Suppose you have some money to invest-for simplicity. $1 - and you are planning

Question

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 0.09 (9%) and standard deviation 0.08. and suppose that R_b is random with mean 0.06 (6%) and standard deviation 0.04. The correlation between R_s and R_b is 0.28. 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 = wR_s + (1 -w)R_b. Suppose that w = 0.56. 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.) Suppose that w = 0.83. 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 = minimizes the standard deviation of R. (Round your response to two decimal places.)

Explanation / Answer

when w=0.56

mean of R =0.56*0.09+0.44*0.06=0.0768

std deviation of R =((0.56*0.08)2+(0.44*0.04)2+2*0.56*0.44*0.28)1/2=0.3746

when w=0.83

mean of R=0.83*0.09+0.17*0.06=0.0849

std deviation of R =((0.83*0.08)2+(0.17*0.04)2+2*0.83*0.17*0.28)1/2=0.2889

mean will be maximum at W=1;

and standard deviation =0.08

std deviation of R will be maximum at w=1/2

hence std deviation =0.3768

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