Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Let \(R\) be binomial with probability \(\pi\) and denominator \(m\), and consider estimators of \(\pi\) of form \(T=(R+a) /(m+b)\), for \(a, b \geq 0\). Find a condition under which \(T\) has lower mean squared error than the maximum likelihood estimator \(R / m\), and discuss which is preferable when \(m=5,10\).

Short Answer

Expert verified
Estimator T can have lower MSE than MLE if the inequality in Step 6 is satisfied. For small sample sizes (e.g., m=5), T can perform better; for larger samples (e.g., m=10), MLE is often preferable.

Step by step solution

01

Define the Estimators

We are given two estimators: the maximum likelihood estimator (MLE) \( \hat{\pi}_{MLE} = \frac{R}{m} \) and the estimator \( T = \frac{R+a}{m+b} \). Both are used to estimate the binomial probability \( \pi \).
02

Calculate Expected Value and Variance of MLE

The expected value of MLE is \( E[\hat{\pi}_{MLE}] = E\left[\frac{R}{m}\right] = \frac{1}{m} E[R] = \frac{1}{m} (m\pi) = \pi \). The variance is \( \text{Var}(\hat{\pi}_{MLE}) = \frac{1}{m^2} \text{Var}(R) = \frac{1}{m^2} (m\pi(1-\pi)) = \frac{\pi(1-\pi)}{m} \).
03

Calculate Expected Value and Variance of Estimator T

The expected value is \( E[T] = E\left[\frac{R+a}{m+b}\right] = \frac{E[R]+a}{m+b} = \frac{m\pi + a}{m+b} \). The variance is \( \text{Var}(T) = \frac{1}{(m+b)^2} \text{Var}(R) = \frac{m\pi(1-\pi)}{(m+b)^2} \).
04

Calculate Mean Squared Error (MSE) of MLE

The MSE of the MLE is \( MSE(\hat{\pi}_{MLE}) = E[(\hat{\pi}_{MLE} - \pi)^2] = \text{Var}(\hat{\pi}_{MLE}) = \frac{\pi(1-\pi)}{m} \) since the MLE is unbiased.
05

Calculate Mean Squared Error (MSE) of T

The MSE of T is given by \( MSE(T) = E[(T - \pi)^2] = (E[T] - \pi)^2 + \text{Var}(T) = \left(\frac{m\pi + a}{m+b} - \pi\right)^2 + \frac{m\pi(1-\pi)}{(m+b)^2} \).
06

Determine Condition for T to have Lower MSE than MLE

Set \( MSE(T) < MSE(\hat{\pi}_{MLE}) \). This gives us the inequality:\[\left(\frac{m\pi + a}{m+b} - \pi\right)^2 + \frac{m\pi(1-\pi)}{(m+b)^2} < \frac{\pi(1-\pi)}{m}\]This inequality must be solved to find values of \( a \) and \( b \) that satisfy the condition for any given \( m \).
07

