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hw4 - MS&E 120 Probabilistic Analysis Fall 2006-2007 Prof...

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MS&E 120: Probabilistic Analysis Fall 2006-2007, Prof. Shachter HOMEWORK #4 SOLUTIONS CHAPTER #5 PROBLEMS 39 y e y X y Y e e F e X P y X P y Y P y F = = = = = 1 ) ( ) ( ) (log ) ( ) ( y y e y e y X y e e e y F dx d y f = = = ) ( ) ( , < < y 40 y y F y X P y e P y Y P y F X X Y log ) (log ) log ( ) ( ) ( ) ( = = = = = y y F dx d y f X y 1 ) ( ) ( = = , 1 y e CHAPTER #5 THEORETICAL EXERCISES 28 )) ( ( ) ( ) ( ) ( 1 y F X P y F P y Y P y F X X Y = = = since X F is invertible y y F F X X = = )) ( ( 1 Note that the above is the distribution function for uniform random variable. Moreover, X F runs from 0 to 1. Hence, y ~ U(0,1) Note: From the above property, we can simulate any random variable X if we can first simulate a Uniform (0,1) random variable and determining the corresponding transformation. 30 ) log ( ) log ( ) ( ) ( σ μ φ = = = x x X P x e P x F X Y ) log ( 1 ) (log ' ) ( σ μ φ = = x x x F x f X y , y > 0 CHAPTER #6 PROBLEMS 1 a) X = largest value, Y = sum
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To find the joint probability mass function, first we note that X can take value between 1 and 6 (inclusive) and Y can take value between 2 and 12 (inclusive. Then we look at the possible outcome(s) for each state of X and Y Outcomes Y 2 3 4 5 6 7 8 9 10 11 1 (1,1) - - - - - - - - - 2 - (1,2),(2,1) (2,2) - - - - - - - X 3 - - (1,3),(3,1) (2,3),(3,2) (3,3) - - - - - 4 - - - (1,4),(4,1) (2,4),(4,2) (3,4),(4,3) (4,4) - - - 5 - - - - (1,5),(5,1) (2,5),(5,2) (5,3),(3,5) (4,5),(5,4) (5,5) - 6 - - - - - (1,6),(6,1) (2,6)(6,2) (3,6),(6,3) (4,6),(6,4) (5,6),(6,5 Now, since the probability of each outcome to appear is 1/36, this is our probability mass function table: Probability Y 2 3 4 5 6 7 8 9 10 11 12 1 1/36 - - - - - - - - - - 2 - 2/36 1/36 - - - - - - - - X 3 - - 2/36 2/36 1/36 - - - - - - 4 - - - 2/36 2/36 2/36 1/36 - - - - 5 - - - - 2/36 2/36 2/36 2/36 1/36 - - 6 - - - - - 2/36 2/36 2/36 2/36 2/36 1/36 Check: the sum all the probabilities above is 1 b) X = value on 1 st die, Y = the larger value Here we assume that if we have the same outcome from the 2 dices, we take that outcome as the “larger value” As above, we note that X can take value from 1 to 6, Y can take value from 1 to 6, and then enumerate the outcomes Outcomes Y 1 2 3 4 5 6 1 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) 2 - (2,2),(2,1) (2,3) (2,4) (2,5) (2,6) X 3 - - (3,3),(3,2),(3,1) (3,4) (3,5) (3,6) 4 - - - (4,4),(4,3),(4,2),(4,1) (4,5) (4,6) 5 - - - - (5,5),(5,4),(5,3),(5,2),(5,1) (5,6) 6 - - - - - (6,6),(6,5),(6,4),(6,3),(6 And the probability mass function table is (check: all must sum to 1) Probability Y 1 2 3 4 5 6 1 1/36 1/36 1/36 1/36 1/36 1/36 2 - 2/36 1/36 1/36 1/36 1/36 X 3 - - 3/36 1/36 1/36 1/36
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4 - - - 4/36 1/36 1/36 5 - - - - 5/36 1/36 6 - - - - - 6/36 c) X = smallest, Y = largest Outcomes Y 1 2 3 4 5 6 1 (1,1) (1,2),(2,1) (1,3),(3,1) (1,4),(4,1) (1,5),(5,1) (1,6),(6,1) 2 - (2,2) (2,3),(3,2) (2,4),(4,2) (2,5),(5,2) (2,6),(6,2) X 3 - - (3,3) (3,4),(4,3) (3,5),(5,3) (3,6),(6,3) 4 - - - (4,4) (4,5),(5,4) (4,6),(6,4) 5 - - - - (5,5) (5,6),(6,5) 6 - - - - - (6,6) Probability Y 1 2 3 4 5 6 1 1/36 2/36 2/36 2/36 2/36 2/36 2 - 1/36 2/36 2/36 2/36 2/36 X 3 - - 1/36 2/36 2/36 2/36 4 - - - 1/36 2/36 2/36 5 - - - - 1/36 2/36 6 - - - - - 1/36 10 a) ( ) ( ) ( ) ( ) dy e e dy dx e e dxdy e dxdy y x f Y X P y y y x y y y x y ∫∫ ∫ ∫ + = = = = < 0 0 0 0 0 1 , [ ] 2 1 0 2 2 1 2 0 = + = = y y y y e e dy e e b) ( ) ( ) ( ) ( ) dy e e dy dx e e dxdy e dxdy y x f a X P a y a x y a y x a ∫∫ ∫ ∫ + = = = = < 0 0 0 0 0 1 , ( ) ( ) a y a e dy e e = = 1 1 0 12 Assume that men and women are equally likely to arrive at the drugstore and they arrive independently.
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