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Homework 4 Discrete Probability Solutions

# Homework 4 Discrete Probability Solutions - Homework 4...

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Homework 4 Solutions Discrete Probability Distribution 1. Consider the following discrete probability distribution = = = = = = otherwise x x x x x X P 0 16 , 4 . 0 8 , 25 . 0 4 , 15 . 0 2 , 2 . 0 ) ( Compute a. E[X] = 0.2*2+0.15*4+0.25*8+0.4*16=9.4 b. E[-2X+3]=-2*E[X]+3=-15.8 c. Var[X] = 33.24 Standard deviation of X= sqrt(Var(X))=5.766 d. Var[-2X+3]=4*Var[X]=132.96 2. Let X be a random variable with distribution function P(X ≤ x) = F(x) and let Y = 1+3X be another random variable. If E[Y] = 10 and Var[Y]=16. Compute the E[X] and Var[X]. Solution: E[Y]=1+3E[X]=10 therefore E[X]=3, Var[Y]=9Var[X]=16 therefore Var[X]=16/9. 3. The price of a stock in a given trading day changes according to the following distribution = = = = = = 2 125 . 0 1 125 . 0 0 5 . 0 1 25 . 0 ) ( x x x x x X P Find the distribution of the change in stock price after two trading days assuming that the changes in two consecutive days are independent. Find the expectation and the variance of the total change in two days.

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Solution: Let the random variable X1 represent the change in the first day and X2 change in the second day. So the total change in two consecutive days will be X1+X2. The distribution of this random variable is = = = = = = = = = + 4 3 64 / 1 64 / 2 2 64 / 9 1 16 / 3 0 16 / 5 1 16 / 4 2 16 / 1 ) 2 1 ( x x x x x x x x X X P Note that E[X] = 0.125 and Var[X] = 0.8594 E[X1+X2] = E[X1] + E[X2] = 2E[X] = 2*0.125=0.25 Var[X1+X2] = Var[X1] + Var [X2] = 2Var[X] = 1.719 4. A die is rolled twice. Let X denotes the sum of two numbers that turn up (specifically
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Homework 4 Discrete Probability Solutions - Homework 4...

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