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Probset4_sol

# Probset4_sol - Probability and Stochastic Processes...

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Unformatted text preview: Probability and Stochastic Processes Homework 4 Solutions 19th Feb,Spring 2010 Problem Solutions : Yates and Goodman, 2.5.2 2.5.5 2.5.9 2.5.10 2.6.3 2.6.5 2.6.6 2.7.2 2.7.3 2.7.5 2.7.7 2.7.8 2.8.5 2.8.8 and 2.8.10 Problem 2.5.2 Solution Voice calls and data calls each cost 20 cents and 30 cents respectively. Furthermore the respective probabilities of each type of call are 0.6 and 0.4. (a) Since each call is either a voice or data call, the cost of one call can only take the two values associated with the cost of each type of call. Therefore the PMF of X is P X ( x ) = . 6 x = 20 . 4 x = 30 otherwise (1) (b) The expected cost, E [ C ], is simply the sum of the cost of each type of call multiplied by the probability of such a call occurring. E [ C ] = 20(0 . 6) + 30(0 . 4) = 24 cents (2) Problem 2.5.5 Solution From the solution to Problem 2.4.3, the PMF of X is P X ( x ) = . 4 x = 3 . 4 x = 5 . 2 x = 7 otherwise (1) The expected value of X is E [ X ] = summationdisplay x xP X ( x ) = 3(0 . 4) + 5(0 . 4) + 7(0 . 2) = 2 . 2 (2) Problem 2.5.9 Solution In this double-or-nothing type game, there are only two possible payoffs. The first is zero dollars, which happens when we lose 6 straight bets, and the second payoff is 64 dollars which happens unless we lose 6 straight bets. So the PMF of Y is P Y ( y ) = (1 / 2) 6 = 1 / 64 y = 0 1 (1 / 2) 6 = 63 / 64 y = 64 otherwise (1) The expected amount you take home is E [ Y ] = 0(1 / 64) + 64(63 / 64) = 63 (2) So, on the average, we can expect to break even, which is not a very exciting proposition. 1 Problem 2.5.10 Solution By the definition of the expected value, E [ X n ] = n summationdisplay x =1 x parenleftbigg n x parenrightbigg p x (1 p ) n x (1) = np n summationdisplay x =1 ( n 1)! ( x 1)!( n 1 ( x 1))! p x 1 (1 p ) n 1 ( x 1) (2) With the substitution x = x 1, we have E [ X n ] = np n 1 summationdisplay x =0 parenleftbigg n 1 x parenrightbigg p x (1 p ) n x bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright 1 = np n 1 summationdisplay x =0 P X n 1 ( x ) = np (3) The above sum is 1 because it is the sum of a binomial random variable for n 1 trials over all possible values. Problem 2.6.3 Solution From the solution to Problem 2.4.3, the PMF of X is P X ( x ) = . 4 x = 3 . 4 x = 5 . 2 x = 7 otherwise (1) (a) The PMF of W = X satisfies P W ( w ) = P [ X = w ] = P X ( w ) (2) This implies P W ( 7) = P X (7) = 0 . 2 P W ( 5) = P X (5) = 0 . 4 P W (3) = P X ( 3) = 0 . 4 (3) The complete PMF for W is P W ( w ) = . 2 w = 7 . 4 w = 5 . 4 w = 3 otherwise (4) (b) From the PMF, the CDF of W is F W ( w ) = w &amp;lt; 7 . 2 7 w &amp;lt; 5 . 6 5 w &amp;lt; 3 1 w 3 (5) (c) From the PMF, W has expected value...
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Probset4_sol - Probability and Stochastic Processes...

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