Probstat/midterm problem 9 solution

This is how a simple charity game is played. There are $2n$ people where $n\geq 1$. There are also $2n$ tickets whose numbers on each of them is either 0 or 1. There are n tickets with number 1 on them, and $n$ tickets with number 0 on them. The ticket are randomly given to the people; each person gets one ticket. (So that $n$ people get tickets number 1 and the other $n$ people get ticket number 0.)

Then the people are randomly paired up into n pairs. In each pair the person who has lower number should donate the difference to the charity fund. The person with higher number does not donate. For example, if two people in a pair have ticket numbers 1 and 0, the person with ticket number 0 has to donate 1 baht. If two people in a pair have the same number, they do not donate.

Note: In this question, do not have to try too hard to reduce your answers to the minimal form. Any reasonable expressions are acceptable.

Two approaches There are basically two approach for analyzing the expectation and the variance. We can look at each person or we can look at each pair. The first solution analyzes each pair and is a bit simpler. The second one analyzes each person.

Solution 1

First note that the assignment of tickets is done independently of the pairing of people. Therefore, we can imaging a game where people start by pairing up and then random tickets are assigned. Note that since the tickets are assigned randomly the way people are paired up is essentially irrelevant; therefore, we can assume that the pairing is fixed. Under this assumption, there are $n$ pairs that we need to consider.

Expectation

(a) Let random variable $X$ denote the amount of money donated. Find $E[X]$. Show your work clearly.

Consider pair $i$. Let random variable

$X_{i}=\left\{{\begin{array}{ll}1&\mathrm {if} \ \mathrm {pair} \ i\ \mathrm {donates} ,\\0&\mathrm {otherwise} \end{array}}\right.$

Note that $E[X_{i}]=Pr\{X_{i}=1\}$. Pair $i$ makes the donation if and only if the two people in the pair have different tickets. Let's distinguish two people in this pair by referring to them as the first person and the second person.

They have different ticket numbers with probability

$Pr\{X_{i}=1\}={\frac {n}{2n-1}},$

because the first person can have any ticket, then the second person must get a different ticket (there are $n$ correct tickets out of $2n-1$ tickets left).

Since

$X=\sum _{i=1}^{n}X_{i},$

using the linearity of expectation we have that

$E[X]=E[\sum _{i=1}^{n}X_{i}]=\sum _{i=1}^{n}E[X_{i}]=\sum _{i=1}^{n}{\frac {n}{(2n-1)}}=n\cdot {\frac {n}{(2n-1)}}={\frac {n^{2}}{2n-1}}.$

Note that this is slightly larger than 1/4 of the people.

Variance

We start by calculating $E[X^{2}]$. Note that

$X^{2}=\left(\sum _{i=1}^{n}X_{i}\right)^{2}=\sum _{i=1}^{n}X_{i}^{2}+\sum _{i=1}^{n}\sum _{j\neq i}X_{i}X_{j}.$

Consider $E[X_{i}^{2}]$. We have that

$E[X_{i}^{2}]=1^{2}\cdot P\{X_{i}^{2}=1\}+0^{2}\cdot P\{X_{i}^{2}=0\}=P\{X_{i}^{2}=1\}=P\{X_{i}=1\},$

and from the previous solution, we conclude that $E[X_{i}^{2}]={\frac {n}{2n-1}}.$

Let's consider the second term, $E[X_{i}X_{j}]$. Again note that the only way $X_{i}X_{j}$ is non zero is when both $X_{i}$ and $X_{j}$ are one. You can check that this occurs with probability

${\frac {n}{2n-1}}\cdot {\frac {n-1}{2n-3}}$.

Therefore, $E[X_{i}X_{j}]={\frac {n}{2n-1}}\cdot {\frac {n-1}{2n-3}}$.

Note that from the linearity of expectation, we have

$E[X^{2}]=E\left[\left(\sum _{i=1}^{n}X_{i}\right)^{2}\right]=E\left[\sum _{i=1}^{n}X_{i}^{2}+\sum _{i=1}^{n}\sum _{j\neq i}X_{i}X_{j}\right]=\sum _{i=1}^{n}E\left[X_{i}^{2}\right]+\sum _{i=1}^{n}\sum _{j\neq i}E\left[X_{i}X_{j}\right].$

Plugging in the calculations yields

$E[X^{2}]=\left(\sum _{i=1}^{n}{\frac {n}{2n-1}}\right)+\left(\sum _{i=1}^{n}\sum _{j\neq i}{\frac {n}{2n-1}}\cdot {\frac {n-1}{2n-3}}\right)=\left({\frac {n^{2}}{2n-1}}\right)+\left({\frac {n(n-1)}{2}}\cdot {\frac {n}{2n-1}}\cdot {\frac {n-1}{2n-3}}\right)$

Therefore,

$Var(X)=E[X^{2}]-E[X]^{2}=\left({\frac {n^{2}}{2n-1}}\right)+\left({\frac {n(n-1)}{2}}\cdot {\frac {n}{2n-1}}\cdot {\frac {n-1}{2n-3}}\right)-\left({\frac {n^{2}}{2n-1}}\right)^{2},$

which is ...

Solution 2

Expectation

(a) Let random variable $X$ denote the amount of money donated. Find $E[X]$. Show your work clearly.

Consider person $i$. Let random variable

$X_{i}=\left\{{\begin{array}{ll}1&\mathrm {if} \ \mathrm {person} \ i\ \mathrm {donates} ,\\0&\mathrm {otherwise} \end{array}}\right.$

Note that $E[X_{i}]=Pr\{X_{i}=1\}$. Person $i$ donates if and only if the person gets ticket number 0, and the person's pair get ticket number 1. This occurs with probability

$Pr\{X_{i}=1\}={\frac {n}{2n}}\cdot {\frac {n}{2n-1}}={\frac {n}{2(2n-1)}}$

Since

$X=\sum _{i=1}^{2n}X_{i},$

using the linearity of expectation we have that

$E[X]=E[\sum _{i=1}^{2n}X_{i}]=\sum _{i=1}^{2n}E[X_{i}]=\sum _{i=1}^{2n}{\frac {n}{2(2n-1)}}=2n\cdot {\frac {n}{2(2n-1)}}={\frac {n^{2}}{2n-1}}.$

Note that this is slightly larger than 1/4 of the people.

Remarks: In this solution, we analyze the experiment in terms of the people, so there are $2n$ people that we need to look at.

Variance

To be added.