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I am struggeling with this example:

We have some money to invest and put a fraction $w$ into a stock market fund and the rest $(1 − w)$ into a bond fund. The stock fund yields $\large{R_s}$ after one year and the bond fund $\large{R_b}$. $\large{R_s}$ is random with mean $0.08$ and standard deviation $0.07$. $\large{R_b}$ is random with mean $0.05$ and standard deviation $0.04$. The correlation between $\large{R_b}$ and $\large{R_s}$ is $0.25$. The return on your investment is $$\large{R = w\times R_s +(1−w)R_b}$$

  • (a) Suppose that $w = 0.5$. Compute the mean and standard deviation of $R$.
  • (b) What value of $w$ makes the mean of $R$ as large as possible? What is the standard deviation of $R$ for this value of $w$.

a) $$R = 0.5 \times R_s + 0.5\times R_b$$ The mean: $$R = 0.5\times 0.08 + 0.5\times 0.05$$ $$R=0.065$$

The standard deviation: $$R = 0.5\times 0.07 + 0.5\times 0.04$$ $$R=0.055$$

b) Please give me a hint to calculate this question?

Are my calculations right?

Thx for your reply!

UPDATE:

After the calculation of the $Cov(X,Y) = 0.0007$

I got: $$\operatorname{Var}(aX+bY)=a^2(0.0049)+b^2(0.0016)+2ab0.0007$$

Therefore the Variance has to be: $$(a+b)*Var(0.0016*0.0049)$$

Are my conclusions right?

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Let us compute the variance of $R$. The calculation is substantially more complicated than the one in the post. We work with variance, not standard deviation. But recall that variance is the square of the standard deviation.

You may recall the formula $$\operatorname{Var}(aX+bY)=a^2\operatorname{Var}(X)+b^2\operatorname{Var}(Y)+2ab\operatorname{Cov}(X,Y).$$ That gives you everything that is needed to calculate the variance of $R$, except that you were given the correlation coefficient $\rho(R_s,R_b)$, not the covariance. However, we have $$\rho(X,Y)=\frac{\operatorname{Cov}(X,Y)}{\sqrt{\operatorname{Var(X)}\operatorname{Var}(Y)}},$$ so now you have all the necessary ingredients. Putting things together, and doing the computations, is left to you.

As to the second question, it is a lot easier than the first. You could compute the mean of $R$ explicitly as a function of $w$, and graph the mean return $y$ as a function of $w$. Note that $w$ ranges between $0$ and $1$. Then the $w$ that yields the greatest mean is visually obvious. But instead, think money. The stock has greater mean yield (but greater volatility). So to maximize your mean, what should you do?

Added: This is about the additional computation that you did. The covariance was calculated correctly. The variance of $aR_s+bR_b$ is therefore $$a^2(0.0049)+b^2(0.0016)+2ab(0.0007).\tag{$1$}$$ Here $a=1-w$ and $b=w$. But you were only asked to compute for $w=\frac{1}{2}$, so we take $a=b=\frac{1}{2}$. The last line "Therefore the Variance $\dots$" is not correct. We calculate, using $(1)$. The variance turns out to be $0.001975$. You were asked for the standard deviation, which is therefore about $0.04444$.

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  • $\begingroup$ Maybe shorting is allowed, so $w \in \mathbb{R}$? $\endgroup$ – copper.hat Mar 9 '13 at 18:09
  • $\begingroup$ Hey Andre, just one last question. Are my assumptions for my mean right? $\endgroup$ – Le Chifre Mar 12 '13 at 9:53
  • $\begingroup$ I plotet it in excel and got the maxmimum return I get is at: $w=0.05$. Could this be true or are my assumptions false? $\endgroup$ – Le Chifre Mar 13 '13 at 14:07
  • $\begingroup$ Had typo problem in previous comment, retyping. The maximum expected value is at $w=0$: all stocks, which is clear without calculation. But if we do want to calculate, expected return is $(1-w)(0.08)+(w)(0.05)=0.08-(0.03)(w)$. This is max at $w=0$. $\endgroup$ – André Nicolas Mar 13 '13 at 16:39
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    $\begingroup$ @maximus: Noticed you had added to the post. You calculated the covariance correctly, but not the variance, so I have completed that calculation. $\endgroup$ – André Nicolas Mar 13 '13 at 18:15

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