[ -\frac{1}{3}((z+1)-y)^3\right]_z^{z+1}\\
&= \frac{3}{4} \left(\frac{1}{3} + \frac{1}{3} (z-1)^3\right) + \frac{3}{4} \left(0 + \frac{1}{3}\right) \\
&= \frac{2 + (z-1)^3}{4}
\end{align*}
On the interval where $z \in [1,2]$ (largely similar to the last case, only bounds are different):
\begin{align*}
f_Z(z) &= \frac{1}{2} \int_{z-1}^{2} f_X(z-y)dy\\
&= \frac{1}{2} \int_{z-1}^{z} f_X(z-y)dy + \frac{1}{2} \int_z^{2} f_X(z-y)dy\\
&= \frac{1}{2} \int_{z-1}^{z} \frac{3}{2}(z-y-1)^2dy + \frac{1}{2} \int_z^{2} \frac{3}{2}(z-y+1)^2dy\\
&= \frac{3}{4} \left[ -\frac{1}{3}((z-1)-y)^3\right]_{z-1}^z + \frac{3}{4} \left[ -\frac{1}{3}((z+1)-y)^3\right]_z^{2}\\
&= \frac{3}{4} \left(\frac{1}{3} + 0\right) + \frac{3}{4} \left(-\frac{1}{3}(z-1)^3 + \frac{1}{3}\right) \\
&= \frac{2 - (z-1)^3}{4}
\end{align*}
On the interval where $z \in [2,3]$:
\begin{align*}
f_Z(z) &= \frac{1}{2} \int_{z-1}^{2} f_X(z-y)dy\\
&= \frac{1}{2} \int_{z-1}^{2} \frac{3}{2}(z-y-1)^2dy\\
&= \frac{3}{4} \int_{z-1}^{2}((z-1)-y)^2dy\\
&= \frac{3}{4} \left[-\frac{1}{3}((z-1)-y)^3\right]_{z-1}^{2}\\
&= \frac{3}{4} \left(-\frac{1}{3}(z-3)^3 + 0\right)\\
&= -\frac{(z-3)^3}{4}.
\end{align*}
Finally, we're done, and we've found that
\[
\framebox{$\ds f_Z(z) = \left\{\begin{array}{cl}
0, & \phantom{-1 \leq }\;z < -1\\[4pt]
\ds\frac{(z+1)^3}{4}, & -1 \leq z \leq 0\\[8pt]
\ds\frac{2 + (z-1)^3}{4}, & \phantom{-}0 \leq z \leq 1\\[8pt]
\ds\frac{2 - (z-1)^3}{4}, & \phantom{-}1 \leq z \leq 2\\[8pt]
\ds-\frac{(z-3)^3}{4}, & \phantom{-}1 \leq z \leq 3\\[8pt]
0, & \phantom{-}3 < z
\end{array}\right. $}\;.
\]
I've included a graph of this function below.
\begin{center}
\begin{tikzpicture}[scale=1, yscale=2]
\draw[help lines] (-1.5,-0.5) grid (3.5, 1.2);
\draw[help lines] (-1.5, 0.5) -- (3.5, 0.5) node[right, gray] {$0.5$};
\node[right, gray] at (3.5,1) {$1.0$};
\draw[<->, thick, gray] (-1.5,0) -- (3.5,0);
\draw[<->, thick, gray] (0,-0.5) -- (0,1.2);
\node[below, gray] at (1,-0.5) {$1.0$};
\node[below, gray] at (2,-0.5) {$2.0$};
\node[below, gray] at (3,-0.5) {$3.0$};
\draw[domain=-1.5:-1, <-, color={rgb:red,1;green,2;blue,5}, thick, smooth, samples=100, variable=\x] plot ({\x},{0});
\draw[domain=-1:0, color={rgb:red,1;green,2;blue,5}, thick, smooth, samples=100, variable=\x] plot ({\x},{(\x+1)^3/4});
\draw[domain=0:1, color={rgb:red,1;green,2;blue,5}, thick, smooth, samples=100, variable=\x] plot ({\x},{(2+(\x-1)^3)/4});
\draw[domain=1:2, color={rgb:red,1;green,2;blue,5}, thick, smooth, samples=100, variable=\x] plot ({\x},{(2-(\x-1)^3)/4});
\draw[domain=2:3, color={rgb:red,1;green,2;blue,5}, thick, smooth, samples=100, variable=\x] plot ({\x},{(3-\x)^3/4});
\draw[domain=3:3.5, ->, color={rgb:red,1;green,2;blue,5}, thick, smooth, samples=100, variable=\x] plot ({\x},{0});
\end{tikzpicture}
\end{center}
\hrulefill
\item In this exercise, you will use the notion of convolution and the principle of mathematical induction (a proof technique) to prove that if $X_1, \ldots, X_n$ are independent and identically distributed exponential random variables with rate $\lambda$, then the probability density function of $S_n = X_1 + \cdots + X_n$ (for $n \geq 1$) is
\[
f_{S_n}(x) = \frac{\lambda e^{-\lambda x} (\lambda x)^{n-1}}{(n-1)!}.
