4 d since p y 1 1 the random variable y is

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4. (d) Since P ( Y = 1) = 1 , the random variable Y is deterministic and always equal to 1. The CDF of Y can be computed as follows: F Y ( y ) = P ( Y y ) = P (1 y ) = u ( y - 1) . This CDF is depicted in Figure 6. 4. (e) Let θ [0 , 1] and define G ( x ) = θF ( x ) + (1 - θ ) F Y ( x ) . The function G is a valid CDF because it is a linear combination with non-negative coefficients (both θ and 1 - θ are non-negative) of CDFs so it is right-continuous and non-decreasing. Also, lim x →-∞ G ( x ) = 0 and lim x →∞ G ( x ) = θ lim x →∞ F ( x ) + (1 - θ ) lim x →∞ F Y ( x ) = θ + 1 - θ = 1 . Since F ( x ) [0 , 1] and F Y ( x ) [0 , 1] , it also follows that 0 G ( x ) 1 . Note that G is the CDF of a discrete random variable Z with PMF P [ Z = 0] = θ 3 , P [ Z = 1] = 1 - θ , and P [ Z = 2] = 2 θ 3 . 5. (a) Let A denote the event that I buy a Cappuccino Crunch milkshake (then A c denotes the event that I do not buy a Cappuccino Crunch milkshake), and let C denote the event that I get a cookie. Then by Total Probability, P ( C ) = P ( C | A ) P ( A ) + P ( C | A c ) P ( A c ) = q 1 p + q 2 (1 - p ) . 5. (b) By Bayes’ Rule, P ( A | C ) = P ( C | A ) P ( A ) P ( C ) = q 1 p q 1 p + q 2 (1 - p ) . 5. (c) By Bayes’ Rule, P ( A c | C c ) = P ( C c | A c ) P ( A c ) P ( C c ) = (1 - P ( C | A c )) P ( A c ) 1 - P ( C ) = (1 - q 2 )(1 - p ) 1 - ( q 1 p + q 2 (1 - p )) . 5. (d) For A and C to be independent, we must have P ( A ) P ( C ) = P ( A C ) , i.e., p ( q 1 p + q 2 (1 - p )) = q 1 p q 1 p 2 + q 2 p (1 - p ) = q 1 p q 1 p - q 1 p 2 - q 2 p (1 - p ) = 0 q 1 p (1 - p ) - q 2 p (1 - p ) = 0 p (1 - p )( q 1 - q 2 ) = 0 . Thus we need p = 0 , or p = 1 , or q 1 = q 2 for independence. The condition q 1 = q 2 is sufficient for independence because then whether or not I buy a Cappuccino Crunch milkshake clearly does not affect the probability that I get a cookie. If p = 0 or p = 1 , then the event A has zero or unit probability, and in either case it is independent of everything. 3
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-0.5 0 0.5 1 1.5 2 2.5 3 0 0.2 0.4 0.6 0.8 1 x F(x) (0,1/3) Figure 4: The function F 0 0.5 1 1.5 2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x PMF of X Length=1/3 Length=2/3 Figure 5: The PMF of X 0 0.5 1 1.5 2 0 0.2 0.4 0.6 0.8 1 y F Y (y) Figure 6: The CDF of Y 7
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6. (a) For p X ( x ) to be a valid PMF, 4 k =0 p X ( k ) = 1 . Thus 1 / 2 + 1 / 4 + 1 / 8 + 1 / 16 + θ = 1 , which gives θ = 1 / 16 .
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  • Spring '05
  • HAAS
  • Probability theory, CDF, first bit

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