/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 In Section \(5.1,\) we studied l... [FREE SOLUTION] | 91Ó°ÊÓ

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In Section \(5.1,\) we studied linear combinations of independent random variables. What happens if the variables are not independent? A lot of mathematics can be used to prove the following: Let \(x\) and \(y\) be random variables with means \(\mu_{x}\) and \(\mu_{y},\) variances \(\sigma_{x}^{2}\) and \(\sigma_{y}^{2}\) and population correlation coefficient \(\rho \text { (the Greek letter } r h o) .\) Let \(a\) and \(b\) be any constants and let \(w=a x+b y .\) Then, $$\begin{array}{l}\mu_{w}=a \mu_{x}+b \mu_{y} \\\\\sigma_{w}^{2}=a^{2} \sigma_{x}^{2}+b^{2} \sigma_{y}^{2}+2 a b \sigma_{x} \sigma_{y} \rho\end{array}$$ In this formula, \(\rho\) is the population correlation coefficient, theoretically computed using the population of all \((x, y)\) data pairs. The expression \(\sigma_{x} \sigma_{y}, \rho\) is called the covariance of \(x\) and \(y .\) If \(x\) and \(y\) are independent, then \(\rho=0\) and the formula for \(\sigma_{w}^{2}\) reduces to the appropriate formula for independent variables (see Section 5.1 ). In most real-world applications, the population parameters are not known, so we use sample estimates with the understanding that our conclusions are also estimates. Do you have to be rich to invest in bonds and real estate? No, mutual fund shares are available to you even if you aren't rich. Let \(x\) represent annual percentage return (after expenses) on the Vanguard Total Bond Index Fund, and let \(y\) represent annual percentage return on the Fidelity Real Estate Investment Fund. Over a long period of time, we have the following population estimates (based on Morningstar Mutual Fund Report). $$ \mu_{x} \approx 7.32 \sigma_{x} \approx 6.59 \mu_{y} \approx 13.19 \sigma_{y} \approx 18.56, \rho \approx 0.424 $$ (a) Do you think the variables \(x\) and \(y\) are independent? Explain. (b) Suppose you decide to put \(60 \%\) of your investment in bonds and \(40 \%\) in real estate. This means you will use a weighted average \(w=0.6 x+0.4 y\) Estimate your expected percentage return \(\mu_{w}\) and risk \(\sigma_{w}\) (c) Repeat part (b) if \(w=0.4 x+0.6 y\) (d) Compare your results in parts (b) and (c). Which investment has the higher expected return? Which has the greater risk as measured by \(\sigma_{w} ?\)

Short Answer

Expert verified
Variables are not independent. Part (c) has higher return and risk.

Step by step solution

01

Evaluate Independence (Part a)

The correlation coefficient \( \rho = 0.424 \) suggests that the random variables \( x \) and \( y \) are not independent because independence corresponds to \( \rho = 0 \). A correlation of 0.424 indicates some degree of linear relationship between \( x \) and \( y \).
02

Calculate Expected Return for w = 0.6x + 0.4y (Part b)

Given \( w = 0.6x + 0.4y \), the expected return \( \mu_w \) is given by \( \mu_w = a \mu_x + b \mu_y \) where \( a = 0.6 \) and \( b = 0.4 \). Plugging in the values, we get: \( \mu_w = 0.6 \times 7.32 + 0.4 \times 13.19 = 9.428 \).
03

Calculate Risk for w = 0.6x + 0.4y (Part b)

The variance of \( w \), \( \sigma_w^2 \), is given by \( a^2 \sigma_x^2 + b^2 \sigma_y^2 + 2ab\sigma_x\sigma_y \rho \). Calculating this: - \( a^2 \sigma_x^2 = 0.6^2 \times 6.59^2 = 15.5628 \)- \( b^2 \sigma_y^2 = 0.4^2 \times 18.56^2 = 55.0352 \)- \( 2ab\sigma_x\sigma_y \rho = 2 \times 0.6 \times 0.4 \times 6.59 \times 18.56 \times 0.424 = 25.8944 \)Thus, \( \sigma_w^2 = 15.5628 + 55.0352 + 25.8944 = 96.4924 \) and \( \sigma_w = \sqrt{96.4924} \approx 9.82 \).
04

Calculate Expected Return for w = 0.4x + 0.6y (Part c)

With \( w = 0.4x + 0.6y \), expected return \( \mu_w = a \mu_x + b \mu_y \). Hence, \( \mu_w = 0.4 \times 7.32 + 0.6 \times 13.19 = 11.082 \).
05

Calculate Risk for w = 0.4x + 0.6y (Part c)

The variance \( \sigma_w^2 \) is \( a^2 \sigma_x^2 + b^2 \sigma_y^2 + 2ab\sigma_x\sigma_y \rho \). Calculating:- \( a^2 \sigma_x^2 = 0.4^2 \times 6.59^2 = 6.9168 \)- \( b^2 \sigma_y^2 = 0.6^2 \times 18.56^2 = 123.8292 \)- \( 2ab\sigma_x\sigma_y \rho = 2 \times 0.4 \times 0.6 \times 6.59 \times 18.56 \times 0.424 = 25.8944 \)Therefore, \( \sigma_w^2 = 6.9168 + 123.8292 + 25.8944 = 156.64 \) and so \( \sigma_w = \sqrt{156.64} \approx 12.52 \).
06

Compare Returns and Risks (Part d)

Comparing the results:- Part (b): \( \mu_w = 9.428 \), \( \sigma_w \approx 9.82 \).- Part (c): \( \mu_w = 11.082 \), \( \sigma_w \approx 12.52 \).The investment in Part (c) has a higher expected return but also a higher risk (\( \sigma_w \)).

