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Chapter 7: Problem 17

Suppose that \(X_{1}\) and \(X_{2}\) are independent random variables having a common mean \(\mu\). Suppose also that \(\operatorname{Var}\left(X_{1}\right)=\sigma_{1}^{2}\) and \(\operatorname{Var}\left(X_{2}\right)=\sigma_{2}^{2} .\) The value of \(\mu\) is unknown and it is proposed to estimate \(\mu\) by a weighted average of \(X_{1}\) and \(X_{2}\). That is, \(\lambda X_{1}+(1-\lambda) X_{2}\) will be used as an estimate of \(\mu\), for some appropriate value of \(\lambda\). Which value of \(\lambda\) yields the estimate having the lowest possible variance? Explain why it is desirable to use this value of \(\lambda\).

Short Answer

Expert verified
The value of \(\lambda\) that yields the lowest possible variance of the weighted average estimator is \(\lambda = \frac{\sigma_{2}^{2}}{\sigma_{1}^{2} + \sigma_{2}^{2}}\). Using this value of \(\lambda\) is desirable because it results in an estimator with the lowest variance, meaning that the estimator is centered more closely around the true value of \(\mu\), making our estimate more accurate.

Step by step solution

01

Formulate the expression for the weighted average estimator's variance

We first need to find the variance of the estimator given by the weighted average of \(X_{1}\) and \(X_{2}\), which is the random variable \[Y = \lambda X_{1} + (1-\lambda) X_{2}\] To find the variance of \(Y\), we'll use the formula: \[\operatorname{Var}(Y)= \operatorname{Var}\left(\lambda X_{1} + (1-\lambda) X_{2}\right)\]
02

Apply properties of variances

Since \(X_1\) and \(X_2\) are independent random variables, we can make use of the properties of variances: \[\operatorname{Var}(Y) = \operatorname{Var}\left(\lambda X_{1}\right) + \operatorname{Var}\left((1-\lambda) X_{2}\right)\] Now, we also know that \(\operatorname{Var}(kX)=k^2 \operatorname{Var}(X)\) for any constant \(k\), so we can write: \[\operatorname{Var}(Y) = \lambda^2 \operatorname{Var}\left(X_{1}\right) + (1-\lambda)^2 \operatorname{Var}\left(X_{2}\right)\] Substitute the given variances of \(X_{1}\) and \(X_{2}\) to get: \[\operatorname{Var}(Y) = \lambda^2 \sigma_{1}^{2} + (1-\lambda)^2 \sigma_{2}^{2}\]
03

Minimize the variance of the estimator

To minimize the variance of the estimator, we need to find the value of \(\lambda\) that minimizes the equation: \[\operatorname{Var}(Y) = \lambda^2 \sigma_{1}^{2} + (1-\lambda)^2 \sigma_{2}^{2}\] To minimize the variance with respect to \(\lambda\), we can take the derivative of the equation and set it to zero: \[\frac{d}{d\lambda}\left(\lambda^2 \sigma_{1}^{2} + (1-\lambda)^2 \sigma_{2}^{2}\right) = 0\] Simplify and solve for \(\lambda\): \[2\lambda \sigma_{1}^{2} - 2(1-\lambda) \sigma_{2}^{2} = 0\]
04

Solve for lambda

Solve the equation for \(\lambda\): \[\lambda (2\sigma_{1}^{2} + 2\sigma_{2}^{2}) = 2\sigma_{2}^{2}\] \[\lambda = \frac{2\sigma_{2}^{2}}{2(\sigma_{1}^{2} + \sigma_{2}^{2})} = \frac{\sigma_{2}^{2}}{\sigma_{1}^{2} + \sigma_{2}^{2}}\]
05

Conclusion and explanation

The value of \(\lambda\) that yields the lowest possible variance of the weighted average estimator is: \[\lambda = \frac{\sigma_{2}^{2}}{\sigma_{1}^{2} + \sigma_{2}^{2}}\] Using this value of \(\lambda\) is desirable because it results in an estimator with the lowest variance. A lower variance means that the estimator is centered more closely around the true value of \(\mu\), making our estimate more accurate. Using this value of \(\lambda\) provides the best trade-off between the variances of \(X_{1}\) and \(X_{2}\) to achieve the most accurate estimation of the common mean \(\mu\).

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Key Concepts

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

Independent Random Variables
Understanding the concept of independent random variables is crucial when working with probability and statistics. Two random variables, such as
\(X_{1}\) and \(X_{2}\) in our exercise, are said to be independent if the occurrence of one does not affect the probability distribution of the other. This is foundational because it allows us to analyze the behavior of these variables separately before combining results.

