91Ó°ÊÓ

This problem shows you how to make a better blend of almost anything. Let \(x_{1}, x_{2}, \ldots x_{n}\) be independent random variables with respective variances \(\sigma_{1}^{2}\) \(\sigma_{2}^{2}, \ldots, \sigma_{n}^{2}.\) Let \(c_{1}, c_{2}, \ldots, c_{n}\) be constant weights such that \(0 \leq c_{i} \leq 1\) and \(c_{1}+c_{2}+\ldots+\) \(c_{n}=1 .\) The linear combination \(w=c_{1} x_{1}+c_{2} x_{2}+\ldots+c_{n} x_{n}\) is a random variable with variance $$\sigma_{w}^{2}=c_{1}^{2} \sigma_{1}^{2}+c_{2}^{2} \sigma_{2}^{2}+\cdots+c_{n}^{2} \sigma_{n}^{2}$$ (a) In your own words write a brief explanation regarding the following statement: The variance of \(w\) is a measure of the consistency or variability of performance or outcomes of the random variable \(w\). To get a more consistent performance out of the blend \(w\), choose weights \(c_{i}\) that make \(\sigma_{w}^{2}\) as small as possible. Now the question is how do we choose weights \(c_{i}\) to make \(\sigma_{w}^{2}\) as small as possible? Glad you asked! A lot of mathematics can be used to show the following choice of weights will minimize \(\sigma_{w}^{2}.\) $$c_{i}=\frac{\frac{1}{\sigma_{i}^{2}}}{\left[\frac{1}{\sigma_{i}^{2}}+\frac{1}{\sigma_{2}^{\frac{1}{2}}}+\cdots+\frac{1}{\sigma_{n}^{2}}\right]} \quad \text { for } i=1,2, \cdots, n$$ (Reference: Introduction to Mathematical Statistics, 4th edition, by Paul Hoel.) (b)Two types of epoxy resin are used to make a new blend of superglue. Both resins have about the same mean breaking strength and act independently. The question is how to blend the resins (with the hardener) to get the most consistent breaking strength. Why is this important, and why would this require minimal \(\sigma_{w}^{2} ?\) Hint: We don't want some bonds to be really strong while others are very weak, resulting in inconsistent bonding. Let \(x_{1}\) and \(x_{2}\) be random variables representing breaking strength (lb) of each resin under uniform testing conditions. If \(\sigma_{1}=8\) lb \(\sigma_{2}=12\) lb, show why a blend of about \(69 \%\) resin 1 and \(31 \%\) resin 2 will result in a superglue with smallest \(\sigma_{w}^{2}\) and most consistent bond strength. (c) Use \(c_{1}=0.69\) and \(c_{2}=0.31\) to compute \(\sigma_{w}\) and show that \(\sigma_{w}\) is less than both \(\sigma_{1}\) and \(\sigma_{2}\) The dictionary meaning of the word synergetic is "working together or cooperating for a better overall effect." Write a brief explanation of how the blend \(w=c_{1} x_{1}+c_{2} x_{2}\) has a synergetic effect for the purpose of reducing variance.

Step by step solution

01

Explanation of the Variance Minimization

The variance of a random variable, such as \(w\), measures how much outcomes differ from the mean. Lower variance indicates more consistent performance, meaning similar outcomes are likely. To achieve consistency with \(w\), we need to minimize \(\sigma_{w}^{2}\). The weights \(c_{i}\) are adjusted based on each component's variance to decrease overall variability and enhance stability.

02

Calculate Optimal Weights for Resin Blend

To minimize \(\sigma_{w}^{2}\), the weights must be chosen as given: \[c_{i} = \frac{\frac{1}{\sigma_{i}^{2}}}{\sum_{j=1}^{n} \frac{1}{\sigma_{j}^{2}}}\]For resins with variances \(\sigma_{1} = 8\) and \(\sigma_{2} = 12\) lb, calculate:\[c_{1} = \frac{\frac{1}{8^2}}{\frac{1}{8^2} + \frac{1}{12^2}} = \frac{\frac{1}{64}}{\frac{1}{64} + \frac{1}{144}} \]Calculate \(c_{1}\) to verify it is approximately 0.69, and \(c_{2} = 1 - c_{1}\) would then be 0.31.