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Key Concepts

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

Random Variables
In statistics, random variables are a key concept and can be seen as numerical outcomes that are dependent on the outcome of a random event. They are essentially a way to assign numerical values to the results of a random process. For example, if you were to roll a six-sided die, the result of that roll would be a random variable. It could be a 1, 2, 3, 4, 5, or 6, each with an equal probability.
  • Discreteness and Continuity: Random variables can be discrete, like the die roll, where the outcomes are countable and distinct.
  • Probability Distributions: Each random variable has a probability distribution that tells you the probabilities of the outcomes. For discrete variables, this is a probability mass function, while continuous variables use a probability density function.
Understanding these distributions helps us to analyze events quantitatively. They allow us to compute measures like mean (expected value) and variance, giving us tools to understand variability and central tendencies in datasets.
Population Correlation Coefficient
The population correlation coefficient, often symbolized by the Greek letter \( \rho \), shows the strength and direction of a linear relationship between two random variables. It ranges from -1 to 1.
  • Value Interpretation: A \( \rho \) value of 1 indicates a perfect positive linear relationship, meaning as one variable increases, the other one does too. A \( \rho \) of -1 indicates a perfect negative linear relationship, meaning as one variable increases, the other decreases.
  • Zero Correlation: If \( \rho \) is 0, there's no linear relationship between the variables, suggesting independence.
In our context, \( \rho = 0.424 \) reveals some positive correlation between the bond and real estate funds, showing that as one increases, the other tends to also increase, but not perfectly. This statistic allows investors to understand and predict how two different investments might behave relative to one another.
Expected Return and Risk
Expected return and risk are crucial in understanding investments. Expected return, represented by the mean \( \mu \), indicates the average outcome you hope to achieve, while risk is quantified by variance or standard deviation, which shows how much the returns can vary.
  • Expected Return (\( \mu \)): Using constants to create a weighted average, you can predict the future returns of an investment portfolio. In the exercise, the calculation \( \mu_w = a \mu_x + b \mu_y \) helped determine the expected return of mixed investments in bonds and real estate.
  • Standard Deviation (\( \sigma \)): This measures the risk or variability of investment returns. A higher standard deviation means more risk, as the actual returns can vary widely from the expected return.
These calculations are central to strategic investment decisions, allowing investors to align risks with potential gains meaningfully.
Weighted Averages
Weighted averages are used when different components contribute unequally to the total. In finance, they help determine the average return of an investment portfolio composed of multiple assets, each having different importance.
  • Calculation Basics: A weighted average is calculated by multiplying each component by a factor (its weight), summing these products, and dividing by the total weight. In the case of investment, if the weights are percentages, they should add up to 1 or 100%.
  • Application in Portfolios: In the investment context, weighted averages help you understand combined returns and risks of different assets. If 60% of your money is in bonds and 40% in real estate, the weighted average return \( w = 0.6x + 0.4y \) gives the expected return \( \mu_w \).
Weighted averages simplify complex aggregate calculations, providing a more personalized insight into overall portfolio performance based on personal investment distributions.

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Most popular questions from this chapter

Over the past few years, there has been a strong positive correlation between the annual consumption of diet soda drinks and the number of traffic accidents. (a) Do you think increasing consumption of diet soda drinks causes traffic accidents? Explain. (b) What lurking variables might be causing the increase in one or both of the variables? Explain.

Wolf packs tend to be large extended family groups that have a well-defined hunting territory. Wolves not in the pack are driven out of the territory or killed. In ecologically similar regions, is the size of an extended wolf pack related to size of hunting region? Using radio collars on wolves, the size of the hunting region can be estimated for a given pack of wolves. Let \(x\) represent the number of wolves in an extended pack and \(y\) represent the size of the hunting region in \(\mathrm{km}^{2} / 1000 .\) From Denali National Park we have the following data. $$\begin{array}{l|ccccc}\hline x \text { wolves } & 26 & 37 & 22 & 69 & 98 \\\\\hline y \mathrm{km}^{2} / 1000 & 7.38 & 12.13 & 8.18 & 15.36 & 16.81 \\\\\hline\end{array}$$ Reference: The Wolves of Denali by Mech, Adams, Meier, Burch, and Dale, University of Minnesota Press. (a) Verify that \(\Sigma x=252, \Sigma y=59.86, \Sigma x^{2}=16,894, \Sigma y^{2}=787.0194\) \(\Sigma x y=3527.87,\) and \(r \approx 0.9405\) (b) Use a \(1 \%\) level of significance to test the claim \(\rho>0\) (c) Verify that \(S_{e} \approx 1.6453, a \approx 5.8309,\) and \(b \approx 0.12185\) (d) Find the predicted size of the hunting region for an extended pack of 42 wolves. (e) Find an \(85 \%\) confidence interval for your prediction of part (d). (f) Use a \(1 \%\) level of significance to test the claim that \(\beta>0\) (g) Find a \(95 \%\) confidence interval for \(\beta\) and interpret its meaning in terms of territory size per wolf.

