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Chapter 6: Problem 8

Let the table $$\begin{array}{c|cccccc}\mathrm{x} & 0 & 1 & 2 & 3 & 4 & 5 \\\\\hline \text { Frequency } & 7 & 14 & 12 & 13 & 6 & 3 \end{array}$$ represent a summary of a random sample of size 55 from a Poisson distribution. Find the maximum likelihood estimate of \(P(X=2)\).

Short Answer

Expert verified
The maximum likelihood estimate of \(P(X=2)\) is 0.27.

Step by step solution

01

Calculate the Sample Mean

The sample mean \(位\) is calculated as: \( 位 = 鈭憍f(x) = (鈭� (x * frequency)) / sample size \). Using the provided table, this evaluates to \( 位 = (0*7 + 1*14 + 2*12 + 3*13 + 4*6 + 5*3) / 55 = 2.0 \)
02

Use the Poisson Probability Mass Function to calculate \(P(X=2)\)

Using the Poisson probability mass function \(P(X=k) = \frac{(位^k)*(e^{-位})}{k!}\), where \(k=2\) and \(位=2.0\), we find that \(P(X=2) = \frac{e^{-2}(2^2)}{2!} = 0.27.\)

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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 a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event. It is named after the French mathematician Sim茅on Denis Poisson.

This distribution is particularly useful for modeling events with low probabilities and short time intervals, such as the number of phone calls a call center receives in an hour or the number of decay events from a radioactive source in a given period of time.

The key parameter of the Poisson distribution is the mean rate of occurrence, denoted by \( \lambda \), which is also the expected number of occurrences within the specified interval. A distinct feature of this distribution is that the variance is equal to the mean, \( \lambda \), indicating that as the mean rate of events increases, their variability also increases.
Probability Mass Function
The probability mass function (PMF) is a mathematical function that gives us the probability that a discrete random variable is exactly equal to some value. For the Poisson distribution, the PMF is defined as:\[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \], where \( k \) is the number of occurrences, \( e \) is Euler's number (approximately equal to 2.71828), and \( k! \) (read as 'k factorial') is the product of all positive integers up to \( k \).

The PMF provides the probabilities for all possible outcomes of our random variable, essentially mapping the distribution. When we know the PMF, it's straightforward to find the probability of our random variable taking on a specific value by substituting that value into the PMF equation.
Sample Mean Calculation
The sample mean, especially in the context of the Poisson distribution, is the average rate at which events occur. It can be computed by adding the product of each value and its frequency from the sample and then dividing by the total number of observations in the sample. The formula is given by:\[ \lambda = \frac{\sum (x \times frequency)}{sample\ size} \]

In the given exercise, we calculated the sample mean using the provided frequency distribution table, resulting in \( \lambda = 2.0 \). This calculation is crucial as it serves as the estimated rate parameter for the Poisson distribution, which we subsequently use to find the maximum likelihood estimate for the probability of observing a specific number of events.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from \(N\left(\mu, \sigma^{2}\right)\). (a) If the constant \(b\) is defined by the equation \(P(X \leq b)=0.90\), find the mle of \(b\). (b) If \(c\) is given constant, find the mle of \(P(X \leq c)\).

Consider a location model (Example \(6.2 .2\) ) when the error pdf is the contaminated normal (3.4.14) with \(\epsilon\) as the proportion of contamination and with \(\sigma_{c}^{2}\) as the variance of the contaminated part. Show that the ARE of the sample median to the sample mean is given by $$e\left(Q_{2}, \bar{X}\right)=\frac{2\left[1+\epsilon\left(\sigma_{c}^{2}-1\right)\right]\left[1-\epsilon+\left(\epsilon / \sigma_{c}\right)\right]^{2}}{\pi}$$ Use the hint in Exercise \(6.2 .5\) for the median. (a) If \(\sigma_{c}^{2}=9\), use \((6.2 .34)\) to fill in the following table: $$\begin{array}{|l|l|l|l|l|}\hline \epsilon & 0 & 0.05 & 0.10 & 0.15 \\ \hline e\left(Q_{2}, X\right) & & & & \\\\\hline\end{array}$$ (b) Notice from the table that the sample median becomes the "better" estimator when \(\epsilon\) increases from \(0.10\) to \(0.15 .\) Determine the value for \(\epsilon\) where this occurs [this involves a third-degree polynomial in \(\epsilon\), so one way of obtaining the root is to use the Newton algorithm discussed around expression \((6.2 .32)]\).

