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Let \(X\), the payoff from playing a certain game, have pmf $$ f(x ; \theta)=\left\\{\begin{array}{cc} \theta & x=-1 \\ (1-\theta)^{2} \theta^{x} & x=0,1,2, \ldots \end{array}\right. $$ a. Verify that \(f(x ; \theta)\) is a legitimate pmf, and determine the expected payoff. [Hint: Look back at the properties of a geometric random variable discussed in Chapter 3.] b. Let \(X_{1}, \ldots, X_{n}\) be the payoffs from \(n\) independent games of this type. Determine the mle of \(\theta\). [Hint: Let \(Y\) denote the number of observations among the \(n\) that equal \(-1\) that is, \(Y=\Sigma I\left(Y_{i}=-1\right)\), where \(I(A)=1\) if the event \(A\) occurs and 0 otherwise \(\\}\), and write the likelihood as a single expression in terms of \(\sum x_{i}\) and \(y\).] c. What is the approximate variance of the mle when \(n\) is large?

Short Answer

Expert verified
The pmf is legitimate, expected payoff \( E(X) = (-1)\theta + \frac{\theta}{1-\theta} \); MLE \( \hat{\theta} = \frac{Y}{Y + \sum x_i} \); variance \( \approx \frac{1}{n} \).

Step by step solution

01

Verify PMF

To check if \( f(x; \theta) \) is a legitimate probability mass function (pmf), we need to ensure that \( \sum_{x} f(x; \theta) = 1 \). So we calculate: \[P(X = -1) = \theta \P(X = x) = (1-\theta)^2 \theta^x, \; x = 0, 1, 2, \ldots\]The infinite sum of the probabilities for \( x = 0, 1, 2, \ldots \) is a geometric series:\[\sum_{x=0}^{\infty} (1-\theta)^2 \theta^x = (1-\theta)^2 \sum_{x=0}^{\infty} \theta^x \]Since \( \sum_{x=0}^{\infty} \theta^x = \frac{1}{1-\theta} \), Substituting these in:\[\theta + (1-\theta)^2 \cdot \frac{1}{1-\theta} = \theta + (1-\theta) = 1,\] Hence, \( f(x; \theta) \) is a legitimate pmf.
02

Determine Expected Payoff

Using the PMF for the expectations,\[E(X) = \sum_{x} x \cdot f(x; \theta) = (-1)\theta + \sum_{x=0}^{\infty} x \cdot (1-\theta)^2 \theta^x \]Using the properties of the geometric distribution, the expected value \( E(X) \) for \( x = 0,1,2,\ldots \) is:\[E(Y) = \frac{\theta}{1-\theta}\]Thus,\[E(X) = (-1)\theta + \frac{(1-\theta)^2 \cdot \theta}{1-\theta}\]
03

Find the MLE of \( \theta \)

To find the maximum likelihood estimator (MLE) of \( \theta \), we use the likelihood function for \( n \) independent games:\[L(\theta) = \prod_{i=1}^{n} f(x_i; \theta)\]Splitting into two parts, based on the hint,\[L(\theta) = \theta^Y \cdot \prod_{x_i eq -1} (1-\theta)^2 \theta^{x_i}\]Simplifies to:\[L(\theta) = \theta^Y \cdot (1-\theta)^{2(n-Y)} \cdot \theta^{\sum x_i}\]Taking the log likelihood:\[l(\theta) = Y \ln \theta + 2(n-Y) \ln (1-\theta) + \sum x_i \ln \theta\]Differentiating and setting the derivative equal to zero, solving gives:\[\hat{\theta} = \frac{Y}{Y + \sum x_i}\]
04

Find the Approximate Variance of the MLE

The approximate variance of the MLE \( \hat{\theta} \) when \( n \) is large can be found using the observed Fisher information, calculated as:\[I(\theta) = -E\left[\frac{\partial^2}{\partial \theta^2} \ln L(\theta)\right]\]The second derivative is:\[-\frac{Y}{\hat{\theta}^2} - \frac{2(n-Y)}{(1-\hat{\theta})^2} - \sum x_i \left(\frac{1}{\hat{\theta}^2}\right)\]Approximate variance:\[Var(\hat{\theta}) \approx \frac{1}{I(\hat{\theta})} = \frac{1}{n}\]

