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Chapter 6: Problem 16

A biased coin is tossed \(N\) times, where \(N\) is a random variable with finite mean. Show that if the numbers of heads and tails are independent, then \(N\) is Poisson. [You may want to use the fact that all continuous solutions of \(f(x+y)=f(x) f(y)\) take the form \(f(x)=e^{\lambda x}\) for some \(\lambda\).]

Short Answer

Expert verified
The random variable \(N\) must be Poisson distributed.

Step by step solution

01

Define Key Variables and Concepts

Let us denote the probability of heads in a single toss as \(p\), where the probability of tails is \(1-p\). The number of heads \(H\) in \(N\) tosses is a binomial random variable \(H \sim \text{Binomial}(N, p)\). Similarly, the number of tails \(T\) is given by \(T = N - H\).
02

Independence Condition Interpretation

The problem states that the numbers of heads and tails are independent. Generally, \(H\) and \(T\) being independent implies that the joint distribution is the product of their individual distributions. In mathematical terms, \(P(H = h, T = t) = P(H = h)P(T = t)\).
03

Applying the Binomial Distribution

Considering \(N\) fixed, the distribution of \(H = h\) is \(P(H = h|N = n) = \binom{n}{h} p^h (1-p)^{n-h}\). For independence between \(H\) and \(T\) to hold for all \(n\), a key feature of the Poisson process must be leveraged.
04

Derive Condition of Poisson Process

Since \(H\) and \(T\) are independent and their sum \(H + T = N\), we exploit the result that continuous solutions of the functional equation \(f(x+y) = f(x)f(y)\) take the form \(f(x) = e^{\lambda x}\). Here, the only discrete distribution consistent with such independence and equivalently with the exponential form is the Poisson distribution.
05

Conclude That N is Poisson

From Steps 1-4, since the defining characteristic for \(N\) in terms of independence of the number of outcomes (heads and tails) aligns with the Poisson distribution, we conclude that \(N\) must be Poisson distributed. This follows from the special property of the Poisson process where the intervals are independent due to its memoryless nature.

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Key Concepts

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

Poisson Distribution
The Poisson distribution is essential in probability theory for modeling the number of events occurring within a fixed interval of time or space. It applies when these events occur randomly, independently, and rarely. Key features include:
  • The average number of events (\( \lambda \)) that happen over the interval.
  • The probability of a given number of events happening within that period is given by the formula:\[P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}\]where \( e \) is the base of the natural logarithm, \( \lambda \) is the mean number of occurrences, \( k \) is the number of occurrences, and \( k! \) is the factorial of \( k \).
The connection to the problem at hand is that when the sum of two independent counting processes (such as heads and tails in coin tosses) must adhere to the properties of independence, it suggests a Poisson distribution for the total number of trials (\( N \)).
The independence and memoryless properties make Poisson processes suitable for modeling scenarios like phone call arrivals at a call center or decay events in a radioactive sample.
Binomial Distribution
In probability theory, a binomial distribution describes the number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success. Key details include:
  • A fixed number of trials (\( n \)).
  • A constant probability of success in each trial (\( p \)).
  • The probability of observing exactly \( k \) successes in \( n \) trials is given by:\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]where \( \binom{n}{k} \) is the binomial coefficient, representing combinations of \( n \) trials taken \( k \) at a time.
In our context, the number of heads, \( H \), in \( N \) coin tosses follows a binomial distribution. The problem looks at how relationships between multiple binomial-distributed variables can reveal deeper distributions, such as Poisson, under certain conditions.
Independence of Events
Independence in probability theory means the occurrence of one event does not affect the probability of another event occurring. For random variables, it indicates that knowing the outcome of one gives no information about the other. Here are some points to consider:
  • If two events, \( A \) and \( B \), are independent, the probability of both occurring is the product of their individual probabilities:\[P(A \cap B) = P(A) \cdot P(B)\]
  • For the exercise, we assess the independence between the number of heads and tails resulting from \( N \) coin tosses. If \( H \) (heads) and \( T \) (tails) are independent, the condition is mathematically represented by the equation \( P(H = h, T = t) = P(H = h) P(T = t) \).
The independence here adds an interesting aspect: if the independent relationship holds under all conditions, it leads us to conclude that the overall number of trials, \( N \), follows a Poisson distribution, a result unique to such independence with additive random variables.

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Most popular questions from this chapter

Let \(X\) and \(Y\) be independent with \(r\) th cumulants \(\kappa_{r}^{(X)}\) and \(\kappa_{r}^{(Y)}\). Show that \(X+Y\) has \(r\) th cumulant \(\kappa_{r}^{(X)}+\kappa_{r}^{(Y)} .\)

Let \(X\) have p.g.f. \(G(s)\), and let \(E\) be the event that \(X\) is even. Show that $$ \mathbf{E}\left(s^{x} \mid E\right)=\frac{G(s)+G(-s)}{G(1)+G(-1)} $$

A loaded die may show different faces with different probabilities. Show that it is not possible to load two traditional cubic dice in such a way that the sum of their scores is uniformly distributed on \(\\{2,3, \ldots, 12\\}\).

In a multiple-choice examination, a student chooses between one true and one false answer to each question. Assume the student answers at random, and let \(N\) be the number of such answers until she first answers two successive questions correctly. Show that \(\mathbf{E}\left(s^{N}\right)=s^{2}\left(4-2 s-s^{2}\right)^{-1}\). Hence, find \(\mathbf{E}(N)\) and \(\mathbf{P}(N=k)\). Now find \(\mathbf{E}(N)\) directly.

Let the number of tosses required for a fair coin to show a head be \(T\). An integer \(X\) is picked at random from \(\\{1, \ldots, T\\}\) with equal probability \(\frac{1}{T}\) of picking any one. Find \(G_{X}(s)\).
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