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Chapter 6: Problem 9

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 50 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 found by first calculating the sample mean and then using that mean in the Poisson formula to solve for \(P(X=2)\). The final solution would be a numerical value.

Step by step solution

01

Calculate the Sample Mean

The first step to solving this problem is to calculate the sample mean. To do this, each value of x is multiplied with its respective frequency, and the results are added together. This total is divided by the total frequency to find the mean, noted as \(\lambda\). The equation for this would look like the following: \(\lambda = \frac{\Sigma(frequency \times x)}{total frequency}\)
02

Use the Poisson Formula

The Poisson formula is used to calculate the maximum likelihood estimate of \(P(X=2)\). The formula for this is as follows: \(P(X=k) = \frac{{e^{-\lambda} \lambda^{k}}}{k!}\), where \(k=2\) in this case and \(\lambda\) is the mean obtained in the first step.
03

Substitute and Calculate

Now substitute the values of \(k\) and \(\lambda\) from the previous steps into the Poisson formula, then solve the equation to find \(P(X=2)\).

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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. The key is that these events happen with a known constant mean rate and independently of the time since the last event. The variable \( X \) represents the number of events and can take values such as 0, 1, 2, etc.
For example, you might use the Poisson distribution to model the number of customer arrivals at a store during an hour.
  • It is characterized by a single parameter \( \lambda \), which is the average number of events in the given time frame or area.
  • The Poisson probability formula is given by \[ P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \] where \( k \) is the number of events of interest and \( e \approx 2.718 \) is the base of the natural logarithm.
  • This distribution is commonly used in various fields such as insurance, finance, and natural sciences for modeling random events.
Sample Mean Calculation
Sample mean calculation is essential to determine the parameter \( \lambda \) of the Poisson distribution when analyzing the frequency of random events. In the context of the given data, we start by determining the sample mean by taking each observed value \( x \), multiplying it by its frequency, and then dividing the total by the total number of observations. This results in an average value, which is equivalent to \( \lambda \) in Poisson terms.
  • The formula used for calculating the sample mean is: \[ \lambda = \frac{\Sigma(frequency \times x)}{total ext{ }frequency} \] where \( \Sigma \) denotes the sum over all values of \( x \).
  • In the exercise, this would be calculated as: \[ \lambda = \frac{7\times0 + 14\times1 + 12\times2 + 13\times3 + 6\times4 + 3\times5}{50} \]
  • Once the calculation is performed, \( \lambda \) represents the mean number of occurrences which is used in the Poisson formula.
Probability Calculation
Probability calculation in the context of the Poisson distribution involves using the sample mean \( \lambda \) to find the probability of a specific number of occurrences. Once you have \( \lambda \), you can calculate any event probability such as \( P(X=2) \) using the Poisson formula.
  • Taking the calculated mean from the sample data as \( \lambda \), substitute \( \lambda \) and \( k=2 \) into the formula: \[ P(X=2) = \frac{e^{-\lambda} \lambda^2}{2!} \]
  • Solve the equation by performing the operations:
    • Compute \( \lambda^2 \) by squaring the sample mean.
    • Compute \( e^{-\lambda} \) using the exponential function.
    • Calculate \( 2! \) which equals 2.
  • The result gives the likelihood of experiencing exactly 2 events in the given conditions.

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Most popular questions from this chapter

Let \(Y_{1}0\). (a) Show that \(\Lambda\) for testing \(H_{0}: \theta=\theta_{0}\) against \(H_{1}: \theta \neq \theta_{0}\) is \(\Lambda=\left(Y_{n} / \theta_{0}\right)^{n}\), \(Y_{n} \leq \theta_{0}\), and \(\Lambda=0\), if \(Y_{n}>\theta_{0}\) (b) When \(H_{0}\) is true, show that \(-2 \log \Lambda\) has an exact \(\chi^{2}(2)\) distribution, not \(\chi^{2}(1) .\) Note that the regularity conditions are not satisfied.

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 \(\operatorname{Pr}(X \leq b)=0.90\), find the mle of \(b\). (b) If \(c\) is given constant, find the mle of \(\operatorname{Pr}(X \leq c)\).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a Bernoulli \(b(1, \theta)\) distribution, where \(0<\theta<1\) (a) Show that the likelihood ratio test of \(H_{0}: \theta=\theta_{0}\) versus \(H_{1}: \theta \neq \theta_{0}\) is based upon the statistic \(Y=\sum_{i=1}^{n} X_{i} .\) Obtain the null distribution of \(Y\). (b) For \(n=100\) and \(\theta_{0}=1 / 2\), find \(c_{1}\) so that the test rejects \(H_{0}\) when \(Y \leq c_{1}\) or \(Y \geq c_{2}=100-c_{1}\) has the approximate significance level of \(\alpha=0.05 .\) Hint: Use the Central Limit Theorem.

Let \(X\) be \(N(0, \theta), 0<\theta<\infty\) (a) Find the Fisher information \(I(\theta)\). (b) If \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample from this distribution, show that the mle of \(\theta\) is an efficient estimator of \(\theta\). (c) What is the asymptotic distribution of \(\sqrt{n}(\widehat{\theta}-\theta) ?\)

Consider a location model $$X_{i}=\theta+e_{i}, \quad i=1, \ldots, n$$ where \(e_{1}, e_{2}, \ldots, e_{n}\) are iid with pdf \(f(z)\). There is a nice geometric interpretation for estimating \(\theta .\) Let \(\mathbf{X}=\left(X_{1}, \ldots, X_{n}\right)^{\prime}\) and \(\mathbf{e}=\left(e_{1}, \ldots, e_{n}\right)^{\prime}\) be the vectors of observations and random error, respectively, and let \(\mu=\theta 1\) where 1 is a vector with all components equal to one. Let \(V\) be the subspace of vectors of the form \(\mu_{i}\) i.e, \(V=\\{\mathbf{v}: \mathbf{v}=a \mathbf{1}\), for some \(a \in R\\} .\) Then in vector notation we can write the model as $$\mathbf{X}=\boldsymbol{\mu}+\mathbf{e}, \quad \boldsymbol{\mu} \in V$$
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