Question

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

Short answer

MLE for \(\theta\) is \(\hat{\theta} = \max\{x_1,...,x_n\}\). The constant \(c\) so that \(E(c \hat{\theta})=\theta\) is \(c = 1/(n+1)\). The MLE for the median of the distribution given the sample is \(\hat{\theta}/2 = \max\{x_1,...,x_n\}/2\), which is a consistent estimator.

Step-by-step solution

Setting Up The Likelihood Function

The likelihood function of the given distribution is \[L(\theta; x)=\prod_{i=1}^{n} f(x_{i} ; \theta)=\prod_{i=1}^{n} \frac{2 x_{i}}{\theta^{2}}, \quad 0<x_{i} \leq \theta\] for each observation \(x_i\) and parameter \(\theta\).

Calculating the Log-Likelihood and its Derivative

Take logarithm of the likelihood function to get the log-likelihood function which simplifies computations: \[\ln L(\theta; x)=\sum_{i=1}^{n} \ln f(x_{i} ; \theta)=\sum_{i=1}^{n} \ln \left(\frac{2 x_{i}}{\theta^{2}}\right)=2n\ln2 + \sum_{i=1}^{n} \ln x_{i} - 2n\ln\theta.\] Then derive this log-likelihood function with respect to \(\theta\) to find the first order condition: \[\frac{d (\ln(L(\theta; x)))}{d\theta}=-\frac{2n}{\theta}.\]

Finding the Maximum Likelihood Estimate

To find the MLE for \(\theta\), set the derivative of the log-likelihood function equal to zero and solve for \(\theta\): \[0=-\frac{2n}{\theta},\] which gives the equation no solution, implying that the log-likelihood function is strictly decreasing in \(\theta\). Therefore, \(\theta\) takes its smallest possible value in the support as \( \hat{\theta} = \max\{x_1,...,x_n\}\).

Finding the Value of \(c\)

In this step, to find the value of the constant \(c\) so that \(E(c \hat{\theta})=\theta\), first observe that for \(0 < x \leq \theta\), the cdf of \( \hat{\theta}\) is \(F_{\hat{\theta}}(x) = P(\hat{\theta} \leq x) = P(\max\{x_1,...,x_n\} \leq x) = (x / \theta)^n\). Its pdf is the derivative of the cdf \(f_{\hat{\theta}}(x) = n (x/\theta)^{n-1} (1/\theta)\). Now calculate the expected value of \(c \hat{\theta}\) and set it equal to \(\theta\). This gives \[E(c \hat{\theta}) = \int_0^\theta c x f_{\hat{\theta}}(x) dx = c \int_0^\theta x n (x/\theta)^{n-1} (1/\theta) dx = c n/\theta \int_0^\theta x (x/\theta)^{n-1} dx = c n/\theta (\theta^{2}/(n+1)) = \theta.\] Solving for \(c\), we get \(c = 1/(n+1)\).

Determining The MLE for The Median

The pdf can be rearranged as \(f(x ; \theta)=2 \theta^{-2} x = 2(1/\theta^2)x\), which is a Uniform(0, 1) scaled by \(\theta\). Therefore, the median is \(\theta/2\) since the median of a Uniform(0,1) distribution is 0.5. Hence, the MLE for the median of the distribution given the sample is \(\hat{\theta}/2 = \max\{x_1,...,x_n\}/2\).

Proving The Consistency of The Estimator

Will show that the MLE for the median is a consistent estimator meaning it approaches the true median as the sample size increases. For large n, \(\hat{\theta}\) converges in probability to \(\theta\) because the maximum of a sample from a continuous distribution converges in probability to that distribution's supremum. Therefore, \(\hat{\theta}/2\) converges in probability to \(\theta/2\), the true median. Hence, the MLE for the median is a consistent estimator.

Key concepts

Probability Density Function

The probability density function (PDF) is a fundamental concept in statistics, essential for understanding how probability is distributed across a range of possible values in continuous random variables. It represents the likelihood of a random variable taking on a specific value.

For example, in the given exercise, the PDF for a random variable X is defined as \(f(x ; \theta)=2x / \theta^{2}\), where \(0
By solving \( E(c \theta) = \theta\), which leads to the conclusion that \(c = 1/(n+1). \theta\). The value of \(c\) acts as a scalar to adjust the estimator so that its expected value matches the parameter \(\theta\), thereby unifying the scale between the sample and the population.

Log-Likelihood

The log-likelihood is a transformation of the likelihood function, which is used to make the process of finding the maximum likelihood estimation (MLE) more mathematically tractable. Taking the natural logarithm of the likelihood function simplifies the product into a sum, making it easier to handle, especially with large sample sizes.

In the exercise's context, the log-likelihood function is obtained by taking the natural log of the likelihood, resulting in the expression: \(\ln L(\theta; x)=2n\ln2 + \sum_{i=1}^{n} \ln x_{i} - 2n\ln\theta.\) Deriving this function with respect to \(\theta\) and setting it to zero would typically provide the MLE. However, as the derivative \(-\frac{2n}{\theta}\) does not equal zero for any value of \(\theta\), it implies the estimate is found at the edge of the parameter space, specifically at the maximum observed value of the sample set.

Expected Value

The expected value, also known as the mean, is a measure of the center of a probability distribution. It is calculated as the weighted average of all possible values that the random variable can take on, with the weights being their corresponding probabilities.

In the exercise, the aim is to find the constant \(c\) that makes the expected value of \(c \theta\), is gathered from solving the integral \(E(c \theta) = \theta\). This computation represents the average of all possible values of \(c \theta\) acts as a scaling factor to align the estimator's expected value with the actual parameter \(\theta\).

Consistency of an Estimator

An estimator is considered consistent if it produces values that converge to the true parameter value as the sample size increases to infinity. Consistency ensures the reliability of an estimator in the long run since estimates will be closer to the actual parameter value with a bigger sample.

In this scenario, the MLE for the median is proven to be consistent, implying that as more data is collected, the estimated median, \(\hat{\theta}/2\), approaches the true median, \(\theta/2\), of the underlying distribution. This is under the concept that the maximum of the sample increasingly reflects the domain of the distribution as more observations are included, which implies that the estimator will provide an accurate value given a sufficiently large sample size.