91Ó°ÊÓ

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 \(a ?\)

Step by step solution

01

Intuition for Unbiasedness of the Estimator

The maximum likelihood estimator \( \hat{a} = \max(X_i) \) cannot be an unbiased estimator for the parameter \( a \) because it is based on the largest observed value, which is always less than or equal to the true maximum \( a \). Thus, \( \hat{a} \) is likely to underestimate \( a \) because there is a very low probability that the sample maximum will exactly equal the boundary \( a \).

02

Understanding Expectation and Bias

Given that \( E(\hat{a}) = \frac{n a}{n+1} \), \( \hat{a}\) is biased since its expectation does not equal \( a \). The expectation suggests \( \hat{a} \) consistently underestimates \( a \). The bias is \( a - \frac{n a}{n+1} = \frac{a}{n+1} \), which approaches zero as \( n \to \infty \). Thus, for large \( n \), the estimator \( \hat{a} \) becomes approximately unbiased.

03

Proposing an Unbiased Estimator

An unbiased estimator for \( a \) can be proposed as \( \hat{a}_{unbiased} = \frac{n+1}{n} \max(X_i) \). This correction factor compensates for the bias, making the expectation of this estimator equal to the true value \( a \).

04

Deriving the Cumulative Distribution Function (CDF) of Y

Since \( Y = \max(X_i) \), we have \( P(Y \leq y) = P(X_1 \leq y) P(X_2 \leq y) \ldots P(X_n \leq y) \). For a uniform distribution, \( P(X_i \leq y) = \frac{y}{a} \). So, \( P(Y \leq y) = \left(\frac{y}{a}\right)^n \).

05

Deriving the Probability Density Function (PDF) of Y

The PDF, \( f(y) \), is the derivative of the CDF: \( f(y) = \frac{d}{dy} \left(\frac{y}{a}\right)^n = \frac{n y^{n-1}}{a^n} \) for \( 0 \leq y \leq a \), and \( 0 \) otherwise.

06

Assessing the Bias with the PDF

Using the derived PDF \( f(y) = \frac{n y^{n-1}}{a^n} \), we find that \( E(Y) = \int_{0}^{a} y \frac{n y^{n-1}}{a^n} \ dy = \frac{n a}{n+1} \), confirming that \( \hat{a} = \max(X_i) \) is indeed biased as its expectation is not \( a \).

07

Comparing Two Unbiased Estimators

Given \( V(\hat{a}_1) = \frac{a^2}{3n} \) and \( V(\hat{a}_2) = \frac{a^2}{n(n+2)} \), \( \hat{a}_2 \) has a smaller variance than \( \hat{a}_1 \) for \( n > 1 \). Since lower variance implies higher precision, \( \hat{a}_2 \) is a better estimator. Thus, \( \hat{a}_2 \) is a better estimator of \( a \) since it estimates \( a \) with less variability and hence, more reliability.

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Key Concepts

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

Understanding Unbiased Estimators

In statistics, an unbiased estimator is a statistical tool that, on average, accurately predicts the true value of a parameter. This means that if you calculated the estimator repeatedly with random samples, the average of your calculations would converge to the true parameter value. However, not all estimators are unbiased. For example, the maximum likelihood estimator (MLE) for the upper bound of a uniform distribution, represented as \( \hat{a} = \max(X_i) \), is biased because it tends to underestimate the true parameter \( a \). This is because the largest observed value in the sample cannot exceed \( a \), making it unlikely to equal \( a \) unless \( a \) is sampled exactly, which is a rare occurrence. Understanding and identifying unbiased estimators is crucial because they give more reliable results compared to biased ones.

Exploring Uniform Distribution

A uniform distribution is a type of probability distribution where all outcomes are equally likely within a given range. When values of \( X \) are uniformly distributed between 0 and \( a \), each value within this range is as probable as any other, meaning the probability density function (PDF) is constant. Therefore, each value \( x \) satisfies \( 0 \leq x \leq a \), and it's equally likely to occur. This property of uniform distribution makes calculations like deriving distribution functions somewhat simpler, as we exploit the symmetry and equal probability of occurrence. In practice, knowing how to deal with uniform distributions helps in constructing and interpreting models that assume such patterns, like when estimating an unknown maximum \( a \).

Demystifying Estimator Bias

Estimator bias is the difference between the expected value of an estimator and the true value of the parameter being estimated. If an estimator consistently overestimates or underestimates the parameter, it’s considered biased. The bias reflects systematic errors in estimation rather than random errors and, in certain scenarios, can be corrected. For example, if the maximum likelihood estimator for \( a \) is \( \hat{a} = \max(X_i) \) and is found to be biased, correcting with a factor like \((n+1)/n\) can yield an unbiased estimator \( \hat{a}_{unbiased} = \frac{n+1}{n} \max(X_i) \). This new estimator adjusts for the tendency to underestimate, bringing its expected value in line with the parameter \( a \), thus mitigating the estimator bias. Identifying and correcting biases in estimators ensures more accurate statistical predictions.

Variance Comparison in Estimators

When evaluating estimators, comparing their variance is a key step. Variance indicates how much the estimated values fluctuate around the expected value—lower variance often implies more reliable and stable estimates. For instance, in the context of estimating \( a \) from a uniform distribution, we have two unbiased estimators: \( \hat{a}_1 = 2\bar{X} \) with variance \( V(\hat{a}_1) = \frac{a^2}{3n} \), and \( \hat{a}_2 = \left[(n+1)/n\right] \max(X_i) \) with variance \( V(\hat{a}_2) = \frac{a^2}{n(n+2)} \). Since \( \hat{a}_2 \) has a smaller variance than \( \hat{a}_1 \) when \( n > 1 \), it is considered to be a more precise estimator. The choice of estimator impacts not just the bias but also how consistently it produces close approximations to the true parameter. Therefore, selecting an estimator with lower variance often leads to improved accuracy and reliability.