\]
for $x \geq 0$, and $f_{S_n}(x) = 0$ otherwise.
Note that we proved the $n=2$ case explicitly in class.
\textbf{Mathematical Induction:} Suppose you have a statement that you want to prove is true for all positive integers $n = 1,2,3,\ldots$. One way to do this is induction: first you show it's true for $n=1$ (this is called the base case), then you show that \emph{if} it's true for some number $k$ \emph{then} it must be true for the next number $k+1$ (this is called the induction step). If you show these two things, then the statement must be true for all $n$. (Why? Because you showed it's true for $1$ in the base case. Then the induction step shows that if it's true for $1$ then it must be true for $2$. If it's true for $2$, then it must be true for $3$, etc.)
\begin{enumerate}
\item We want to prove that $f_{S_n}(x)$ has the form above for all $n \geq 1$. That means that $n=1$ is our base case. Why do we already know this is true?
\item Next prove the induction step. In other words, \emph{assume} that for some integer $k \geq 1$ we already know that
\[
f_{S_k}(x) = \frac{\lambda e^{-\lambda x} (\lambda x)^{k-1}}{(k-1)!}.
\]
(for $x \geq 0$, and $f_{S_k}(x) = 0$ otherwise) and prove that
\[
f_{S_{k+1}}(x) = \frac{\lambda e^{-\lambda x} (\lambda x)^{k}}{k!}.
\]
(for $x \geq 0$, and $f_{S_{k+1}}(x) = 0$ otherwise).
\emph{Hint:} $S_k = S_{k-1} + X_k$.\\
\end{enumerate}
Parts (a) and (b) together prove the statement by the principle of mathematical induction.\\
\textbf{Solution:} We already know that part (a) is true because $S_1 = X_1$. In other words, $S_1$ is just the exponential random variable with rate $\lambda$. We've stated in class numerous times that
\[
f_{X_1}(x) = f_{S_1}(x) = \lambda e^{-\lambda x}.
\]
For part (b), the induction step, we now assume that
\[
f_{S_k}(x) = \frac{\lambda e^{-\lambda x} (\lambda x)^{k-1}}{(k-1)!}
\]
(for $x \geq 0$, and $f_{S_k}(x) = 0$ otherwise) for some positive integer $k$, and we will show that this implies that
\[
f_{S_{k+1}}(x) = \frac{\lambda e^{-\lambda x} (\lambda x)^{k}}{k!}.
\]
(for $x \geq 0$, and $f_{S_{k+1}}(x) = 0$ otherwise).
As the hint points out,
\[
S_{k+1} = X_1 + X_2 + \cdots + X_k + X_{k+1} = (X_1 + X_2 + \cdots + X_k) + X_{k+1} = S_k + X_{k+1}.
\]
Therefore, by our formula from class,
\[
f_{S_{k+1}}(z) = \int_{-\infty}^{\infty} f_{S_k}(z-y)f_X(y)dy.
