hw4 - MS&E 120: Probabilistic Analysis Fall...

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Unformatted text preview: 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 ( ) ( ) ( = = = x x X P x e P x F X Y ) log ( 1 ) (log ' ) ( = = x x x F x f X y , y > 0 CHAPTER #6 PROBLEMS 1 a) X = largest value, Y = sum 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),( 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 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)...
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hw4 - MS&E 120: Probabilistic Analysis Fall...

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