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Chapter 7: Problem 3

Let \(X\) be a random variable with the following probability distribution: $$ f(x)=\left\\{\begin{array}{ll} (\theta+1) x^{9}, & 0 \leq x \leq 1 \\ 0, & \text { otherwise } \end{array}\right. $$ Find the maximum likelihood estimator of \(\theta\) based on a random sample of size \(n\).

Short Answer

Expert verified
The MLE suggests checking around endpoint boundaries for a viable \(\theta\), typically near \(-1\).

Step by step solution

01

Understand the Likelihood Function

Given a random sample \(X_1, X_2, \ldots, X_n\) from the probability distribution, the likelihood function is formed by the product of the probability density functions of each observation. For this problem, the likelihood function is \(L(\theta) = \prod_{i=1}^{n} f(X_i) = \prod_{i=1}^{n} (\theta + 1) X_i^9\).
02

Simplify the Likelihood Function

Notice that since each \(X_i\) contributes \((\theta+1)X_i^9\), the product simplifies to \(L(\theta) = (\theta + 1)^n \prod_{i=1}^{n} X_i^9\). This represents the likelihood function in a simplified form.
03

Derive the Log-Likelihood Function

The log-likelihood is normally easier to work with. Take the natural logarithm of the likelihood function: \( \log L(\theta) = n\log(\theta + 1) + 9\sum_{i=1}^{n} \log X_i\).
04

Differentiate and Solve for Critical Values

Differentiate the log-likelihood function \( \log L(\theta)\) with respect to \(\theta\) and set the derivative to zero to find the critical points. The derivative is \( \frac{n}{\theta + 1} = 0\). This equation simplifies to solving \(n = 0\), which doesn't provide a viable solution directly in terms of \(\theta\).
05

Examine the Boundary for Maximum Likelihood

Given the form of our equation, consider endpoints or possible behaviors around boundaries. Here, examine the endpoint based on the assumption that \(x_i\) terms are bound between 0 and 1. The calculus indicates critical value behaviors around boundary situations such as \(-1\). Evaluate \(\theta = -1\) in this context, deriving from valid probability density attributes, to suggest the value may not move far from theoretical minimums in bounded definition.

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Key Concepts

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

Probability Distribution
A probability distribution describes how the values of a random variable are spread or distributed. In the case of our exercise, the random variable, \(X\), can take values between 0 and 1. The function given, \(f(x)\), specifies the probability density function (PDF), which further described how likely different outcomes are. Often, this is presented as a function that integrates (sums) to 1 over its defined range, ensuring it is a valid distribution.In this specific problem, the PDF is: - \((\theta+1)x^9\) for \(0 \leq x \leq 1\) - 0 otherwise The role of \(\theta\) is crucial as it affects the shape and behavior of the distribution. A good understanding of probability distribution helps in detecting how data surrounds the mean and suggesting inferential statistics conclusions.
Log-Likelihood Function
The log-likelihood function is derived from the likelihood and serves a crucial role in statistical estimation, especially in finding estimates like Maximum Likelihood Estimators (MLE). It transforms products of probabilities into sums which are more convenient to manage mathematically.In this exercise, we begin with the likelihood function constructed from a sample of size \(n\):\[L(\theta) = \prod_{i=1}^{n} (\theta + 1) X_i^9 = (\theta + 1)^n \prod_{i=1}^{n} X_i^9\]The log-likelihood function \(\log L(\theta)\) simplifies our task:\[\log L(\theta) = n\log(\theta + 1) + 9 \sum_{i=1}^{n} \log X_i\]Logarithms help in simplifying multiplication and division by turning them into addition and subtraction, making it easier to differentiate and solve for maximum values. This characteristic ultimately aids in estimating the parameter \(\theta\) more easily.Understanding and utilizing the log-likelihood function helps address how well the model aligns with the given data.
Random Variable
A random variable is a fundamental concept in probability and statistics, providing a numerical description of the outcome of a random phenomenon. In this context, our random variable \(X\) represents the outcomes which can take any value within a specified range.For our problem, \(X\) is defined with:- Range: 0 to 1- Distribution: Defined by \(f(x)\), the function given in the problemA critical aspect of random variables is their ability to handle real-world uncertainties by quantifying the effects of random processes in a structured manner. Through their usage, we can interpret complex data and find relationships and inferential statistics (e.g., statistical tests). In estimating \(\theta\), understanding attribute ranges and behavior is vital to obtaining an accurate result.
Probability Density Function
The probability density function (PDF) is a function that illustrates the likelihood of a random variable to take on a particular value. A PDF is mainly applicable when dealing with continuous random variables, where probabilities are determined over a range of values rather than for specific outcomes. In our exercise, the PDF for the random variable \(X\) is given by:\[f(x) = \begin{cases} (\theta+1)x^9, & 0 \leq x \leq 1 \0, & \text{otherwise}\end{cases}\]This function describes how likely different values of \(X\) are, given the parameter \(\theta\). It effectively fuels Maximum Likelihood Estimation (MLE) by providing the probability basis for the likelihood function.PDFs are critical because they provide insights into the intervals where values are more probable, plus they ensure that the integral over all possible values equals 1, aligning with the core requirement of probability distributions. This property is utilized in forming estimators and understanding distribution shapes.

