HW 02: More Bayes, Data, and Sampling

Solution

Due date

This assignment is due on Wednesday, September 10 before class (hard deadline so I can post the solution in time for you to study for midterm). To be considered on time, the following must be done by the due date:

  • Final .pdf file submitted on Gradescope

Learning goals

In this assignment, you will…

  • practice the Bayes’ theorem again
  • practice determining types of data
  • relates random sampling to probability concept

Conceptual exercises

Instructions

The conceptual exercises focus on explaining concepts and showing results mathematically. Show your work for each question. I expect you not to skip any steps and carefully understand the thinking process when working on these exercises.

You may write the answers and associated work for conceptual exercises by hand or type them in your Quarto document.

Exercise 1: The Game Show Host or “Monty Hall” Problem (14 points)

Watch a clip from a movie titled 21 (2008) in this Youtube video. The same problem is also posted in Parade Magazine in 1990 :

“Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, ‘Do you want to pick door No. 2?’ Is it to your advantage to switch your choice?”

Answer these question to help you make the decision:

  1. Before the host open door no. 3:

    • what is the probability that the car is in door No. 1?
    • what is the probability that the car is in door No. 2?
    • what is the probability that the car is in door No. 3?

    Solution:

    Before the host opens door no.3, all these probabilities are the same:

    \[ P(\text{Car in 1})=P(\text{Car in 2})=P(\text{Car in 3})=1/3. \]

  2. May be it doesn’t seems too obvious, but the host will never open a door concealing the car. Now, compute the following conditional probability:

    • Probability that the host open door No. 3 given that the car is in door No.1

    • Probability that the host open door No. 3 given that the car is in door No.2

    • Probability that the host open door No. 3 given that the car is in door No.3

    Solution:

    We assume the host will never open a door concealing a car. You’ve also picked door No. 1 at this point.

    If the car is in door No.1, then the host could have opened either door No.2 or door No. 3. Thus, \[P(\text{Open 3}|\text{Car in 1})=1/2.\]

    If the car is in door No.2, then the host must have chosen door No. 3. Thus, \[P(\text{Open 3}|\text{Car in 2})=1.\]

    If the car is in door No.3, then the host will not open door No. 3. Thus, \[P(\text{Open 3}|\text{Car in 3})=0.\]

  3. Using the law of total probability, can you compute the probability that the host open door No. 3?

    Solution:

    The event that the car is either in door No.1, No.2, and No.3 are mutually exclusive. Thus by the law of total probability:

    \[P(\text{Open 3})=P(\text{Open 3}|\text{Car in 1})P(\text{Car in 1})+P(\text{Open 3}|\text{Car in 2})P(\text{Car in 2})+P(\text{Open 3}|\text{Car in 3})P(\text{Car in 3}).\]

    \[P(\text{Open 3})=(1/2)(1/3)+(1)(1/3)+(0)(1/3)=1/2.\]

  4. Compute the following posterior probability using Bayes’ theorem:

    • Probabiliy that the car is in door No.1 given that the host open door No. 3
    • Probabiliy that the car is in door No.2 given that the host open door No. 3
    • Probabiliy that the car is in door No.3 given that the host open door No. 3

    Solution:

    Recall the Bayes’ Theorem, we have:

    • \(P(\text{Car in 1}|\text{Open 3})=\frac{P(\text{Open 3}|\text{Car in 1})P(\text{Car in 1})}{P(\text{Open 3})}=\frac{(1/2)(1/3)}{(1/2)}=1/3.\)
    • \(P(\text{Car in 2}|\text{Open 3})=\frac{P(\text{Open 3}|\text{Car in 2})P(\text{Car in 1})}{P(\text{Open 3})}=\frac{(1)(1/3)}{(1/2)}=2/3.\)
    • \(P(\text{Car in 3}|\text{Open 3})=\frac{P(\text{Open 3}|\text{Car in 3})P(\text{Car in 1})}{P(\text{Open 3})}=\frac{(0)(1/3)}{(1/2)}=0.\)

  5. Compare the prior probability (in point 1) and the posterior probability (in point 4), is it better to keep door No. 1 or switch to door No. 2?

    Solution:

    Comparing the prior probability (1/3) for all doors. Now, if you switch to door No. 2, the posterior probability is (2/3). It is in your best interests to switch. It doesn’t necessarily mean you always win, but if the game is repeated many times, you’re more likely to get a car using this strategy.

Exercise 2: Data and Sampling (6 + 9 points)

  1. For each of the following examples, indicate whether the data are cross-sectional, time-series, or panel data.
    1. An avid runner records the number of miles that she runs every day for 100 straight days: Time-series data
    2. A random sample of 100 older adults, aged 65 and older, were surveyed in January 2023 about their health status and medical expenditures: Cross-sectional data
    3. A financial analyst randomly picks 20 companies that are listed on the New York Stock Exchange and gathers data on their profits and sales for the year 2022: Cross-sectional data
    4. A Canadian economist gathers annual data on each of Canada’s 13 administrative divisions, including the population and the unemployment rate for each division, for each year between 2000 and 2020: Panel data
    5. A random sample of 100 college students is asked whether they have received an influenza vaccine in the last year: Cross-sectional data
    6. A random sample of 100 college seniors is asked for their semester grade point average for each of their first six semesters at the university: Panel data
  2. Consider drawing a simple random sample of \(n=10\) observations from a population consisting of \(100\) units.

    1. How many possible ways are there to draw the simple random sample?

      Solution:

      There are \(\begin{pmatrix}100\\10\end{pmatrix}\) possible ways to draw a simple random sample.

    2. What is the probability that a given observation from the population is in the simple random sample that is drawn?

      Solution:

      The probability that a given observation is drawn is \(\frac{10}{100}=0.1\).

    3. What is the probability that any two given observations from the population are in the simple random sample that is drawn?

      Solution:

      The probability that any two given observations are drawn is \(\frac{10}{100}\cdot\frac{9}{99}\approx 0.0091\).

Grading (30 points)

Component Points
Ex 1 14
Ex 2 15
Formatting 1

“Formatting” grade is to assess whether or not you mark specific page of the pdf to a specific question to help the TA when grading.