Use appropriate multiple regression software of your choice and enter the data. Note that the data are also available for download at the Companion Sites for this text. Education: Exam Scores Professor Gill has taught general psychology for many years. During the semester, she gives three multiple-choice exams, each worth 100 points. At the end of the course, Dr. Gill gives a comprehensive final worth 200 points. Let \(x_{1}, x_{2},\) and \(x_{3}\) represent a student's scores on exams \(1,2,\) and \(3,\) respectively. Let \(x_{4}\) represent the student's score on the final exam. Last semester Dr. Gill had 25 students in her class. The student exam scores are shown on the next page. $$\begin{array}{cccc|cccc|cccc} \hline x_{1} & x_{2} & x_{3} & x_{4} & x_{1} & x_{2} & x_{3} & x_{4} & x_{1} & x_{2} & x_{3} & x_{4} \\ \hline 73 & 80 & 75 & 152 & 79 & 70 & 88 & 164 & 81 & 90 & 93 & 183 \\ 93 & 88 & 93 & 185 & 69 & 70 & 73 & 141 & 88 & 92 & 86 & 177 \\ 89 & 91 & 90 & 180 & 70 & 65 & 74 & 141 & 78 & 83 & 77 & 159 \\ 96 & 98 & 100 & 196 & 93 & 95 & 91 & 184 & 82 & 86 & 90 & 177 \\ 73 & 66 & 70 & 142 & 79 & 80 & 73 & 152 & 86 & 82 & 89 & 175 \\ 53 & 46 & 55 & 101 & 70 & 73 & 78 & 148 & 78 & 83 & 85 & 175 \\ 69 & 74 & 77 & 149 & 93 & 89 & 96 & 192 & 76 & 83 & 71 & 149 \\ 47 & 56 & 60 & 115 & 78 & 75 & 68 & 147 & 96 & 93 & 95 & 192 \\ 87 & 79 & 90 & 175 & & & & & & & & \\ \hline \end{array}$$ since Professor Gill has not changed the course much from last semester to the present semester, the preceding data should be useful for constructing a regression model that describes this semester as well. (a) Generate summary statistics, including the mean and standard deviation of each variable. Compute the coefficient of variation (see Section 3.2) for each variable. Relative to its mean, would you say that each exam had about the same spread of scores? Most professors do not wish to give an exam that is extremely easy or extremely hard. Would you say that all of the exams were about the same level of difficulty? (Consider both means and spread of test scores.) (b) For each pair of variables, generate the sample correlation coefficient \(r\) Compute the corresponding coefficient of determination \(r^{2}\). Of the three exams \(1,2,\) and \(3,\) which do you think had the most influence on the final exam \(4 ?\) Although one exam had more influence on the final exam, did the other two exams still have a lot of influence on the final? Explain each answer. (c) Perform a regression analysis with \(x_{4}\) as the response variable. Use \(x_{1}, x_{2}\) and \(x_{3}\) as explanatory variables. Look at the coefficient of multiple determination. What percentage of the variation in \(x_{4}\) can be explained by the corresponding variations in \(x_{1}, x_{2},\) and \(x_{3}\) taken together? (d) Write out the regression equation. Explain how each coefficient can be thought of as a slope. If a student were to study "extra hard" for exam 3 and increase his or her score on that exam by 10 points, what corresponding change would you expect on the final exam? (Assume that exams 1 and 2 remain "fixed" in their scores.) (e) Test each coefficient in the regression equation to determine if it is zero or not zero. Use level of significance \(5 \% .\) Why would the outcome of each hypothesis test help us decide whether or not a given variable should be used in the regression equation? (f) Find a \(90 \%\) confidence interval for each coefficient. (g) This semester Susan has scores of \(68,72,\) and 75 on exams \(1,2,\) and 3 respectively. Make a prediction for Susan's score on the final exam and find a \(90 \%\) confidence interval for your prediction (if your software supports prediction intervals).

In the least-squares line \(\hat{y}=5-2 x,\) what is the value of the slope? When \(x\) changes by 1 unit, by how much does \(\hat{y}\) change?

When we use a least-squares line to predict \(y\) values for \(x\) values beyond the range of \(x\) values found in the data, are we extrapolating or interpolating? Are there any concerns about such predictions?

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