University of Washington
Department of Mathematics
Homework 6
MATH 394: Probability I
Instructor: Arman Jahangiri
Submission: Single PDF on Gradescope
Deadline: 10:00 PM Pacific Time on the listed due date
This homework is worth 50 points.
Across the quarter, there are eight homework assignments, worth 400 points total. Homework assignments together account for 40% of the final course grade. Further course policies can be seen in the MATH 394 syllabus.
Please:
Each student is allotted six late days for the quarter. A late day extends a homework deadline by up to 24 hours without penalty. For example, submitting an assignment anytime between 10:01 PM on the due date and 10:00 PM the following day counts as one late day.
The following rules apply:
Once all late days have been exhausted, additional late submissions will incur a penalty of 10% per day, up to a maximum deduction of 50%. Assignments submitted more than five days late, or after solutions have been released, will not be accepted.
If a serious technical issue prevents a timely Gradescope submission, students may temporarily submit their assignment by email to the instructor at armanjg@uw.edu. In such cases:
Email submissions are intended only for genuine technical emergencies and should not be used as a substitute for timely Gradescope submission.
Problem 1.
Source: Blitzstein and Hwang, Introduction to Probability, Chapter 5, Problem 14.
Let \(U_1,\dots ,U_n\) be i.i.d. \(\operatorname {Unif}(0,1)\), and \[ X=\max (U_1,\dots ,U_n). \] What is the PDF of \(X\)? What is \(E(X)\)?
Hint: Find the CDF of \(X\) first, by translating the event \(X\le x\) into an event involving \(U_1,\dots ,U_n\).
Problem 2.
Source: Blitzstein and Hwang, Introduction to Probability, Chapter 5, Problem 36.
A post office has 2 clerks. Alice enters the post office while 2 other customers, Bob and Claire, are being served by the 2 clerks. She is next in line. Assume that the time a clerk spends serving a customer has an \(\operatorname {Exp}(\lambda )\) distribution.
Hint: No integrals are needed.
Problem 3.
Source: Blitzstein and Hwang, Introduction to Probability, Chapter 5, Problem 43.
The Exponential is the analog of the Geometric in continuous time. This problem explores the connection between Exponential and Geometric in more detail, building up on what was taught in class, asking what happens to a Geometric in a limit where the Bernoulli trials are performed faster and faster but with smaller and smaller success probabilities.
Suppose that Bernoulli trials are being performed in continuous time; rather than only thinking about first trial, second trial, etc., imagine that the trials take place at points on a timeline. Assume that the trials are at regularly spaced times \[ 0,\Delta t,2\Delta t,\dots , \] where \(\Delta t\) is a small positive number. Let the probability of success of each trial be \(\lambda \Delta t\), where \(\lambda \) is a positive constant. Let \(G\) be the number of failures before the first success in discrete time, and let \(T\) be the time of the first success in continuous time.
Hint: Draw a timeline and try out a simple example.
Hint: First find \(P(T>t)\).
Problem 4.
Source: Blitzstein and Hwang, Introduction to Probability, Chapter 6, Problem 1–2.
Problem 5.
Source: Blitzstein and Hwang, Introduction to Probability, Chapter 6, Problem 21.
Let \[ X_n\sim \operatorname {Bin}(n,p_n) \] for all \(n\ge 1\), where \(np_n\) is a constant \(\lambda >0\) for all \(n\), so that \[ p_n=\frac {\lambda }{n}. \] Let \[ X\sim \operatorname {Pois}(\lambda ). \]
Show that the MGF of \(X_n\) converges to the MGF of \(X\). This gives another way to see that the \(\operatorname {Bin}(n,p)\) distribution can be well-approximated by the \(\operatorname {Pois}(\lambda )\) distribution when \(n\) is large, \(p\) is small, and \(\lambda =np\) is moderate.
Problem 6.
A student has developed a highly scientific method for deciding whether to complete an optional online practice module. They go to the roof of their apartment building and throw one of their shoes toward the ground.
Let \(X\in [0,1]\) be the proportion of the module that the student completes. Assume that, with probability \(p\), the shoe lands in a tree and becomes permanently stuck. In that case, the student spends the rest of the evening trying to recover it and completes none of the module, so \(X=0\).
With probability \(1-p\), the shoe does not get stuck in the tree. The student then attempts the module, and the proportion \(X\) that they complete is uniformly distributed on \((0,1)\).
Let \[ I= \begin {cases} 1, & \text {if the shoe gets stuck in the tree},\\ 0, & \text {otherwise}, \end {cases} \qquad P(I=1)=p, \] where \(0<p<1\).
Let \[ U\sim \operatorname {Unif}(0,1), \] independently of \(I\). In this case, we have \[ X= \begin {cases} 0, & if \ I=1,\\ U, & if \ I=0. \end {cases} \]