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Show that a binomial random variable \(R\) with denominator \(m\) and probability \(\pi\) has cumulant-generating function \(K(t)=m \log \left(1-\pi+\pi e^{t}\right)\). Find \(\lim K(t)\) as \(m \rightarrow \infty\) and \(\pi \rightarrow 0\) in such a way that \(m \pi \rightarrow \lambda=0\). Show that $$ \operatorname{Pr}(R=r) \rightarrow \frac{\lambda^{r}}{r !} e^{-\lambda} $$ and hence establish that \(R\) converges in distribution to a Poisson random variable. This yields the Poisson approximation to the binomial distribution, sometimes called the law of small numbers. For a numerical check in the S language, try \(\mathrm{y}<-0: 10 ;\) lambda \(<-1 ; \mathrm{m}<-10 ; \mathrm{p}<-\) lambda/m round(cbind(y, pbinom \((\mathrm{y}\), size \(=\mathrm{m}, \mathrm{prob}=\mathrm{p})\), ppois \((\mathrm{y}\), lambda) \()\), digits \(=3\) ) with various other values of \(m\) and \(\lambda\).

Short Answer

Expert verified
The binomial distribution converges to a Poisson distribution under specified limits, verifying the Poisson approximation.

Step by step solution

01

Understanding the Binomial Cumulant-Generating Function

The cumulative generating function (CGF) for a binomial random variable \( R \) with parameters \( m \) and \( \pi \) is given by \( K(t) = m \log \left(1 - \pi + \pi e^t \right) \). This arises because the CGF of a single Bernoulli trial with success probability \( \pi \) is \( \log(1 - \pi + \pi e^t) \). Since \( R \) is the sum of \( m \) independent Bernoulli trials, we multiply by \( m \).
02

Taking the Limit as Parameters Change

For \( m \to \infty \) and \( \pi \to 0 \) with \( m \pi = \lambda = 0 \), we expand \( K(t) = m \log(1 + \pi(e^t - 1)) \). Using the approximation \( \log(1 + x) \approx x \) for small \( x \), \( K(t) \approx m \cdot \pi (e^t - 1) \), and with \( m \pi = \lambda \), we get \( \lim_{m \to \infty, \pi \to 0} K(t) = \lambda (e^t - 1) \).
03

Convergence to Poisson Distribution

The result \( K(t) = \lambda (e^t - 1) \) corresponds to the CGF of a Poisson distribution with parameter \( \lambda \). The Probability Mass Function of a Poisson random variable is \( \operatorname{Pr}(R=r) = \frac{\lambda^r}{r!} e^{-\lambda} \). Thus, as \( m \to \infty \) and \( \pi \to 0 \) such that \( m\pi \to \lambda \), the binomial random variable \( R \) converges in distribution to a Poisson(\( \lambda \)) random variable.
04

Numerical Verification in S Language

Using S language, we check with \( \mathrm{y}<-0:10 \), \( \lambda<-1 \), \( \mathrm{m}<-10 \), and \( \mathrm{p}<-(\lambda/\mathrm{m}) \). Then evaluate \( \text{round(cbind(y, pbinom(y, size=m, prob=p), ppois(y, lambda)), digits=3)} \). This code compares cumulative binomial and Poisson probabilities for different values of \( m \) and \( \lambda \), affirming the Poisson approximation.

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Key Concepts

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

Cumulant-Generating Function
The cumulant-generating function (CGF) is a powerful tool in probability theory. It is used to characterize the distribution of a random variable. For a binomial random variable, with parameters \(m\) (the number of trials) and \(\pi\) (the success probability), the CGF is expressed as \(K(t) = m \log \left(1 - \pi + \pi e^t \right)\).
This expression arises from the CGF of a single Bernoulli trial, given by \(\log(1 - \pi + \pi e^t)\). As a binomial random variable is the sum of \(m\) independent Bernoulli trials, the CGF multiplies by \(m\).
The CGF is not only a tool for characterizing distributions but also plays a key role in the process of finding limits and approximations, such as the transition from binomial to Poisson distribution.
Binomial Distribution
The binomial distribution is one of the foundational distributions in statistics. It models the number of successes in a fixed number of independent Bernoulli trials.
  • The parameter \(m\) represents the number of trials.
  • \(\pi\) denotes the probability of success in each trial.

A binomial distribution can model many real-world situations, such as the number of heads obtained when flipping a coin \(m\) times. The probability mass function for a binomially distributed random variable \(R\) is given by:
\(\Pr(R = r) = \binom{m}{r} \pi^r (1 - \pi)^{m-r}\)
This function is crucial in calculating probabilities for binomially distributed events. However, when \(m\) is large and \(\pi\) is small, a binomial can be approximated using a Poisson distribution, simplifying calculations.
Poisson Distribution
The Poisson distribution serves as an approximation for the binomial when certain conditions are met, specifically when the number of trials \(m\) is large, and the success probability \(\pi\) is small but such that their product \(m\pi\) (mean) is constant.
This distribution is parameterized by \(\lambda\), the average number of successes in a fixed region or interval. The probability mass function for a Poisson random variable is:
  • \( \operatorname{Pr}(R = r) = \frac{\lambda^r}{r!} e^{-\lambda} \)

