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Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from the Poisson distribution with \(0<\theta \leq 2\). Show that the mle of \(\theta\) is \(\widehat{\theta}=\min \\{\bar{X}, 2\\}\).

Short Answer

Expert verified
The maximum likelihood estimator \(\widehat{\theta}\) for the Poisson distribution with the given condition is \(\min \{\bar{X}, 2\}\).

Step by step solution

01

Define the log likelihood function

From the definition of the Poisson probability mass function (PMF), the likelihood function can be written as \( L(\theta) = \prod_{i=1}^{n} \frac{\theta^{x_i}e^{-\theta}}{x_i!}\). The log-likelihood function is then \(l(\theta)= \log(L(\theta)) = \sum_{i=1}^{n} [x_i \log(\theta) - \theta - \log(x_i!)]\).
02

Compute the derivative of the log likelihood function

To find the maximum likelihood estimate (MLE), we take the derivative of the log-likelihood with respect to \(\theta\) and set it equal to 0. The derivative of \(l(\theta)\) is \(\frac{d}{d\theta}l(\theta) = \sum_{i=1}^{n} [x_i/\theta - 1]\).\n
03

Solve for theta

Setting the derivative equal to zero gives \(0 = \sum_{i=1}^{n} x_i/\theta - n\). Solving for \(\theta\) we obtain \(\theta = \bar{X}\).
04

Incorporate given condition

However, from given condition, \(0 < \theta \leq 2\), so the maximum likelihood estimate of \(\theta\) under this condition is \(\widehat{\theta} = \min \{\bar{X}, 2\}\).

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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 fundamental concept in probability theory and statistics. It describes the probability of a given number of events happening in a fixed interval of time or space. These events must occur with a known constant mean rate and be independent of the time since the last event. A classic example is the number of emails received in an hour.
The main characteristics of the Poisson distribution include:
  • **Parameter (\(\theta\)):** This is the mean number of events in the interval. It is also equal to the variance.
  • **Discrete Nature:** The distribution applies to non-negative integer counts (e.g., 0, 1, 2, ...).
  • **Skewness:** Typically right-skewed, but as \(\theta\) increases, it becomes more symmetric.
Understanding how the Poisson distribution works is essential for deriving estimates like the maximum likelihood estimation (MLE) and for making informed decisions based on data, such as setting limits on expected values.
Log-Likelihood Function
The log-likelihood function is a transformation of the likelihood function. Instead of working directly with the product of probabilities, the log-likelihood simplifies calculations by summing the logarithms of probabilities. This is particularly helpful when dealing with the Poisson distribution, where likelihood involves products of terms.
For a random sample \(X_1, X_2, \ldots, X_n\) from a Poisson distribution, the likelihood function is given by \(L(\theta) = \prod_{i=1}^{n} \frac{\theta^{x_i}e^{-\theta}}{x_i!}\). Applying the log transformation, the log-likelihood becomes:\[l(\theta) = \sum_{i=1}^{n} [x_i \log(\theta) - \theta - \log(x_i!)]\]This form is linear in terms of its constituents, making derivation simpler. The maximum value of the log-likelihood corresponds to the most plausible parameter value given the data, which is central to the concept of maximum likelihood estimation (MLE).
Derivative Calculation
Derivative calculation is critical for finding the maximum of the log-likelihood function, which gives us the MLE of the parameter. In the context of a Poisson distribution, we apply calculus to find where this derivative equals zero, pointing to a local maximum.
Taking the derivative of the log-likelihood function \(l(\theta) = \sum_{i=1}^{n} [x_i \log(\theta) - \theta - \log(x_i!)]\) with respect to \(\theta\), we get:\[\frac{d}{d\theta}l(\theta) = \sum_{i=1}^{n}\left( \frac{x_i}{\theta} - 1 \right)\]To find the MLE, set this derivative equal to zero:\[0 = \sum_{i=1}^{n} \frac{x_i}{\theta} - n\]Solving this equation for \(\theta\) determines the value that maximizes the likelihood function. In this case, it reveals \(\theta = \bar{X}\), the sample mean, before applying problem-specific constraints.
Statistical Inference
Statistical inference involves making deductions about a population parameter based on a sample. Maximum likelihood estimation (MLE) is a prominent method in statistical inference for estimating parameters.
In the given exercise, after deriving that \(\theta = \bar{X}\), we must reconcile this with the constraint \(0 < \theta \leq 2\). This is where statistical inference informs decision-making by considering additional information:
  • **Constraint Application:** Adjust the MLE considering practical or theoretical limits, here resulting in \(\hat{\theta} = \min\{\bar{X}, 2\}\)
  • **Parameter Interpretation:** Ensures the parameter estimates not only fit the data but also comply with known restrictions, enhancing relevance and accuracy.
Thus, statistical inference guides not just estimation but also evaluation of estimates in context-appropriate scenarios, crucial for precise and meaningful data analysis.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) represent a random sample from each of the distributions having the following pdfs: (a) \(f(x ; \theta)=\theta x^{\theta-1}, 0

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.

Suppose \(X_{1}, \ldots, X_{n}\) are iid with pdf \(f(x ; \theta)=2 x / \theta^{2}, \quad 0

Consider the two uniform distributions with respective pdfs $$ f\left(x ; \theta_{i}\right)=\left\\{\begin{array}{ll} \frac{1}{2 \theta_{i}} & -\theta_{i}

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 that has one positive and one negative root.)

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