91Ó°ÊÓ

The concept of expected value is fundamental in statistics and probability. It is essentially the weighted average of all possible outcomes of a random variable, giving us a measure of the center of a probability distribution. In our exercise, the expected value is used to determine where the average measurement will likely fall when using two different instruments, described as the random variables \(X_1\) and \(X_2\). Given that the expectation is linear, when we form the weighted average \( \bar{\mu} = w X_1 + (1-w) X_2 \), we calculate its expected value as:

• \[ E(\bar{\mu}) = wE(X_1) + (1-w)E(X_2) \]
• Since both instruments estimate the true value \( \mu \), we have \(E(X_1) = \mu\) and \(E(X_2) = \mu\).
• This leads to \( E(\bar{\mu}) = w\mu + (1-w)\mu = \mu \), meaning that as we expect, our weighted estimate remains unbiased.

Understanding expected value helps us ensure that our statistical methods or instruments do not systematically overestimate or underestimate the true value.

Variance refers to how much the values of a random variable differ from its expected value. It is a measure of the spread of a set of values. When we talk about minimizing variance, we're attempting to decrease the uncertainty in our measurements so our results are closer to the expected value, which, in this scenario, is \(\mu\). Our goal in this exercise is to adjust the weight \(w\) so that the variance of the weighted average \( \bar{\mu} \) is as small as possible. Calculating variance for \( \bar{\mu} \) we have:

• \[ \text{Var}(\bar{\mu}) = w^2 \text{Var}(X_1) + (1-w)^2 \text{Var}(X_2) \]
• By substituting the known variances \( \sigma^2_1 \) and \( \sigma^2_2 \) for \(X_1\) and \(X_2\), we find:
• \[ \text{Var}(\bar{\mu}) = w^2 \sigma_1^2 + (1-w)^2 \sigma_2^2 \]

To find the optimal \(w\) that minimizes this variance, we differentiate the variance expression with respect to \(w\), then set it to zero. Solving gives the weight that reduces the variance the most:

• \[ w = \frac{\sigma_2^2}{\sigma_1^2 + \sigma_2^2} \]

This optimal choice of \(w\) balances the variances, making use of the instrument with lower variance more heavily.

The concept of independence is a critical assumption in probability that greatly simplifies calculations. Independent random variables are those whose outcomes do not affect each other. In our scenario, \(X_1\) and \(X_2\) measure the same distance but independently, implying no information overlaps or shares between both instruments.This independence provides a significant simplification when working with statistical properties such as variance, where the variance of two independent random variables \(A\) and \(B\) satisfies:

• \[ \text{Var}(A+B) = \text{Var}(A) + \text{Var}(B) \]

In this problem, because \(X_1\) and \(X_2\) are independent, we simplify the formula for the variance of \( \bar{\mu} \):

• \[ \text{Var}(\bar{\mu}) = w^2 \text{Var}(X_1) + (1-w)^2 \text{Var}(X_2) \]

Overall, the assumption of independence ensures that our computations of variance are straightforward and precise, allowing us to focus on variance minimization with confidence knowing no hidden dependencies are skewing the results.