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Chapter 6: Problem 32

a. Let \(X_{1}, \ldots, X_{n}\) be a random sample from a uniform distribution on \([0, \theta]\). Then the mle of \(\theta\) is \(\hat{\theta}=Y=\max \left(X_{i}\right)\). Use the fact that \(Y \leq y\) iff each \(X_{i} \leq y\) to derive the cdf of \(Y\). Then show that the pdf of \(Y=\max \left(X_{i}\right)\) is $$ f_{Y}(y)=\left\\{\begin{array}{cl} \frac{n y^{n-1}}{\theta^{n}} & 0 \leq y \leq \theta \\ 0 & \text { otherwise } \end{array}\right. $$ b. Use the result of part (a) to show that the mle is biased but that \((n+1) \max \left(X_{i}\right) / n\) is unbiased.

Short Answer

Expert verified
The MLE \(\hat{\theta} = Y\) is biased, but \((n+1) \max(X_i) / n\) is unbiased for \(\theta\).

Step by step solution

01

CDF Derivation of Y

To find the CDF of the maximum of a set of i.i.d random variables \(Y = \max(X_1, X_2, ..., X_n)\), we start with:\[ P(Y \leq y) = P(\max(X_1, X_2, ..., X_n) \leq y). \]This implies that each \(X_i \leq y\). For a uniform distribution on \([0, \theta]\):\[ P(X_i \leq y) = \frac{y}{\theta}\] for \(0 \leq y \leq \theta\). As the \(X_i\) are independent, we have:\[ P(Y \leq y) = P(X_1 \leq y, X_2 \leq y, ..., X_n \leq y) = \left(\frac{y}{\theta}\right)^n.\]
02

PDF Derivation of Y

Once the CDF \(F_Y(y)\) is known, the PDF \(f_Y(y)\) can be derived by differentiating the CDF:\[ f_Y(y) = \frac{d}{dy} F_Y(y) = \frac{d}{dy} \left(\frac{y}{\theta}\right)^n = n \frac{y^{n-1}}{\theta^n}.\]Therefore, the PDF is \[ f_Y(y) = \begin{cases} \frac{n y^{n-1}}{\theta^n} & \text{if } 0 \leq y \leq \theta, \ 0 & \text{otherwise.} \end{cases} \]
03

Checking for Bias in MLE

The MLE estimator is \(\hat{\theta} = Y\). To check if it's biased, we compute \(E[\hat{\theta}]\):\[ E[Y] = \int_0^\theta y \cdot \frac{n y^{n-1}}{\theta^n} \, dy = \frac{n}{\theta^n} \int_0^\theta y^n \, dy.\]Evaluating the integral:\[ \int_0^\theta y^n \, dy = \frac{\theta^{n+1}}{n+1}.\]Thus, \[ E[Y] = \frac{n}{\theta^n} \cdot \frac{\theta^{n+1}}{n+1} = \frac{n\theta}{n+1}.\]This shows that \(E[Y] eq \theta\), indicating a bias.
04

Unbiased Estimator

To find an unbiased estimator, we set it such that its expected value equals \(\theta\). Consider \(E\left[\frac{(n+1)Y}{n}\right]\):\[ E\left[\frac{(n+1)Y}{n}\right] = \frac{n+1}{n} E[Y] = \frac{n+1}{n} \cdot \frac{n\theta}{n+1} = \theta.\]Thus, \((n+1) \max(X_i) / n\) is unbiased for \(\theta\).

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Key Concepts

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

Uniform Distribution
The concept of a uniform distribution is one of simplicity. Imagine you have a range of possible outcomes which all have an equal chance of happening. This is what a uniform distribution represents. It鈥檚 like rolling a fair die where each number from 1 to 6 has an equal probability of 1/6 of showing up.

In the context of a continuous uniform distribution, the outcomes don't just take on discrete values like 1, 2, or 3; they instead cover an entire range between a lower bound and an upper bound, denoted by 0 and \(\theta\). This is particularly relevant in statistics when you are trying to understand maximum likelihood estimation (MLE) from this type of distribution.

With a uniformly distributed sample \(X_1, X_2, ..., X_n\), all values are taken from the interval \([0, \theta]\). When dealing with the maximum of such random samples, the properties of the uniform distribution become crucial to derive estimators like the MLE.
Probability Density Function
The probability density function (PDF) is a key tool in statistics to describe the likelihood of a continuous random variable taking on specific values. Unlike the probability mass function, which you might be familiar with when dealing with discrete random variables, the PDF doesn鈥檛 give probabilities straight away but instead helps in calculating them.

For a uniform distribution on \([0, \theta]\), the PDF helps us understand the concentration of values across this range. When interested in the maximum of a sample from this distribution, knowing the PDF is essential.

