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?
$w = ◻$ maximizes $μ$ . (Round your response to two decimal places.)
What is the standard deviation of $R$ for this value of $w$ ?
$σ = ◻$ for this value of $w$ . (Round your response to two decimal places.)
What is the value of $w$ that minimizes the standard deviation of $R$ ?
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The value of $w$ that minimizes the standard deviation of $R$ is $0.14$ .

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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 $μ = w μ_{s} + (1 - w) μ_{b}$ , where $μ_{s}$ and $μ_{b}$ are the means of $R_{s}$ and $R_{b}$ respectively.

Step 3 ：The standard deviation of $R$ is given by the equation $σ = \sqrt{(w σ_{s})^{2} + ((1 - w) σ_{b})^{2} + 2 w (1 - w) ρ σ_{s} σ_{b}}$ , where $σ_{s}$ and $σ_{b}$ are the standard deviations of $R_{s}$ and $R_{b}$ respectively, and $ρ$ 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 $1.00$ .

Step 7 ：The standard deviation of $R$ for this value of $w$ is $0.09$ .

Step 8 ：The value of $w$ that minimizes the standard deviation of $R$ is $0.14$ .

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