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Most popular questions from this chapter

You plan to use a rod to lay out a square, each side of which is the length of the rod. The length of the rod is \(\mu\). which is unknown. You are interested in estimating the area of the square, which is \(\mu^{2}\). Because \(\mu\) is unknown, you measure it \(n\) times, obtaining observations \(X_{1}, X_{2} \ldots, X_{n}\). Suppose that each measurement is unbiased for \(\mu\) with variance \(\sigma^{2}\) (a) Show that \(\bar{X}^{2}\) is a biased estimate of the area of the square. (b) Suggest an estimator that is unbiased.

Let \(X\) be a geometric random variable with parameter \(p\). Find the maximum likelihood estimator of \(p\) based on a random sample of size \(n\).

Suppose that \(X\) is a normal random variable with unknown mean \(\mu\) and known variance \(\sigma^{2}\). The prior distribution for \(\mu\) is a normal distribution with mean \(\mu_{0}\) and variance \(\sigma_{0}^{2}\). Show that the Bayes estimator for \(\mu\) becomes the maximum likelihood estimator when the sample size \(n\) is large.

PVC pipe is manufactured with a mean diameter of 1.01 inch and a standard deviation of 0.003 inch. Find the probability that a random sample of \(n=9\) sections of pipe will have a sample mean diameter greater than 1.009 inch and less than 1.012 inch.

Let \(X_{1}, X_{2}, \ldots, X\) be uniformly distributed on the interval 0 to \(a\). Recall that the maximum likelihood estimator of \(a\) is \(\hat{a}=\max \left(X_{i}\right)\) (a) Argue intuitively why \(\hat{a}\) cannot be an unbiased estimator for \(a\). (b) Suppose that \(E(\hat{a})=n a /(n+1)\). Is it reasonable that \(\hat{a}\) consistently underestimates \(a\) ? Show that the bias in the estimator approaches zero as \(n\) gets large. (c) Propose an unbiased estimator for \(a\). (d) Let \(Y=\max \left(X_{i}\right)\). Use the fact that \(Y \leq y\) if and only if each \(X_{i} \leq y\) to derive the cumulative distribution function of \(Y\). Then show that the probability density function of \(Y\) is $$ f(y)=\left\\{\begin{array}{ll} \frac{m y^{z-1}}{a^{x}}, & 0 \leq y \leq a \\\ 0, & \text { otherwise } \end{array}\right. $$ Use this result to show that the maximum likelihood estimator for \(a\) is biased. (e) We have two unbiased estimators for \(a\) : the moment estimator \(\hat{a}_{1}=2 \bar{X}\) and \(\hat{a}_{2}=[(n+1) / n] \max \left(X_{i}\right),\) where \(\max \left(X_{i}\right)\) is the largest observation in a random sample of size \(n\). It can be shown that \(V\left(\hat{a}_{1}\right)=a^{2} /(3 n)\) and that \(V\left(\hat{a}_{2}\right)=a^{2} /[n(n+2)]\). Show that if \(n>1, \hat{a}_{2}\) is a better estimator than \(\hat{a}\). In what sense is it a better estimator of \(a ?\)
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