Rao (page 368,1973 ) considers a problem in the estimation of linkages in genetics. McLachlan and Krishnan (1997) also discuss this problem and we present their model. For our purposes, it can be described as a multinomial model with the four categories \(C_{1}, C_{2}, C_{3}\), and \(C_{4} .\) For a sample of size \(n\), let \(\mathbf{X}=\left(X_{1}, X_{2}, X_{3}, X_{4}\right)^{\prime}\) denote the observed frequencies of the four categories. Hence, \(n=\sum_{i=1}^{4} X_{i} .\) The probability model is $$\begin{array}{|c|c|c|c|}\hline C_{1} & C_{2} & C_{3} & C_{4} \\ \hline \frac{1}{2}+\frac{1}{4} \theta & \frac{1}{4}-\frac{1}{4} \theta & \frac{1}{4}-\frac{1}{4} \theta & \frac{1}{4} \theta \\\\\hline\end{array}$$ where the parameter \(\theta\) satisfies \(0 \leq \theta \leq 1 .\) In this exercise, we obtain the mle of \(\theta\). (a) Show that likelihood function is given by $$L(\theta \mid \mathbf{x})=\frac{n !}{x_{1} ! x_{2} ! x_{3} ! x_{4} !}\left[\frac{1}{2}+\frac{1}{4} \theta\right]^{x_{1}}\left[\frac{1}{4}-\frac{1}{4} \theta\right]^{x_{2}+x_{3}}\left[\frac{1}{4} \theta\right]^{x_{4}}$$ (b) Show that the log of the likelihood function can be expressed as a constant (not involving parameters) plus the term $$x_{1} \log [2+\theta]+\left[x_{2}+x_{3}\right] \log [1-\theta]+x_{4} \log \theta$$ (c) Obtain the partial derivative with respect to \(\theta\) of the last expression, set the result to 0, and solve for the mle. (This will result in a quadratic equation which has one positive and one negative root.)

Let \(\left(X_{1}, Y_{1}\right),\left(X_{2}, Y_{2}\right), \ldots,\left(X_{n}, Y_{n}\right)\) be a random sample from a bivariate normal distribution with \(\mu_{1}, \mu_{2}, \sigma_{1}^{2}=\sigma_{2}^{2}=\sigma^{2}, \rho=\frac{1}{2}\), where \(\mu_{1}, \mu_{2}\), and \(\sigma^{2}>0\) are unknown real numbers. Find the likelihood ratio \(\Lambda\) for testing \(H_{0}: \mu_{1}=\mu_{2}=0, \sigma^{2}\) unknown against all alternatives. The likelihood ratio \(\Lambda\) is a function of what statistic that has a well- known distribution?

The Pareto distribution is a frequently used model in the study of incomes and has the distribution function $$F\left(x ; \theta_{1}, \theta_{2}\right)=\left\\{\begin{array}{ll} 1-\left(\theta_{1} / x\right)^{\theta_{2}} & \theta_{1} \leq x \\ 0 & \text { elsewhere }\end{array}\right.$$ where \(\theta_{1}>0\) and \(\theta_{2}>0 .\) If \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample from this distribution, find the maximum likelihood estimators of \(\theta_{1}\) and \(\theta_{2}\). (Hint: This exercise deals with a nonregular case.)
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