Discuss Preferable Estimator for m=5,10

For small \( m \), introducing \( a \) and \( b \) can stabilize the estimator T by reducing its variance, potentially lowering MSE. When \( m=5 \), using \( T \) with optimal \( a \) and \( b \) may yield lower MSE if chosen properly. However, for larger \( m \) like \( m=10 \), MLE becomes more reliable due to its unbiased nature, and differences in MSE become smaller, thus MLE can be preferable.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Maximum Likelihood Estimation (MLE)
Maximum Likelihood Estimation, often abbreviated as MLE, is a fundamental statistical method used to estimate the parameters of a model. The core idea of MLE is to select the parameter values that maximize the likelihood of observing the given data. In the context of a binomial distribution, like the one described in our exercise with the estimator \( \hat{\pi}_{MLE} = \frac{R}{m} \), the MLE aims to provide the value of \( \pi \) that makes the observed binomial random variable \( R \) most probable.
MLE is a popular choice because:
  • It often provides unbiased parameter estimates, meaning it reflects the true parameter value on average over many samples.
  • The variance of MLE decreases as the sample size \( m \) increases, making it more precise with larger datasets.
  • In many scenarios, MLE remains consistent and efficient, especially when data approaches normality.
However, MLE might not always be the best, particularly with small sample sizes, due to potential instability and high variance.
Mean Squared Error (MSE)
Mean Squared Error (MSE) is a crucial criterion used to evaluate the quality of an estimator. It combines both the variance of the estimator and its bias to indicate how far estimator predictions deviate from the true parameter value. The MSE is calculated using the formula:
\[ MSE(T) = (E[T] - \pi)^2 + \text{Var}(T) \]
where \( E[T] \) is the expected value of the estimator and \( \text{Var}(T) \) is its variance. Essentially, the MSE can be broken down into:
  • The square of the bias (if the estimator is not unbiased).
  • The variance, showing the dispersion of estimator values.
A lower MSE indicates a more accurate and reliable estimator. In our exercise, we evaluate the condition for which the estimator \( T \) has a lower MSE compared to the MLE \( \hat{\pi}_{MLE} = \frac{R}{m} \). Attention to MSE is essential when deciding which estimator to use, especially when balancing between bias and variance tradeoffs.
Statistical Estimators
Statistical estimators are functions of sample data used to infer unknown population parameters. In the provided exercise, we explore two different estimators: the MLE \( \hat{\pi}_{MLE} = \frac{R}{m} \) and the estimator form \( T = \frac{R+a}{m+b} \) with \( a, b \geq 0 \). While the primary goal of any estimator is to make the best guess of a population parameter, they can have different properties in terms of bias and variance.
Estimators have certain desirable properties:
  • Unbiasedness: An estimator is unbiased if its expected value equals the true parameter value.
  • Consistency: As the sample size increases, the estimator should converge to the true parameter value.
  • Efficiency: Among unbiased estimators, the one with the smallest variance is considered more efficient.
Understanding these properties helps in choosing the right estimator for specific scenarios, like whether to introduce parameters \(a\) and \(b\) to potentially reduce variance in smaller samples.
Bias-Variance Tradeoff
The bias-variance tradeoff is a fundamental concept in statistics and machine learning that explores the balance between two sources of error. An estimator's total prediction error comprises bias, variance, and irreducible error. When choosing an estimator, one must consider:
  • Bias: The error introduced by approximating the real-world problem with a simplified model. High bias can cause an estimator to miss the relevant relations between features and targets. For example, an estimator \(T\) with added constants \(a\) and \(b\) may introduce bias.
  • Variance: The amount by which the estimate would change if we used a different dataset. Estimators with high variance may vary wildly from one sample to the next. Introducing \(a\) and \(b\) in estimator \(T\) can reduce variance, especially for small \(m\).
The tradeoff involves finding the right balance where neither bias nor variance dominates, resulting in a lower MSE, as seen in our exercise. Particularly, estimator T is examined to potentially have a lower MSE than MLE for smaller values of \(m\), like \(m=5\), by tweaking \(a\) and \(b\) to stabilize predictions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(\bar{Y}\) be the average of a random sample from the uniform density on \((0, \theta)\). Show that \(2 \bar{Y}\) is unbiased for \(\theta\). Find a sufficient statistic for \(\theta\), and obtain an estimator based on it which has smaller variance. Compare their mean squared errors.

One natural transformation of a binomial variable \(R\) is reversal of 'success' and 'failure'. Show that this maps \(R\) to \(m-R\), where \(m\) is the denominator, and that the induced transformation on the parameter space maps \(\pi\) to \(1-\pi\). Which of the critical regions (a) \(\mathcal{Y}_{1}=\\{0,1,20\\}\), (b) \(\mathcal{Y}_{2}=\\{0,1,19,20\\}\), (c) \(\mathcal{Y}_{3}=\\{0,1,10,19,20\\}\) (d) \(\mathcal{Y}_{4}=\\{8,9,10,11,12\\}\), is invariant for testing \(\pi=\frac{1}{2}\) when \(m=20 ?\) Which is preferable and why?

Let \(T=a \sum\left(Y_{j}-\bar{Y}\right)^{2}\) be an estimator of \(\sigma^{2}\) based on a normal random sample. Find values of \(a\) that minimize the bias and mean squared error of \(T\).

(a) Let \(Y_{1}, \ldots, Y_{n}\) be a random sample from the exponential density \(\lambda e^{-\lambda y}, y>0, \lambda>0\) Say why an unbiased estimator \(W\) for \(\lambda\) should have form \(a / S\), and hence find \(a\). Find the Fisher information for \(\lambda\) and show that \(\mathrm{E}\left(W^{2}\right)=(n-1) \lambda^{2} /(n-2)\). Deduce that no unbiased estimator of \(\lambda\) attains the Cramér-Rao lower bound, although \(W\) does so asymptotically. (b) Let \(\psi=\operatorname{Pr}(Y>a)=e^{-\lambda a}\), for some constant \(a\). Show that $$ I\left(Y_{1}>a\right)= \begin{cases}1, & Y_{1}>a \\ 0, & \text { otherwise }\end{cases} $$ is an unbiased estimator of \(\psi\), and hence obtain the minimum variance unbiased estimator. Does this attain the Cramér-Rao lower bound for \(\psi\) ?

Let \(Y_{1}, \ldots, Y_{n}\) be a random sample from an unknown density \(f\). Let \(I_{j}\) indicate whether or not \(Y_{j}\) lies in the interval ( \(\left.a-\frac{1}{2} h, a+\frac{1}{2} h\right]\), and consider \(R=\sum I_{j}\). Show that \(R\) has a binomial distribution with denominator \(n\) and probability $$ \int_{a-\frac{1}{2} h}^{a+\frac{1}{2} h} f(y) d y $$ Hence show that \(R /(n h)\) has approximate mean and variance \(f(a)+\frac{1}{2} h^{2} f^{\prime \prime}(a)\) and \(f(a) / n h\), where \(f^{\prime \prime}\) is the second derivative of \(f\). What implications have these results for using the histogram to estimate \(f(a)\) ?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free