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sol1 - ACTSC 372 Assignment 1 Solutions 1[3 points Suppose...

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ACTSC 372 – Assignment 1 – Solutions 1. [3 points] Suppose you can invest in two stocks, A and B. The returns R A , R B , on these stocks depend on the state of the economy and are modelled as follows (where returns are expressed in percentages): state probability R A R B Recession 0.2 -5 -8 Normal 0.5 10 14 Boom 0.3 20 21 (a) Compute the expected return for each stock. Solution: E( R A ) = 0 . 2 × ( - 0 . 05) + 0 . 5 × 0 . 1 + 0 . 3 × 0 . 2 = 0 . 10 = 10% and E( R B ) = 0 . 2 × - 0 . 08 + 0 . 5 × 0 . 14 + 0 . 3 × 0 . 21 = 0 . 117 = 11 . 7%. (b) Compute the standard deviation of the return for each stock. Solution: σ A = (0 . 2( - 0 . 05 - 0 . 1) 2 +0 . 5(0 . 1 - 0 . 1) 2 +0 . 3(0 . 2 - 0 . 1) 2 ) 1 / 2 = 8 . 66% and σ B = (0 . 2( - 0 . 08 - 0 . 117) 2 + 0 . 5(0 . 14 - 0 . 117) 2 + 0 . 3(0 . 21 - 0 . 117) 2 ) 1 / 2 = 10 . 31%. (c) Compute the covariance and correlation between the two stocks. Solution: Cov( R A , R B ) = 0 . 2( - 0 . 05 - 0 . 1)( - 0 . 08 - 0 . 117) + 0 . 5(0 . 1 - 0 . 1)(0 . 14 - 0 . 117) + 0 . 3(0 . 2 - 0 . 1)(0 . 21 - 0 . 117) = 0 . 0087 and ρ AB = Cov( R A , R B ) / ( σ A σ B ) = 0 . 974778. 2. [3 points] Suppose there are N stocks in the economy, each with an expected return of 9% and a standard deviation of the return equal to 12%. The covariance between the returns is the same for all pairs of (distinct) stocks. (a) Suppose N = 100. If an equally weighted portfolio composed of these N = 100 assets has a standard deviation of 9%, what is the covariance between the returns (for distinct pairs of stocks)? Solution: Let c be the common covariance between two stocks. We have that σ 2 R = 0 . 09 2 = 100 i =1 1 100 2 0 . 12 2 + 2 99 i =1 100 j = i +1 1 100 1 100 c Therefore 0 . 09 2 = 0 . 01 × 0 . 12 2 + 2 × c × 0 . 01 2 99 × 100 2 and so c = 0 . 008036. (b) Suppose the covariance is equal to the value found in (a) (for all N ). What becomes the variance of the portfolio as N goes to infinity ? Solution: We have that σ 2 R = 1 N 0 . 12 2 +2 × 1 N 2 × ( N - 1) × N 2 0 . 008036 = 1 N 0 . 12 2 +(1 - 1 /N )0 . 008036 0 . 008036 as N → ∞ .
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