\Question{A Meeting of Three}
James, Jon, and Mike each arrive at the movie theater at times uniformly distributed in the interval $(\text{3:00}, \text{3:20})$. Their arrival times are independent. Each person waits 5 minutes after their arrival before heading into the theater. What is the probability they all see each other before going into the theater?
\Question{Exponential Practice}
\begin{Parts}
\Part Let $X_1, X_2 \sim \operatorname{Exponential}(\lambda)$ be independent, $\lambda > 0$.
Calculate the density of $Y := X_1 + X_2$.
[\textit{Hint}: One way to approach this problem would be to compute the CDF of $Y$ and then differentiate the CDF.]
\Part Let $t > 0$.
What is the density of $X_1$, conditioned on $X_1 + X_2 = t$?
[\textit{Hint}: Once again, it may be helpful to consider the CDF $\Pr(X_1 \le x \mid X_1 + X_2 = t)$.
To tackle the conditioning part, try conditioning instead on the event $\{X_1 + X_2 \in [t, t + \varepsilon]\}$, where $\varepsilon > 0$ is small.]
\end{Parts}
\Question{Normal Darts?}
Alex and John are playing a game of darts. Let $(X_a, Y_a)$ and $(X_j, Y_j)$ denote the coordinates of Alex's and John's darts on the board and are distributed in the following way:
\begin{enumerate}
\item $X_a, Y_a \sim \mathbb{N} (0, 1)$ independently
\item $X_j, Y_j$ are distributed uniformly in a circle of radius $3$
\end{enumerate}
The winner of the game is determined by whoever's darts is closer to the center of the board at $(0,0)$. In this question, we will compute the probability that Alex wins the game. We will denote the squared distances of the darts from the center by $r_a = X_a^2 + Y_a^2$ and by $r_j = X_j^2 + Y_j^2$.
\begin{Parts}
\Part What is the distribution of $r_a$? \\
Hint: Consider the joint distribution and the following change of variables formula: Suppose we want to integrate the function $f(x,y)$ over the circle $\left(\sqrt{x^2 + y^2} \leq R\right)$. Then, we have the following change of variables formula:
\begin{equation*}
\int_{\sqrt{x^2 + y^2} \leq R} f(x,y) dx dy = \int_{0}^{R} \int_{0}^{2\pi} f(r\cos \theta, r \sin \theta) r d\theta dr
\end{equation*}
You may find the identity $\sin(\theta)^2 + \cos(\theta)^2 =1$ useful.
\Part What is the distribution of $r_j$? (Hint: Try computing the CDF first)
\Part What is the probability that Alex wins the game?
\end{Parts}
% why_is_it_gaussian.text
\Question{Why Is It Gaussian?}
Let $X$ be a normally distributed random variable with mean $\mu$ and variance $\sigma^2$. Let $Y = aX+b$, where $a$ and $b$ are non-zero real numbers.
Show explicitly that $Y$ is normally distributed with mean $a\mu + b$ and variance $a^2\sigma^2$. The PDF for the Gaussian Distribution is $\frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}$.
\nosolspace{7.5cm}
% (Your proof should be more explicit than what's in the class notes.
% One approach is to start with the cumulative distribution function of $Y$ and use it to derive the probability density function of $Y$.)
\Question{Moments of the Gaussian}
$\E[X^k]$, where $k \in \N$, is called the \textit{$k$th moment} of the distribution.
In this problem, we will calculate the moments of a standard normal distribution.
\begin{Parts}
\Part Prove the identity
\begin{align} \label{eq:gaussian-identity}
\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \exp\p*{ - \frac{tx^2}{2} } dx
&= t^{-1/2}
\end{align}
for $t > 0$. \\
Hint: Consider a normal distribution with variance $\frac{1}{t}$ and mean $0$.
\Part For the rest of the problem, $X$ is a standard normal distribution (with mean $0$ and variance $1$).
Use part a to compute $\E[X^{2k}]$ for $k \in \N$.
[\textit{Hint}: Try differentiating with respect to $t$ $k$ times.]
\Part Compute $\E[X^{2k+1}]$ for $k \in \N$.
\end{Parts}
\Question{Noisy Love}
%Due to the Central Limit Theorem, the Gaussian distribution is often used as a model for noise.
%In this problem, we will see how to perform calculations with Gaussian noise models.
Suppose you have confessed to your love interest on Valentine's Day and you are waiting to hear back.
Your love interest is trying to send you a binary message: ``$0$'' means that your love interest is not interested in you, while ``$1$'' means that your love interest reciprocates your feelings.
Let $X$ be your love interest's message for you.
Your current best guess of $X$ has $\Pr(X = 0) = 0.7$ and $\Pr(X = 1) = 0.3$.
Unfortunately, your love interest sends you the message through a noisy channel, and instead of receiving the message $X$, you receive the message $Y = X + \varepsilon$, where $\varepsilon$ is independent Gaussian noise with mean $0$ and variance $0.49$.
\begin{Parts}
\Part First, you decide upon the following rule: if you observe $Y > 0.5$, then you will assume that your love interest loves you back, whereas if you observe $Y \le 0.5$, then you will assume that your love interest is not interested in you.
What is the probability that you are correct using this rule?
(Express your answer in terms of the CDF of the standard Gaussian distribution $\Phi(z) = \Pr(\mc{N}(0, 1) \le z)$, and then evaluate your answer numerically.)
\Part Suppose you observe $Y = 0.6$.
What is the probability that your love interest loves you back?
[\textit{Hint}: This problem requires conditioning on an event of probability $0$, namely, the event $\{Y = 0.6\}$.
To tackle this problem, think about conditioning on the event $\{Y \in [0.6, 0.6 + \delta]\}$, where $\delta > 0$ is small, so that $f_Y(0.6) \cdot \delta \approx \Pr(Y \in [0.6, 0.6 + \delta])$, and then apply Bayes Rule.]
\Part Suppose you observe $Y = y$.
For what values is it more likely than not that your love interest loves you back?
[\textit{Hint}: As before, instead of considering $\{Y = y\}$, you can consider the event $\{Y \in [y, y + \delta]\}$ for small $\delta > 0$.
So, when is $\Pr(X = 1 \mid Y \in [y, y + \delta]) \ge \Pr(X = 0 \mid Y \in [y, y + \delta])$?]
\Part Your new rule is to assume that your love interest loves you back if (based on the value of $Y$ that you observe) it is more likely than not that your love interest loves you back.
Under this new rule, what is the probability that you are correct?
\end{Parts}