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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.)
10 of 10 questions
Suppose that $w=0.92$. Compute the mean and standard deviation of $R$.
The mean is . (Round your response to three decimal places.)
The standard deviation is (Round your response to three decimal places.)
What value of $w$ makes the mean of $R$ as large as possible?
\[
w=
\]
maximizes $\mu$. (Round your response to two decimal places.)
What is the standard deviation of $R$ for this value of $w$ ?
\[
\sigma=
\]
for this value of $w$. (Round your response to two decimal places.)
What is the value of $w$ that minimizes the standard deviation of $R$ ? $w=1$ minimizes the standard deviation of $R$. (Round vour resoonse to two decimal olaces.)
Answer
Final Answer: The mean of $R$ when $w=0.92$ is \(\boxed{0.097}\) and the standard deviation of $R$ when $w=0.92$ is \(\boxed{0.084}\).
Step 1 ：Given that the fraction of money invested in the stock fund is $w=0.92$, the mean return of the stock fund is $0.1$, the standard deviation of the stock fund is $0.09$, the mean return of the bond fund is $0.06$, the standard deviation of the bond fund is $0.05$, and the correlation between the stock and bond returns is $0.31$.
Step 2 ：We can calculate the mean return on the investment $R$ using the formula $R=w R_{s}+(1-w) R_{b}$. Substituting the given values, we get $R=0.92*0.1+(1-0.92)*0.06=0.0968$.
Step 3 ：We can calculate the standard deviation of the return on the investment $R$ using the formula $\sqrt{w^2 \sigma_{s}^2 + (1-w)^2 \sigma_{b}^2 + 2w(1-w)\rho\sigma_{s}\sigma_{b}}$. Substituting the given values, we get $\sqrt{0.92^2 * 0.09^2 + (1-0.92)^2 * 0.05^2 + 2*0.92*(1-0.92)*0.31*0.09*0.05}=0.084126$.
Step 4 ：Rounding to three decimal places, the mean return on the investment $R$ when $w=0.92$ is $0.097$ and the standard deviation of the return on the investment $R$ when $w=0.92$ is $0.084$.
Step 5 ：Final Answer: The mean of $R$ when $w=0.92$ is \(\boxed{0.097}\) and the standard deviation of $R$ when $w=0.92$ is \(\boxed{0.084}\).
