Question

Assume that stock market returns have the market index as a common factor, and that all stocks in the economy have a beta of 1.9 on the market index. Firm-specific returns all have a standard deviation of 40%. Suppose that an analyst studies 20 stocks and finds that one-half of them have an alpha of +2.8%, and the other half have an alpha of −2.8%. Suppose the analyst invests $1.0 million in an equally weighted portfolio of the positive alpha stocks, and shorts $1 million of an equally weighted portfolio of the negative alpha stocks. a. What is the expected profit (in dollars) and standard deviation of the analyst’s profit? (Do not round intermediate calculations. Round your answers to the nearest whole dollar amount.) b. How does your answer change if the analyst examines 60 stocks instead of 20 stocks? 120 stocks?

Answer 1

a). Shorting equally the 10 negative-alpha stocks and investing the proceeds equally in the 10 positive-alpha stocks eliminates the market exposure and creates a zero-investment portfolio.

denoting the market factor as rM,the expected dollar return is

= $1,000,000 x [0.028 + 1.9 x rM] - $1,000,000 x [-0.028 + 1.9 x rM]

= $1,000,000 x 0.1064 = $106,400

The sensitivity of the payoff of this portfolio to the market factor is zero because the exposures of the positive alpha and negative alpha stocks cancels out. (Notice that the terms involving RMsum to zero.) Thus, the systematic component of total risk also is zero. The variance of the analyst's profit is not zero, however, since this portfolio is not well diversified.

For n = 20 stocks (i.e., long 10 stocks and short 10 stocks) the investor will have a $100,000 position (either long or short) in each stock. Net market exposure is zero, but firm-specific risk has not been fully diversified. The S.D. of dollar returns from the positions in the 20 firms is

= [20 x ($100,000 x 0.40)²]^1/2 = [$32,000,000,000]^1/2 = $178,885.44

b). If n = 60 stocks (30 long and 30 short), $33,333 is placed in each position, and the S.D. of dollar returns is

= [60 x ($33,333.33x0.40)²]^1/2 = [$10,666,666,667]^1/2 = $103,279.56

Similarly, if n = 120 stocks (60 long and 60 short), $16,667 is placed in each position, and the S.D. of dollar returns is

= [100 x ($16,666.67 x 0.40)²]^1/2 = [$4,444,444,444]^1/2 = $66,666.67

Notice that when the number of stocks increased by a factor of 6, from 20 to 120, standard deviation fell by a factor of = 2.683, from $178,885.44 to $66,666.67