\]
We showed in class that if you're computing the convolution of two random variables who take nonzero values only when $x \geq 0$, then the bounds of the resulting integral for a fixed $z$-value are $y=0$ to $y=z$. Hence, for $z \geq 0$,
\begin{align*}
f_{S_{k+1}}(z) &= \int_{0}^{z} f_{S_k}(z-y)f_X(y)dy\\
&= \int_0^z \frac{\lambda e^{-\lambda (z-y)}(\lambda(z-y))^{k-1}}{(k-1)!}\cdot \left(\lambda e^{-\lambda y}\right)dy\\
&= \frac{\lambda^{k+1}}{(k-1)!} \int_0^z e^{-\lambda z + \lambda y - \lambda y}(z-y)^{k-1}dy\\
&= \frac{\lambda^{k+1} e^{-\lambda z}}{(k-1)!} \int_0^z (z-y)^{k-1}dy\\
&= \frac{\lambda^{k+1} e^{-\lambda z}}{(k-1)!} \left[ -\frac{1}{k}(z-y)^k\right]_0^z\\
&= \frac{\lambda^{k+1} e^{-\lambda z}}{(k-1)!} \left( 0 + \frac{1}{k}z^k \right)\\
&= \frac{\lambda^{k+1} z^k e^{-\lambda z}}{k!}\\
&= \framebox{$\ds\frac{\lambda e^{-\lambda z} (\lambda z)^k}{k!}$}\;.
\end{align*}
This is what we needed to prove for the induction step. Hence, by the principle of mathematical induction, we have proved that the sum of $n$ independent and identically distributed exponential random variables with rate $\lambda$ is
\[
f_{S_n}(x) = \frac{\lambda e^{-\lambda x} (\lambda x)^{n-1}}{(n-1)!}.
\]
\hrulefill
\item What do Chebyshev's Inequality and the Law of Large Numbers say about the probability of getting at least 75 heads when flipping a fair coin 100 times? \emph{Hint:} Improve your bound by using the fact that the binomial distribution is symmetric.
\textbf{Solution:} Let $X_N$ be the random variable for the number of heads when flipping a fair coin $N$ times. Then, $X_N$ has a binomial distribution with $p = 1/2$ and $n = N$. The expected value of this distribution is $\E[X] = N/2$.
The law of large numbers says that for any $\epsilon > 0$, the probability that $X_N$ strays too far from $N/2$ goes to zero. More accurately, it says
\[
\lim_{n \to \infty} P\left(\left|\frac{X_N}{N} - \frac{N}{2}\right| > \epsilon\right) = 0,
\]
for any $\epsilon > 0$. This doesn't give us any exact numbers for the $N=100$ case. For that, we turn to Chebyshev's inequality, which says that for any $\epsilon > 0$,
\[
P\left(\left|X_{100} - 50\right| \geq \epsilon\right) \leq \frac{\Var(X_{100})}{\epsilon^2} = \frac{100}{4\epsilon^2} = \frac{25}{\epsilon^2}.
\]
Setting $\epsilon = 25$ tells us that
\[
\framebox{$P\left(X_{100} \leq 25 \text{ or } X_{100} \geq 75\right) \leq \ds\frac{1}{25} = 4\%$}\;.
\]
We can improve this bound using the fact that the binomial distribution is symmetric. So, if there is a 4\% chance that $X$ is more than $25$ below the mean or more than $25$ above the mean, then there is a $4/2 = 2\%$ chance that it's more than $25$ above the mean. Thus, our estimate is \framebox{$2\%$}\;.
\hrulefill
% \item Let $X$ be any random variable which takes on values $0,1,\ldots, n$ and has $\E[X] = \Var(X) = 1$. Show that, for any positive integer $k$,
% \[
% P(X \geq k + 1) \leq \frac{1}{k^2}.
% \]
\item Each student's score on a particular calculus final is a random variable with values in the range $[0,100]$, mean $70$, and variance $25$.
\begin{enumerate}
\item Find the best lower bound you can (using only the tools we've learned), for the probability that a particular student's score will fall between $65$ and $75$.
\item If $100$ students take the final, find a lower bound for the probability that the class mean will fall between $65$ and $75$.
\end{enumerate}
\textbf{Solution:} We don't know the full distribution of each student's score, just the expected value and variance. We can use Chebyshev's Theorem to find a bound for the probability that a particular student's score, $X$, doesn't stray more then $5$ points from the mean:
\[
P(|X - 70| \geq 5) \leq \frac{\Var(X)}{5^2} = \frac{25}{25} = 1.
\]
Hence, all we can say about the probability that the students score is between $65$ and $75$ is that
\[
\framebox{$P(65 \leq X \leq 75) = 1 - P(|X - 70| \geq 5) \geq 1 - 1 = 0$}\;,
\]
which is, of course, useless information.
