Probset5_sol

# Probset5_sol - Probability and Stochastic Processes A...

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Unformatted text preview: Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers Edition 2 Roy D. Yates and David J. Goodman Problem Solutions : Yates and Goodman, 2.9.3 2.9.6 2.10.1 2.10.4 2.10.5 3.1.3 3.1.4 3.2.1 3.2.3 3.2.4 3.2.5 3.3.2 and 3.3.4 Problem 2.9.3 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) The event B = { X > } has probability P [ B ] = P X (5)+ P X (7) = 0 . 6. From Theorem 2.17, the conditional PMF of X given B is P X | B ( x ) = braceleftBigg P X ( x ) P [ B ] x ∈ B otherwise = 2 / 3 x = 5 1 / 3 x = 7 otherwise (2) The conditional first and second moments of X are E [ X | B ] = summationdisplay x xP X | B ( x ) = 5(2 / 3) + 7(1 / 3) = 17 / 3 (3) E bracketleftbig X 2 | B bracketrightbig = summationdisplay x x 2 P X | B ( x ) = 5 2 (2 / 3) + 7 2 (1 / 3) = 33 (4) The conditional variance of X is Var[ X | B ] = E bracketleftbig X 2 | B bracketrightbig − ( E [ X | B ]) 2 = 33 − (17 / 3) 2 = 8 / 9 (5) Problem 2.9.6 Solution (a) Consider each circuit test as a Bernoulli trial such that a failed circuit is called a success. The number of trials until the first success (i.e. a failed circuit) has the geometric PMF P N ( n ) = braceleftbigg (1 − p ) n − 1 p n = 1 , 2 ,... otherwise (1) (b) The probability there are at least 20 tests is P [ B ] = P [ N ≥ 20] = ∞ summationdisplay n =20 P N ( n ) = (1 − p ) 19 (2) 1 Note that (1 − p ) 19 is just the probability that the first 19 circuits pass the test, which is what we would expect since there must be at least 20 tests if the first 19 circuits pass. The conditional PMF of N given B is P N | B ( n ) = braceleftBigg P N ( n ) P [ B ] n ∈ B otherwise = braceleftbigg (1 − p ) n − 20 p n = 20 , 21 ,... otherwise (3) (c) Given the event B , the conditional expectation of N is E [ N | B ] = summationdisplay n nP N | B ( n ) = ∞ summationdisplay n =20 n (1 − p ) n − 20 p (4) Making the substitution j = n − 19 yields E [ N | B ] = ∞ summationdisplay j =1 ( j + 19)(1 − p ) j − 1 p = 1 /p + 19 (5) We see that in the above sum, we effectively have the expected value of J + 19 where J is geometric random variable with parameter p . This is not surprising since the N ≥ 20 iff we observed 19 successful tests. After 19 successful tests, the number of additional tests needed to find the first failure is still a geometric random variable with mean 1 /p . Problem 2.10.1 Solution For a binomial ( n,p ) random variable X , the solution in terms of math is P [ E 2 ] = ⌊ √ n ⌋ summationdisplay x =0 P X ( x 2 ) (1) In terms of Matlab , the efficient solution is to generate the vector of perfect squares x = [0 1 4 9 16 ...] and then to pass that vector to the binomialpmf.m . In this case, the values of the binomial PMF are calculated only once. Here is the code:the values of the binomial PMF are calculated only once....
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Probset5_sol - Probability and Stochastic Processes A...

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