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Chapter 3: Problem 17

A secretary drops \(n\) matching pairs of letters and envelopes down the stairs, and then places the letters into the envelopes in a random order. Use indicators to show that the number \(X\) of correctly matched pairs has mean and variance 1 for all \(n \geq 2\). Show that the mass function of \(X\) converges to a Poisson mass function as \(n \rightarrow \infty\).

Short Answer

Expert verified
The mean is 1, the variance is 1, and the pmf converges to a Poisson(1) distribution as \( n \rightarrow \infty \).

Step by step solution

01

Define the Indicator Variables

Let鈥檚 introduce indicator variables. For each pair, let \( I_k \) be an indicator variable such that \( I_k = 1 \) if the \( k \)-th letter is placed in the correct envelope, and \( I_k = 0 \) otherwise. Thus, the total number of correctly matched pairs \( X \) can be expressed as \( X = \sum_{k=1}^{n} I_k \).
02

Calculate the Expectation

The expected value of \( X \) is given by the sum of the expectations of the \( I_k \). Since each letter has a \( \frac{1}{n} \) probability of being placed in the correct envelope, we have \( E[I_k] = \frac{1}{n} \). Therefore, \( E[X] = \sum_{k=1}^{n} E[I_k] = n \times \frac{1}{n} = 1 \).
03

Calculate the Variance

The variance of \( X \) is calculated using the variance of the indicators. Since \( I_k \) are independent, \( \text{Var}(X) = \sum_{k=1}^{n} \text{Var}(I_k) \). For an indicator variable, \( \text{Var}(I_k) = E[I_k^2] - (E[I_k])^2 = \frac{1}{n} - (\frac{1}{n})^2 = \frac{n-1}{n^2} \). Then, \( \text{Var}(X) = n \times \frac{n-1}{n^2} = \frac{n-1}{n} \). As \( n \rightarrow \infty \), this approaches 1.
04

Conclude Mean and Variance

Thus, for the given setup, both the mean and the variance of \( X \) are 1 for all \( n \geq 2 \).
05