In our exercise, the maximum value \(Y = \max(X_1, X_2, ..., X_n)\) from a sample of uniformly distributed values possesses its own PDF, derived from the cumulative distribution function (CDF). This CDF is \(F_Y(y) = \left(\frac{y}{\theta}\right)^n\) for \(0 \leq y \leq \theta\), and through differentiation, the PDF becomes \(f_Y(y) = \frac{n y^{n-1}}{\theta^n}\) for the same interval. This expression tells us how likely different maximum values are to occur, showing where outcomes are concentrated.
Bias of Estimator
Imagine you're trying to estimate a parameter, like \(\theta\), from your data using a statistic. The bias of an estimator refers to the deviation of the expected value of this estimator from the true value of the parameter. Ideally, an unbiased estimator has an expected value equal to the parameter itself.

In the context of MLE and our problem at hand, the MLE estimator \(\hat{\theta} = Y\) is not unbiased. Calculations show that the expected value \(E[Y] = \frac{n\theta}{n+1}\), indicating that \(E[Y] eq \theta\). Thus, \(\hat{\theta}\) underestimates \(\theta\), as \(E[Y]\) is typically less than \(\theta\).

However, we can recalibrate this estimator to remove bias. By taking \((n+1)Y/n\), this becomes an unbiased estimator since its expected value matches \(\theta\). Such adjustment techniques are common in statistics to increase the precision of parameter estimation.

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Most popular questions from this chapter

An investigator wishes to estimate the proportion of students at a certain university who have violated the honor code. Having obtained a random sample of \(n\) students, she realizes that asking each, "Have you violated the honor code?" will probably result in some untruthful responses. Consider the following scheme, called a randomized response technique. The investigator makes up a deck of 100 cards, of which 50 are of type I and 50 are of type II. Type I: Have you violated the honor code (yes or no)? Type II: Is the last digit of your telephone number a 0,1 , or 2 (yes or no)? Each student in the random sample is asked to mix the deck, draw a card, and answer the resulting question truthfully. Because of the irrelevant question on type II cards, a yes response no longer stigmatizes the respondent, so we assume that responses are truthful. Let \(p\) denote the proportion of honor- code violators (i.e., the probability of a randomly selected student being a violator), and let \(\lambda=P(\) yes response). Then \(\lambda\) and \(p\) are related by \(\lambda=.5 p+(.5)(.3)\). a. Let \(Y\) denote the number of yes responses, so \(Y \sim\) Bin \((n, \lambda)\). Thus \(Y / n\) is an unbiased estimator of \(\lambda\). Derive an estimator for \(p\) based on \(Y\). If \(n=80\) and \(y=20\), what is your estimate? [Hint: Solve \(\lambda=.5 p+.15\) for \(p\) and then substitute \(Y / n\) for \(\lambda .]\) b. Use the fact that \(E(Y / n)=\lambda\) to show that your estimator \(\hat{p}\) is unbiased. c. If there were 70 type I and 30 type II cards, what would be your estimator for \(p\) ?

The shear strength of each of ten test spot welds is determined, yielding the following data (psi): $$ \begin{array}{llllllllll} 392 & 376 & 401 & 367 & 389 & 362 & 409 & 415 & 358 & 375 \end{array} $$ a. Assuming that shear strength is normally distributed, estimate the true average shear strength and standard deviation of shear strength using the method of maximum likelihood. b. Again assuming a normal distribution, estimate the strength value below which \(95 \%\) of all welds will have their strengths.

Using a long rod that has length \(\mu\), you are going to lay out a square plot in which the length of each side is \(\mu\). Thus the area of the plot will be \(\mu^{2}\). However, you do not know the value of \(\mu\), so you decide to make \(n\) independent measurements \(X_{1}, X_{2}, \ldots, X_{n}\) of the length. Assume that each \(X_{i}\) has mean \(\mu\) (unbiased measurements) and variance \(\sigma^{2}\). a. Show that \(\bar{X}^{2}\) is not an unbiased estimator for \(\mu^{2}\). b. For what value of \(k\) is the estimator \(\bar{X}^{2}-k S^{2}\) unbiased for \(\mu^{2}\) ?

Two different computer systems are monitored for a total of \(n\) weeks. Let \(X_{i}\) denote the number of breakdowns of the first system during the \(i\) th week, and suppose the \(X_{i}\) 's are independent and drawn from a Poisson distribution with parameter \(\mu_{1}\). Similarly, let \(Y_{i}\) denote the number of breakdowns of the second system during the \(i\) th week, and assume independence with each \(Y_{i}\) Poisson with parameter \(\mu_{2}\). Derive the mle's of \(\mu_{1}, \mu_{2}\), and \(\mu_{1}-\mu_{2}\).

a. A random sample of 10 houses in a particular area, each of which is heated with natural gas, is selected and the amount of gas (therms) used during the month of January is determined for each house. The resulting observations are \(103,156,118,89,125,147,122,109,138,99\). Let \(\mu\) denote the average gas usage during January by all houses in this area. Compute a point estimate of \(\mu\). b. Suppose there are 10,000 houses in this area that use natural gas for heating. Let \(\tau\) denote the total amount of gas used by all of these houses during January. Estimate \(\tau\) using the data of part (a). What estimator did you use in computing your estimate? c. Use the data in part (a) to estimate \(p\), the proportion of all houses that used at least 100 therms. d. Give a point estimate of the population median usage (the middle value in the population of all houses) based on the sample of part (a). What estimator did you use?
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