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Consider the Poisson distribution with parameter \(\lambda\). Find the maximum likelihood estimator of \(\lambda,\) based on a random sample of size \(n\).

Short Answer

Expert verified
The MLE for \(\lambda\) is \(\hat{\lambda} = \frac{1}{n} \sum_{i=1}^{n} x_i\).

Step by step solution

01

Understanding the Poisson Likelihood Function

To find the maximum likelihood estimator (MLE) of \(\lambda\), begin by considering the probability mass function of a Poisson-distributed random variable \(X_i\) given by \(P(X_i = x_i) = \frac{{\lambda^{x_i} e^{-\lambda}}}{{x_i!}}\). The likelihood function for a random sample \(X_1, X_2, \ldots, X_n\) is the product of individual probabilities: \(L(\lambda) = \prod_{i=1}^{n} \frac{{\lambda^{x_i} e^{-\lambda}}}{{x_i!}}\).
02

Simplifying the Likelihood Function

The likelihood function can be rewritten as \(L(\lambda) = \lambda^{\sum x_i} e^{-n\lambda} \prod_{i=1}^{n} \frac{1}{x_i!}\). For MLE, we can ignore the constant \(\prod_{i=1}^{n} \frac{1}{x_i!}\) since it does not depend on \(\lambda\). Thus, the simplified likelihood function is \(L(\lambda) \propto \lambda^{\sum x_i} e^{-n\lambda}\).
03

Finding the Log-Likelihood Function

To make the differentiation process easier, take the natural logarithm of the likelihood function: \(\log L(\lambda) = \sum x_i \log \lambda - n\lambda + C\), where \(C\) is a constant term from the factorials and can be ignored.
04

Differentiating the Log-Likelihood Function

Differentiate the log-likelihood function with respect to \(\lambda\): \(\frac{d}{d\lambda} \log L(\lambda) = \frac{\sum x_i}{\lambda} - n\).
05

Setting the Derivative to Zero

Set the derivative of the log-likelihood equal to zero to find the critical points: \(\frac{\sum x_i}{\lambda} - n = 0\). Simplifying this gives \(\sum x_i = n\lambda\).
06

Solving for Lambda

Solve the equation \(\sum x_i = n\lambda\) for \(\lambda\). Thus, the maximum likelihood estimator of \(\lambda\) is \(\hat{\lambda} = \frac{1}{n} \sum_{i=1}^{n} x_i\).

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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. These events must happen with a known constant mean rate and independently of the time since the last event. The distribution is particularly useful in situations where events occur randomly over time, yet the average number of events is consistent.- **Key formula**: The probability of observing exactly \( x \) events is given by the probability mass function \( P(X = x) = \frac{\lambda^x e^{-\lambda}}{x!} \), where \( \lambda \) is the average rate (mean) of occurrence within the interval.- **Applications**: It's widely used in various fields such as call center arrivals, number of decay events per unit time in radioactive materials, and modeling traffic flow over a network.The Poisson distribution is characterized by its single parameter \( \lambda \), which serves both as its mean and variance. Understanding this distribution is crucial when dealing with cases involving random events fixed in an interval.
Log-Likelihood Function
When performing maximum likelihood estimation, a crucial step is to construct the log-likelihood function. This function provides a simplified and often more computationally efficient way to find the maximum likelihood estimate of our parameter of interest, in this case, \( \lambda \) for the Poisson distribution.- **Initial Form**: For a random sample, the likelihood function is the product of all probabilities: \( L(\lambda) = \prod_{i=1}^{n} \frac{\lambda^{x_i} e^{-\lambda}}{x_i!} \). - **Log Transformation**: Computing the logarithm of the likelihood, known as the log-likelihood, transforms complex products into simpler sums. For the Poisson distribution, this becomes \( \log L(\lambda) = \sum x_i \log \lambda - n\lambda + C \), with \( C \) as a constant ignoring terms irrelevant to \( \lambda \).By taking the log, we manage the numerical intricacies and enhance the stability of computations, crucial when handling large datasets or complex models. It also simplifies the differentiation needed to find the MLE.
Parameter Estimation
Maximum Likelihood Estimation (MLE) is a powerful statistical method used for estimating the parameters of a statistical model. The primary goal is to find the parameter values that maximize the likelihood of the data given the model. In the context of a Poisson distribution, we seek to determine \( \lambda \) that makes the observed data most probable.- **Step in MLE**: Differentiate the log-likelihood function with respect to \( \lambda \), resulting in \( \frac{d}{d\lambda} \log L(\lambda) = \frac{\sum x_i}{\lambda} - n \).- **Solving for \( \lambda \)**: Setting the derivative equal to zero \( \frac{\sum x_i}{\lambda} - n = 0 \) and solving leads us to \( \hat{\lambda} = \frac{1}{n} \sum_{i=1}^{n} x_i \). This is the maximum likelihood estimator, representing the sample mean.The beauty of MLE lies in its versatility and efficiency. It harnesses the data's power to yield estimators that typically possess desirable properties, such as consistency and asymptotic normality. In practical situations, achieving a deeper understanding of MLE can significantly enhance the reliability of statistical analyses.

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Most popular questions from this chapter

Let three random samples of sizes \(n_{1}=20, n_{2}=10\), and \(n_{3}=8\) be taken from a population with mean \(\mu\) and variance \(\sigma^{2}\). Let \(S_{1}^{2}, S_{2}^{2},\) and \(S_{3}^{2}\) be the sample variances. Show that \(S^{2}=\left(20 S_{1}^{2}+10 S_{2}^{2}+8 S_{3}^{2}\right) / 38\) is an unbiased estimator of \(\sigma^{2}\).

Data on pull-off force (pounds) for connectors used in an automobile engine application are as follows: 79.3,75.1 , 78.2,74.1,73.9,75.0,77.6,77.3,73.8,74.6,75.5,74.0,74.7 75.9,72.9,73.8,74.2,78.1,75.4,76.3,75.3,76.2,74.9,78.0 75.1,76.8 (a) Calculate a point estimate of the mean pull-off force of all connectors in the population. State which estimator you used and why. (b) Calculate a point estimate of the pull-off force value that separates the weakest \(50 \%\) of the connectors in the population from the strongest \(50 \%\). (c) Calculate point estimates of the population variance and the population standard deviation. (d) Calculate the standard error of the point estimate found in part (a). Provide an interpretation of the standard error. (e) Calculate a point estimate of the proportion of all connectors in the population whose pull-off force is less than 73 pounds.

A random sample of \(n=9\) structural elements is tested for compressive strength. We know that the true mean compressive strength \(\mu=5500\) psi and the standard deviation is \(\sigma=100\) psi. Find the probability that the sample mean compressive strength exceeds 4985 psi.

The elasticity of a polymer is affected by the concentration of a reactant. When low concentration is used, the true mean elasticity is \(55,\) and when high concentration is used the mean elasticity is \(60 .\) The standard deviation of elasticity is 4 regardless of concentration. If two random samples of size 16 are taken, find the probability that \(\bar{X}_{\text {high }}-\bar{X}_{\text {low }} \geq 2\).

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