The Poisson model is often used in situations where events occur independently and at a constant average rate. Examples include modeling the number of phone calls received by a call center per hour or the number of decay events per unit time from a radioactive source.
Convergence in Distribution
Convergence in distribution is a concept dealing with how a sequence of random variables behaves as an index (often representing size or time) approaches infinity. For a binomial random variable \(R\) with parameters \(m\) and \(\pi\), under certain conditions, it converges in distribution to a Poisson random variable.
Specifically, as \(m \to \infty\) and \(\pi \to 0\) such that \(m\pi \to \lambda\), the binomial distribution converges to a Poisson distribution:
\(\operatorname{Pr}(R=r) \rightarrow \frac{\lambda^r}{r!} e^{-\lambda}\).
This concept provides the mathematical foundation for the Poisson approximation of the binomial distribution, showing how one type of distribution can approximate another under specific conditions.
Law of Small Numbers
The law of small numbers refers to the phenomenon where, under certain conditions, the distribution of rare or infrequent events, modeled by a Poisson distribution, approximates the distribution of these events when modeled directly by a binomial distribution.
This law is particularly useful when dealing with large numbers of trials where the probability of success is very low, making direct calculation using the binomial formula cumbersome.
Applications of the law of small numbers are vast, spanning fields like epidemiology, telecommunications, and traffic flow, where occurrences are rare, but require modeling over large populations or long periods.

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Most popular questions from this chapter

The Cauchy density (2.16) has no moment-generating function, but its characteristic function is \(\mathrm{E}\left(e^{i t Y}\right)=\exp (i t \theta-|t|)\), where \(i^{2}=-1\). Show that the average \(\bar{Y}\) of a random sample \(Y_{1}, \ldots, Y_{n}\) of such variables has the same characteristic function as \(Y_{1}\). What does this imply?

Let \(Y\) have continuous distribution function \(F\). For any \(\eta\), show that \(X=|Y-\eta|\) has distribution \(G(x)=F(\eta+x)-F(\eta-x), x>0\). Hence give a definition of the median absolute deviation of \(F\) in terms of \(F^{-1}\) and \(G^{-1}\). If the density of \(Y\) is symmetric about the origin, show that \(G(x)=2 F(x)-1\). Hence find the median absolute deviation of the Laplace density \((2.5)\).

Let \(Y=\exp (X)\), where \(X \sim N\left(\mu, \sigma^{2}\right) ; Y\) has the log-normal distribution. Use the moment-generating function of \(X\) to show that \(\mathrm{E}\left(Y^{r}\right)=\exp \left(r \mu+r^{2} \sigma^{2} / 2\right)\), and hence find \(\mathrm{E}(Y)\) and \(\operatorname{var}(Y)\). If \(Y_{1}, \ldots, Y_{n}\) is a log-normal random sample, show that both \(T_{1}=\bar{Y}\) and \(T_{2}=\exp (\bar{X}+\) \(\left.S^{2} / 2\right)\) are consistent estimators of \(\mathrm{E}(Y)\), where \(X_{j}=\log Y_{j}\) and \(S^{2}\) is the sample variance of the \(X_{j} .\) Give the corresponding estimators of \(\operatorname{var}(Y)\). Are the estimators based on the \(Y_{j}\) or on the \(X_{j}\) preferable? Why?

If \(T_{1}\) and \(T_{2}\) are two competing estimators of a parameter \(\theta\), based on a random sample \(Y_{1}, \ldots, Y_{n}\), the asymptotic efficiency of \(T_{1}\) relative to \(T_{2}\) is \(\lim _{n \rightarrow \infty} \operatorname{var}\left(T_{2}\right) / \operatorname{var}\left(T_{1}\right) \times\) \(100 \%\). If \(n=2 m+1\), find the asymptotic efficiency of the sample median \(Y_{(m+1)}\) relative to the average \(\bar{Y}=n^{-1} \sum_{j} Y_{j}\) when the density of the \(Y_{j}\) is: (a) normal with mean \(\theta\) and variance \(\sigma^{2} ;\) (b) Laplace, \((2 \sigma)^{-1} \exp \\{-|y-\theta| / \sigma\\}\) for \(-\infty

If \(Y_{1}, \ldots, Y_{n} \stackrel{\text { iidd }}{\sim} N\left(\mu, \sigma^{2}\right)\), show that \(n^{1 / 2}(\bar{Y}-\mu)^{2} \stackrel{P}{\longrightarrow} 0\) as \(n \rightarrow \infty\). Given that \(\operatorname{var}\left\\{\left(Y_{j}\right.\right.\) \(\left.\mu)^{2}\right\\}=2 \sigma^{4}\), deduce that \(\left(S^{2}-\sigma^{2}\right) /\left(2 \sigma^{4} / n\right)^{1 / 2} \stackrel{D}{\longrightarrow} Z\), where \(Z \sim N(0,1)\). When is this true for non-normal data?

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