For example, consider rolling two dice. The result of the first die doesn鈥檛 influence what will happen with the second鈥攊t鈥檚 equally probable to roll any number combination. This property drastically simplifies calculations involving joint probability distributions, variance, and other statistics. In the context of the given problem, independence posits that knowing the value of \(X_{1}\) gives no information about \(X_{2}\), and vice versa. When calculating the variance of a weighted average of independent variables, we can treat their variances additively, leading us to the next key concept.
Variance of Estimator
The variance of an estimator is a measure of how much the estimated values will spread around the expected value. Think of it as an indicator of precision; a smaller variance means the estimates are more consistently close to the actual value we're trying to predict.

In our exercise, the estimator is the weighted average of two random variables, \( \( \lambda X_{1} + (1-\lambda) X_{2} \) \). The variance of this estimator is given by \( \lambda^{2}\sigma_{1}^{2} + (1-\lambda)^{2}\sigma_{2}^{2} \), derived using the independence of \(X_{1}\) and \(X_{2}\) and the property that \( \operatorname{Var}(kX) = k^2 \operatorname{Var}(X) \) for any constant \(k\). Higher variance corresponds to less confidence in the estimator鈥檚 accuracy, which is why one of our key goals in statistical estimation is to find ways to minimize variance, thus enhancing the estimator's reliability.
Minimizing Variance
Minimizing the variance of an estimator is synonymous with increasing its accuracy. A lower variance indicates that the estimator is more consistently near the actual parameter we're trying to estimate; it鈥檚 reliable. When we have a weighted average estimator like \( \( \lambda X_{1} + (1-\lambda) X_{2} \) \), we aim to optimize \( \lambda \) so that the estimator's variance is as small as possible.

In the given problem, this optimization involves calculus, specifically finding the derivative of the variance with respect to \( \lambda \) and setting it to zero to find a minimum. This crucial step ensures that we have selected the most statistically efficient estimate of \( \mu \), which is the common mean of the independent random variables. The resulting value of \( \lambda \) provides a method to weigh the random variables鈥� variances against one another, thus achieving the balance that produces the minimum variance鈥攎eaning the most precise estimate of \( \mu \).

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Most popular questions from this chapter

A fair die is successively rolled. Let \(X\) and \(Y\) denote, respectively, the number of rolls necessary to obtain a 6 and a 5. Find (a) \(E[X]\) (b) \(E[X \mid Y=1]\); (c) \(E[X \mid Y=5]\).

Suppose that \(A\) and \(B\) each randomly, and independently, choose 3 of 10 objects. Find the expected number of objects (a) chosen by both \(A\) and \(B\); (b) not chosen by either \(A\) or \(B\); (c) chosen by exactly one of \(A\) and \(B\).

The game of Clue involves 6 suspects, 6 weapons, and 9 rooms. One of each is randomly chosen and the object of the game is to guess the chosen three. (a) How many solutions are possible? In one version of the game, after the selection is made each of the players is then randomly given three of the remaining cards. Let \(S, W\), and \(R\) be, respectively, the numbers of suspects, weapons, and rooms in the set of three cards given to a specified player. Also, let \(X\) denote the number of solutions that are possible after that player observes his or her three cards. (b) Express \(X\) in terms of \(S, W\), and \(R\). (c) Find \(E[X]\).

A player throws a fair die and simultaneously flips a fair coin. If the coin lands heads, then she wins twice, and if tails, then one-half of the value that appears on the die. Determine her expected winnings.

A deck of \(n\) cards, numbered 1 through \(n\), is thoroughly shuffled so that all possible \(n !\) orderings can be assumed to be equally likely. Suppose you are to make \(n\) guesses sequentially, where the \(i\) th one is a guess of the card in position \(i\). Let \(N\) denote the number of correct guesses. (a) If you are not given any information about your earlier guesses show that, for any strategy, \(E[N]=1\). (b) Suppose that after each guess you are shown the card that was in the position in question. What do you think is the best strategy? Show that under this strategy $$ \begin{aligned} E[N] &=\frac{1}{n}+\frac{1}{n-1}+\cdots+1 \\ & \approx \int_{1}^{n} \frac{1}{x} d x=\log n \end{aligned} $$ (c) Suppose that you are told after each guess whether you are right or wrong. In this case it can be shown that the strategy that maximizes \(E[N]\) is one which keeps on guessing the same card until you are told you are correct and then changes to a new card. For this strategy show that $$ \begin{aligned} E[N] &=1+\frac{1}{2 !}+\frac{1}{3 !}+\cdots+\frac{1}{n !} \\ &=e-1 \end{aligned} $$
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