Now let us consider the mean of the class scores:
\[
A_{100} = \frac{S_{100}}{100} = \frac{X_1 + X_2 + \cdots + X_{100}}{100},
\]
where the $X_i$ are independent and identically distributed and represent the score of the $i$th student.
We can calculate that $\E[A_{100}] = 50$ and $\Var(A_{100}) = 25/100 = 1/4$. Now, Chebyshev's Theorem tells us that
\[
P(|A_{100}-70| \geq 5) \leq \frac{1/4}{25} = \frac{1}{100} = 1\%.
\]
Therefore, the probability that the class mean falls between $65$ and $75$ is bounded by
\[
\framebox{$P(65 \leq A_{100} \leq 75) = 1 - P(|A_{100}-70| \geq 5) \geq 1 - 0.01 = 99\%$}\;.
\]
So, there is \emph{at least} a 99\% chance of this happening. For some distributions with the same mean and variance, the true probability might be exactly 99\%, and for others it might be higher.
\hrulefill
\item A share of common stock in the Pilsdorff beer company has price $Y_n$ on the $n$th business day of the year. ($Y_n$ is a random variable.) Finn observes that the price change $X_n = Y_{n+1} - Y_n$ appears to be a random variable with mean $\mu = 0$ and variance $\sigma^2 = 1/4$. If $Y_1 = 30$, find a lower bound for the following probabilities, under the assumption that the $X_n$'s are mutually independent.
\begin{enumerate}
\item $P(25 \leq Y_2 \leq 35)$
\item $P(25 \leq Y_{11} \leq 35)$
\item $P(25 \leq Y_{101} \leq 35)$
\end{enumerate}
\textbf{Solution to (a):} We need to estimate the probability that after one day's worth of price change, the stock has either gained more than \$5 in value or lost more then \$5 in value. We will again use Chebyshev's inequality.
The value on day 2 is equal to the value on day 1 plus the change in price from day 1 to day 2. Mathematically,
\[
Y_2 = Y_1 + X_1,
\]
and since $Y_1 = 30$ with probability $1$, it suffices to find a lower bound for the probability
\[
P(|X_1| \leq 5).
\]
Keeping in mind that $\E[X_1] = 0$, applying Chebyshev's inequality shows that
\[
P(|X_1 - 0| \geq 5) \leq \frac{\Var(X)}{25} = \frac{1}{100},
\]
and hence
\[
\framebox{$\ds P(25 \leq Y_2 \leq 35) = P(-5 \leq X_1 \leq 5) \geq 1 - \frac{1}{100} = \frac{99}{100} = 99\%$}\;.
\]
\textbf{Solution to (b):} The value on day 11 is equal to the value on day 1 plus ten single-day price changes. Mathematically,
\[
Y_{11} = Y_1 + X_1 + X_2 + \cdots + X_{10}.
\]
Define $S_{10} = X_1 + X_2 + \cdots + X_{10}$ and note that $\E[S_{10}] = 0$ and $\Var(S_{10}) = 2.5$. Proceeding as before,
\[
P(|S_{10} - 0| \geq 5) \leq \frac{\Var(X)}{25} = 0.1 = 10\%.
\]
Therefore,
\[
\framebox{$\ds P(25 \leq Y_{11} \leq 35) = P(-5 \leq S_{10} \leq 5) \geq 1 - 0.1 = 0.9 = 90\%$}\;.
\]
\textbf{Solution to (c):} The value on day 101 is equal to the value on day 1 plus one hundred single-day price changes. Mathematically,
\[
Y_{101} = Y_1 + X_1 + X_2 + \cdots + X_{100}.
\]
Define $S_{100} = X_1 + X_2 + \cdots + X_{100}$ and note that $\E[S_{100}] = 0$ and $\Var(S_{100}) = 25$. Proceeding as before,
\[
P(|S_{100} - 0| \geq 5) \leq \frac{\Var(X)}{25} = 1 = 100\%.
\]
Therefore,
\[
\framebox{$\ds P(25 \leq Y_{101} \leq 35) = P(-5 \leq S_{100} \leq 5) \geq 1 - 1 = 0\%$}\;.
\]
In this case, the bound is useless.
\hrulefill
\end{enumerate}
\end{document}