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Key Concepts

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

Geometric Distribution
The geometric distribution is a probability distribution that models the number of trials needed to get the first success in a sequence of independent and identically distributed Bernoulli trials. Each Bernoulli trial has two possible outcomes: success or failure. This distribution is particularly useful for modeling situations where you are interested in the number of attempts required to achieve the first success.
With the probability mass function (PMF) given by:
  • For the probability of failure: \( f(x) = (1-\theta)^2 \theta^x, \text{ for } x=0,1,2,\ldots \)
  • And for a special case where no successes occur: \( f(x) = \theta, x=-1 \)
The geometric distribution helps in classifying repeated independent experiments until the first success, making it a key concept in probability theory. Its expected value is calculated using the formula \( E(X) = \frac{1}{\theta} \, \text{ for regular geometric variables, adjusted here accordingly.} \).Understanding the geometric distribution paves the way to tackle related problems, making it easier to analyze trials and errors processes.
Maximum Likelihood Estimation
Maximum Likelihood Estimation (MLE) is a statistical method used to estimate the parameters of a probability distribution. The goal of MLE is to find the parameter values that maximize the likelihood function, which measures how likely it is to observe the given data under different parameter values. In essence, MLE selects the parameter set that makes the observed data most probable.
In our exercise, to find the MLE for \( \theta \), we define the likelihood function based on our PMF. We multiply the probabilities of Y observations equaling \(-1\) and the rest being non-negative outcomes. This results in a likelihood expression:
  • Likelihood function: \( L(\theta) = \theta^Y \cdot (1-\theta)^{2(n-Y)} \cdot \theta^{\sum x_i} \)
  • Log-likelihood: \( l(\theta) = Y \ln \theta + 2(n-Y) \ln (1-\theta) + \sum x_i \ln \theta \)
Solving this, by differentiating and equating to zero, the estimate for \( \theta \) (\( \hat{\theta} \)) is given by the observed proportion of negative payoffs compared to total observations.
MLE is invaluable because it provides a framework for estimating parameters that are consistent and efficient, often yielding results with desirable properties such as unbiasedness and minimum variance.
Fisher Information
Fisher Information is a concept in statistical estimation that measures the amount of information a random sample provides about an unknown parameter. It gives an idea about the precision of parameter estimates. The more Fisher information available, the more accurately you can estimate the parameter values.
In terms of mathematical representation, Fisher information for a parameter \(\theta\) is often denoted as \( I(\theta) \), and is calculated using the negative expected value of the second derivative of the log-likelihood function.
From the problem, for large samples, you apply:
  • Second derivative: \( -\frac{Y}{\hat{\theta}^2} - \frac{2(n-Y)}{(1-\hat{\theta})^2} - \sum x_i \left(\frac{1}{\hat{\theta}^2}\right) \)
  • Approximate variance: \( Var(\hat{\theta}) \approx \frac{1}{I(\hat{\theta})} = \frac{1}{n} \)
Understanding Fisher information is crucial as it not only affects how we gauge the reliability of the MLE, but also outlines the theoretical limits on the variance of unbiased estimators, providing an upper bound performance measure known as the Cramér-Rao lower bound.
Mathematical Expectation
Mathematical Expectation, or the expected value, is the weighted average of all possible values that a random variable can take. It plays a fundamental role in probability and statistics, offering a single summary measure of a random variable's long-term average outcome. This concept helps in understanding the central tendency and provides a means of predicting future values.
For discrete random variables like our geometric distribution, the expectation is computed by summing over all possible values of the random variable, each weighted by its probability:
  • Expected value formula: \( E(X) = \sum_{x} x \cdot f(x; \theta) \)
  • Utilized within the core exercise task to determine: \( E(X) = (-1)\theta + \frac{(1-\theta)^2 \cdot \theta}{1-\theta} \)
This captures an average tendency, even though individual outcomes may differ.
In practical scenarios, learning to calculate expected values offers insights into diverse fields from finance to insurance, where predicting likely outcomes supports decision making and risk assessment.

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Most popular questions from this chapter

At time \(t=0\), there is one individual alive in a certain population. A pure birth process then unfolds as follows. The time until the first birth is exponentially distributed with parameter \(\lambda\). After the first birth, there are two individuals alive. The time until the first gives birth again is exponential with parameter \(\lambda\), and similarly for the second individual. Therefore, the time until the next birth is the minimum of two exponential \((\lambda)\) variables, which is exponential with parameter 2\lambda. Similarly, once the second birth has occurred, there are three individuals alive, so the time until the next birth is an exponential rv with parameter \(3 \lambda\), and so on (the memoryless property of the exponential distribution is being used here). Suppose the process is observed until the sixth birth has occurred and the successive birth times are \(25.2,41.7,51.2,55.5,59.5,61.8\) (from which you should calculate the times between successive births). Derive the mle of \(\lambda\).

The fraction of a bottle that is filled with a particular liquid is a continuous random variable \(X\) with pdf \(f(x ; \theta)=\theta x^{\theta-1}\) for \(00\) ). a. Obtain the method of moments estimator for \(\theta\). b. Is the estimator of (a) a sufficient statistic? If not, what is a sufficient statistic, and what is an estimator of \(\theta\) (not necessarily unbiased) based on a sufficient statistic?

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from the normal distribution with known standard deviation \(\sigma\). a. Find the mle of \(\mu\). b. Find the distribution of the mle. c. Is the mle an efficient estimator? Explain. d. How does the answer to part (b) compare with the asymptotic distribution given by the second theorem?

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from the normal distribution with known mean \(\mu\) but with the variance \(\sigma^{2}\) as the unknown parameter. a. Find the information in a single observation and the Cramér-Rao lower bound. b. Find the mle of \(\sigma^{2}\). c. Find the distribution of the mle. d. Is the mle an efficient estimator? Explain. e. Is the answer to part (c) in conflict with the asymptotic distribution of the mle given by the second theorem? Explain.

Assume that the number of defects in a car has a Poisson distribution with parameter \(\lambda\). To estimate \(\lambda\) we obtain the random sample \(X_{1}\), \(X_{2}, \ldots, X_{n}\). a. Find the Fisher information in a single observation using two methods. b. Find the Cramér-Rao lower bound for the variance of an unbiased estimator of \(\lambda\). c. Use the score function to find the mle of \(\lambda\) and show that the mle is an efficient estimator. d. Is the asymptotic distribution of the mle in accord with the second theorem? Explain.

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