Question

Return again to the previous problem. Now suppose that the manager misestimates thebeta of Waterworks stock, believing it to be .50 instead of .75. The standard deviation ofthe monthly market rate of return is 5%.

a. What is the standard deviation of the (now improperly) hedged portfolio?

b. What is the probability of incurring a loss over the next month if the monthly marketreturn has an expected value of 1% and a standard deviation of 5%? Compare youranswer to the probability you found in Problem 16.

c. What would be the probability of a loss using the data in the previous problem if themanager similarly misestimated beta as .50 instead of .75? Compare your answer tothe probability you found in the previous problem.

d. Why does the misestimation of beta matter somuch more for the 100-stock portfoliothan it does for the 1-stock portfolio?

Question: The following is part of the computer output from a regression of monthly returns on Waterworks stock against the S&P 500 Index. A hedge fund manager believes that Waterworks is underpriced, with an alpha of 2% over the coming month.

a. If he holds a \(3 million portfolio of Waterworks stock and wishes to hedge market exposure for the next month using one-month maturity S&P 500 futures contracts, how many contracts should he enter? Should he buy or sell contracts? The S&P 500 currently is at 1,000 and the contract multiplier is \)250.

b. What is the standard deviation of the monthly return of the hedged portfolio?

c. Assuming that monthly returns are approximately normally distributed, what is the probability that this market-neutral strategy will lose money over the next month?

Assume the risk-free rate is .5% per month.

Short answer

a. 6.129%

b. Unlikely

c. Greater

d. Because the diversified portfolio is very small

Step-by-step solution

Calculation of standard deviation of (now improperly) hedged portfolio ‘a’

$\begin{array}{c}\text{Variance+\hspace{0.17em}=\hspace{0.17em}}{\beta }^2+\text{Var}\left(r_m\right)+\text{Var}\left(e\right)\\ =({0.25}^2\times {0.05}^2)+{0.06}^2\\ =0.00375625\end{array}$

$\begin{array}{c}\sigma =\sqrt{\text{Variance}}\\ =\sqrt{0.00375625}\\ =0.06129\\ =6.129\%\end{array}$

Calculation of the probability of loss ‘b’ 

Since the manager has misestimated the beta, hence he will six contracts =

$\begin{array}{c}\text{Hedgeratio\hspace{0.17em}}=\frac{\$3,000,000\times 0.50}{\$250}\times 2,000\\ =6\text{\hspace{0.17em}contracts}\end{array}$

The dollar value of stock portfolio = $3,000,000 x (1 + r _Portfolio) (because ERR is no longer 2.5% owing to not completely hedged portfolio)

$\begin{array}{l}=\$3,000,000\times (1+r_f)+\beta ({r_M}_{}-r_f)+\alpha +e\\ =\$3,000,000\times (1+.005)+.75({r_M}_{}-.005)+.02+e\\ =\$3,000,000\times (1+.005)+.75({r_M}_{}-.005)+.02+e\\ =\$3,063,750+(\$4500000\times r_M)+\$3,000,000\times e\end{array}$

The dollar proceeds from the futures position=

$\begin{array}{l}=6contracts\times \$250\times (F_0-F_1)\\ =\$1,500\times \left[\right(S_0\left[1.005\right)–S_1]\\ =\$1,500\times S_0[1.005–(1+r_M)]\\ =\$1,500\times [2,000\times (.005–r_M)]\\ =\$15,000-(\$3,000,000\times r_M)\end{array}$

Hence the total value of the stock plus futures position at the end of the month

$\begin{array}{l}=\$3,063,750+(\$4,500,000\times r_M)+(\$3,000,000\times e)+\$15,000-(\$3,000,000\times r_M)\\ =\$3,063,750+(\$1,500,000\times r_M)+(\$3,000,000\times e)\\ =\$3,063,760+(\$1,500,000\times 0.01)+(\$3,000,000\times e)\\ =\$4,563,750+(\$3,000,000)\end{array}$

The expected rate of return for the (improperly) hedged portfolio

$\begin{array}{l}=\left(\frac{\$4,563,750}{\$3,000,000}\right)–1\\ =.02625\\ =2.625\%\end{array}$

The z value for a rate of return of zero:

$\begin{array}{l}=X-\mu /\sigma \\ =0–\frac{0.2625}{0.06129}\\ =-0.4283\end{array}$

Therefore the probability of a negative return = N(-.4167) = .3342

This implies that negative return is unlikely.

Calculation of the probability of loss ‘c’

$\begin{array}{c}\text{Variance+\hspace{0.17em}=\hspace{0.17em}}{\beta }^2+\text{Var}\left(r_m\right)+\text{Var}\left(e\right)\\ =({0.25}^2\times {0.05}^2)+{0.006}^2\\ =1.9225\end{array}$

$\begin{array}{c}\sigma =\sqrt{\text{Variance}}\\ =\sqrt{1.9225}\\ =1.3865\%\end{array}$

The z value for a rate of return of zero:

$\begin{array}{l}=X-\frac{\mu }{\sigma }\\ =0–\frac{0.2625}{0.013685}\\ =-1.8933\end{array}$

Therefore the probability of a negative return = N(-.1.8933) = .0292

This implies that negative return is far greater.

 Step 4: Explanation on mis-estimation of beta ‘d’

Since the idiosyncratic risk of the diversified portfolio is very small, the market exposure becomes very important in contributing to total volatility.