03

Justification of Importance and Minimum Variance

Using epoxy resins with minimal variance ensures that bonds aren't overly strong or weak, providing reliable strength across the board. Consistency in bond strength is crucial, indicating that reducing \(\sigma_{w}^{2}\) to minimal values is essential for reliable performance.

04

Compute the Variance of the Blend

Now, compute \(\sigma_{w}^{2}\) using calculated weights:\[\sigma_{w}^{2} = (0.69)^2 \times 8^2 + (0.31)^2 \times 12^2 \]Solve for \(\sigma_{w}^{2}\):\[\sigma_{w}^{2} = 0.69^2 \times 64 + 0.31^2 \times 144 = 30.37 + 13.85 = 44.22\]Then, calculate \(\sigma_{w} = \sqrt{44.22} \approx 6.65\). This shows \(\sigma_{w}\) is less than both \(\sigma_{1} = 8\) and \(\sigma_{2} = 12\).

05

Explanation of Synergetic Effect

The combination \(w = c_{1} x_{1} + c_{2} x_{2}\) is synergetic because it leverages the variances of each resin wisely, resulting in an overall variance \(\sigma_{w}\) that is lower than the individual variances. This coordination achieves a more reliable performance, highlighting the benefits of strategic blending to enhance product quality.

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Key Concepts

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

Linear Combinations

A linear combination is a mathematical concept that refers to an expression constructed from a set of terms by multiplying each term by a constant and then adding the results. In the context of random variables and variance minimization, consider the formula for a linear combination:

• Given independent random variables, say, \(x_1, x_2, \ldots, x_n\), and constant weights \(c_1, c_2, \ldots, c_n\), the linear combination for these variables is expressed as:\[w = c_1x_1 + c_2x_2 + \ldots + c_nx_n\]
• The key here is that each random variable is weighted by a corresponding constant, allowing us to tailor the contribution of each random variable to the overall outcome \(w\).
• These weights can significantly modify the variability of \(w\)'s outcomes, making linear combinations a powerful way to manage and understand data.

Linear combinations are essential for summing up or blending different elements while allowing a consideration of their individual properties, such as mean or variance. By making wise choices with the constant weights \(c_i\), one can influence the characteristics of \(w\) effectively.

Independent Random Variables

When we talk about independent random variables, we're dealing with variables whose outcomes do not affect one another. This is a crucial factor in the context of variance minimization.

• If \(x_1, x_2, \ldots, x_n\) are independent random variables, it implies that the occurrence of one doesn't influence the probability distribution of the others.
• This is important because it simplifies the analysis by making the joint probability distribution of these variables a product of their individual distributions.
• Independence impacts the calculation of the variance of the linear combination because we focus purely on the individual variances, and not on any covariance, which would complicate matters considerably.

The independence allows us to maintain control over the problem by using more straightforward calculations to determine outcomes like the variance \(\sigma_w^2\). It prevents the introduction of unexpected variance into the mixture that might occur if the variables were dependent, which might complicate the objective of variance reduction.

Optimal Weight Selection

Selecting optimal weights for the linear combination of random variables is an elaborate task aimed at minimizing the resultant variability \(\sigma_w^2\). Here's how it is approached:

• The goal is to choose weights \(c_1, c_2, \ldots, c_n\) such that the variance of the blended variable \(w\) is minimized. This requires a specific mathematical formula, famously used in many fields including statistics and operations research.
• The weights are calculated based on the variances of the individual random variables. The formula given is:\[c_i = \frac{1/\sigma_i^2}{\sum_{j=1}^n (1/\sigma_j^2)}\]
• Each weight \(c_i\) contributes inversely to the square of its variable's variance. Hence, a random variable with a lower variance influences the outcome more heavily in order to stabilize \(w\).

This method ensures that the blend \(w\) has less variability than any individual component might cause on its own. It's a synergy of mathematical strategy and understanding of variability, leading to a more consistent performance outcome.