Show Convergence to Poisson Distribution

The probability mass function (pmf) of \( X \) is driven by the sum of independent, rare events. By the Poisson limit theorem, as \( n \rightarrow \infty \), the distribution of \( X \), with \( E[X] = \lambda = 1 \), converges to a Poisson distribution with parameter \( \lambda = 1 \). Therefore, the pmf of \( X \) approaches \( P(X = k) = \frac{e^{-1}}{k!} \) as \( n \rightarrow \infty \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Indicator Variables
Indicator variables are special tools in probability that help us count specific events. Here, each pair of letter and envelope is assigned an indicator variable, denoted as \( I_k \). This variable takes the value of 1 if the corresponding letter is placed into the correct envelope and 0 otherwise. Think of it as a simple switch with two states: matched or not matched.
  • The sum of all these indicators gives us the total number of correctly matched pairs \( X \).
  • Mathematically, this is written as \( X = \sum_{k=1}^{n} I_k \), where \( n \) is the number of letters (or envelopes).
Why Indicator Variables?
These variables simplify complex problems by breaking them down into simple on/off states. They play a crucial role in determining expected values and variances, making our calculations manageable, especially in problems involving large datasets or random assignments.
Expectation
Expectation, or expected value, is a fundamental concept in probability. It provides a measure of the 'average' or 'mean' outcome of a random event. In our problem, we are interested in the expected number of correctly matched pairs, \( E[X] \).
  • Each letter has a \( \frac{1}{n} \) chance of being placed correctly in its envelope because we are distributing \( n \) letters randomly into \( n \) envelopes.
  • Therefore, the expectation of each indicator variable is \( E[I_k] = \frac{1}{n} \).
  • Since there are \( n \) such independent indicators, we compute \( E[X] = \sum_{k=1}^{n} E[I_k] = n \times \frac{1}{n} = 1 \).
The expected value of exactly 1 means that no matter how many letters \( n \) there are, on average, one letter lands in the right place. This concept helps us understand large-scale patterns from many random acts.
Variance Calculation
Variance offers insight into how much dispersion or variability exists from the expected value. For the correctly matched pairs \( X \), knowing the variance helps us gauge the "spread" of potential outcomes around 1.
  • For each indicator variable \( I_k \), the variance is \( \text{Var}(I_k) = E[I_k^2] - (E[I_k])^2 \), which simplifies to \( \frac{1}{n} - (\frac{1}{n})^2 = \frac{n-1}{n^2} \).
  • Since indicators are independent, the variance of the total number of matches \( X \) is the sum: \( \text{Var}(X) = n \times \frac{n-1}{n^2} = \frac{n-1}{n} \).
As \( n \) becomes larger, \( \frac{n-1}{n} \) approaches 1. This indicates that the variance stabilizes just like the mean. This predictability is vital in understanding the distribution of \( X \), especially as \( n \) grows.
Probability Mass Function
The Probability Mass Function (PMF) provides the probabilities of a discrete random variable taking specific values. As \( n \) increases, the behavior of our random variable \( X \), the number of correctly matched pairs, evolves according to the Poisson distribution.
  • The Poisson distribution is ideal for modeling the number of events in a fixed interval of time or space, given these events occur with a known constant mean rate and independently of the time since the last event.
  • For \( X \) with mean \( \lambda = 1 \) as \( n \rightarrow \infty \), the PMF converges to that of a Poisson distribution: \( P(X = k) = \frac{e^{-\lambda}}{k!} \), where \( \lambda = 1 \).
  • This shift illustrates how independent rare events (like randomly matching a letter with an envelope) can aggregate into a Poisson process when the number of trials becomes very large.
Understanding the PMF's Poisson convergence helps unravel complex probability behavior into more familiar uniform patterns, suitable for applications in various fields like telecommunications and biology.

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Most popular questions from this chapter

Let \(T\) be the time which elapses before a simple random walk is absorbed at either of the absorbing barriers at 0 and \(N\), having started at \(k\) where \(0 \leq k \leq N\). Show that \(P(T<\infty)=1\) and \(\mathbb{E}\left(T^{k}\right)<\infty\) for all \(k \geq 1\).

We toss \(n\) coins, and each one shows heads with probability \(p\), independently of each of the others. Each coin which shows heads is tossed again. What is the mass function of the number of heads resulting from the second round of tosses?

In 1710, J. Arbuthnot observed that male births had exceeded female births in London for 82 successive years. Arguing that the two sexes are equally likely, and \(2^{-82}\) is very small, he attributed this run of masculinity to Divine Providence. Let us assume that each birth results in a girl with probability \(p=0.485\), and that the outcomes of different confinements are independent of each other. Ignoring the possibility of twins (and so on), show that the probability that girls outnumber boys in \(2 n\) live births is no greater than \(\left(\begin{array}{c}2 \pi \\ n\end{array}\right) p^{n} q^{n}\\{q /(q-p)\\}\), where \(q=1-p\). Suppose that 20,000 children are bom in each of 82 successive years. Show that the probability that boys outnumber girls every year is at least \(0.99\). You may need Stirling's formula.

(a) If \(X\) takes non-negative integer values show that \(\mathbb{E}(X)=\sum_{n=0}^{\infty} \mathrm{P}(X>n)\) (b) An um contains \(b\) blue and \(r\) red balls. Balls are removed at random until the first blue ball is drawn. Show that the expected number drawn is \((b+r+1) /(b+1)\). (c) The balls are replaced and then removed at random until all the remaining balls are of the same colour. Find the expected number remaining in the urn.

If \(X\) is geometric, show that \(\mathrm{P}(X=n+k \mid X>n)=\mathbb{P}(X=k)\) for \(k, n \geq 1\). Why do you think that this is called the 'lack of memory' property? Does any other distribution on the positive integers have this property?
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