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Solutions to Problem Set 1

# Therefore we have varya2varx c suppose that

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Unformatted text preview: aE [ X ]} 2 2 Ⱥ Ⱥ = a 2 Ⱥ p1{x 1 − E [ X ]} + p2 { x 2 − E [ X ]} Ⱥ Ⱥ Ⱥ € Note that the term inside the first brackets is the variance of X. Therefore, we have € Var(Y)=a2Var(X) c) Suppose that x1=0 and x2=1. Show that Var (X) ≤ E[X]. Using the formula for the expected value, we obtain E [ X ] = p1 × 0 + p2 × 1 = p2 . 2 2 The variance of X is Var( X ) = p1(0 − p2 ) + p2 (1 − p2 ) . € Because p1+p2=1, we have that p1=1 p2. Substituting in the expression for the variance, we obtain € 2 Var( X ) = (1 − p2 ) p2 2 + p2 (1 − p2 ) = p2 (1 − p2 )( p2 + 1 − p2 ) = p2 (1 − p2 ). Because p2 is between 0 and 1, we have that Var (X) ≤ E[X]. € Question 2. Consider an investment opportunity that generates a profit (net of all investment costs) of either +1 dollar with probability ¾ or  1 dollar with probability ¼. a) What are the mean and the variance of the profit from the investment? Mean = 1/2. Variance = 3/4 b) If yo...
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