Suppose you have some money to invest-for simplicity, \$1 -and you are planning to put a fraction $w$ into a stock market mutual fund and the rest, $1-w$, into a bond mutual fund. Suppose that $\$ 1$ invested in a stock fund yields $R_{s}$ after 1 year and that $\$ 1$ invested in a bond fund yields $R_{b}$, suppose that $R_{s}$ is random with mean $0.1(10 \%)$ and standard deviation 0.09 , and suppose that $R_{b}$ is random with mean $0.06(6 \%)$ and standard deviation 0.05 . The correlation between $R_{s}$ and $R_{b}$ is 0.31 . If you place a fraction $w$ of your money in the stock fund and the rest, $1-w$, in the bond fund, then the return on your investment is $R=w R_{s}+(1-w) R_{b}$. Suppose that $w=0.61$. Compute the mean and standard deviation of $R$. The mean is 0.084 . (Round your response to three decimal places.) The standard deviation is 0.064 . (Round your response to three decimal places.) Suppose that $w=0.92$. Compute the mean and standard deviation of $R$. The mean is 0.097 . (Round your response to three decimal places.) The standard deviation is 0.084 . (Round your response to three decimal places.) What value of $w$ makes the mean of $R$ as large as possible? $w=\square$ maximizes $\mu$. (Round your response to two decimal places.) What is the standard deviation of $R$ for this value of $w$ ? $\sigma=\square$ for this value of $w$. (Round your response to two decimal places.) What is the value of $w$ that minimizes the standard deviation of $R$ ? 0 of 10 questions
Solution
Step 1 ：Given that the return on investment is given by the equation \(R=w R_{s}+(1-w) R_{b}\), where \(R_{s}\) and \(R_{b}\) are the yields of the stock and bond funds respectively, and \(w\) is the fraction of money invested in the stock fund.
Step 2 ：The mean of \(R\) is given by the equation \(\mu = w \mu_{s} + (1-w) \mu_{b}\), where \(\mu_{s}\) and \(\mu_{b}\) are the means of \(R_{s}\) and \(R_{b}\) respectively.
Step 3 ：The standard deviation of \(R\) is given by the equation \(\sigma = \sqrt{(w \sigma_{s})^2 + ((1-w) \sigma_{b})^2 + 2w(1-w)\rho\sigma_{s}\sigma_{b}}\), where \(\sigma_{s}\) and \(\sigma_{b}\) are the standard deviations of \(R_{s}\) and \(R_{b}\) respectively, and \(\rho\) is the correlation between \(R_{s}\) and \(R_{b}\).
Step 4 ：To find the value of \(w\) that maximizes the mean of \(R\), we need to take the derivative of the mean equation with respect to \(w\) and set it equal to zero. This will give us the value of \(w\) that maximizes the mean.
Step 5 ：To find the value of \(w\) that minimizes the standard deviation of \(R\), we need to take the derivative of the standard deviation equation with respect to \(w\) and set it equal to zero. This will give us the value of \(w\) that minimizes the standard deviation.
Step 6 ：The value of \(w\) that makes the mean of \(R\) as large as possible is \(\boxed{1.00}\).
Step 7 ：The standard deviation of \(R\) for this value of \(w\) is \(\boxed{0.09}\).
Step 8 ：The value of \(w\) that minimizes the standard deviation of \(R\) is \(\boxed{0.14}\).
