hw1soln - Solution 1 Mulin Cheng mulinch@caltech.edu...

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Unformatted text preview: Solution 1 Mulin Cheng mulinch@caltech.edu October 18, 2007 1 Bread & Butter: Distribution functions. ( 10 Points ) Express respectively the distribution of 1. X + = max { , X } ( 3 Points ) 2. X- =- min { , X } ( 3 Points ) 3. | X | = X + + X- ( 2 Points ) 4. -X ( 2 Points ) in terms of the distribution function F of the random variable X. Solution: 1. Clearly, the event { max { , X } x } = { x, X x } = if x < , { X x } otherwise . So P ( X + x ) = if x < , F ( x ) otherwise. 2. Similiarly, the event {- min { , X } x } = { - x, X - x } = if x < , { X - x } otherwise . So P ( X- x ) = if x < , 1- lim y - x F ( y ) otherwise. 1 ACM116 Introduction to Stochastic Processes and Modeling 3. P ( | X | x ) = P (- x X + x ) = P ( X x )- P ( X <- x ) if x 0. Therefore P ( | X | x ) = if x < , F ( x )- lim y - x F ( y ) otherwise. 4. P (- X x ) = P ( X - x ), so P (- X x ) = 1- lim y - x F ( y ) 2 Appetizers: Dies ( 10 Points ) On a modern die, the face value 6 is opposite to the face value 1, the face value 5 to the face value 2, and the face value 4 to the face value 3. Now three fair modern dies are rolled one by one. 1. Describe the probability space ( , U , P ) in this problem. ( 2 Points ) 2. The totals X is a random variable. Describe the -algebra U ( X ). ( 2 Points ) 3. Compute P ( X = 9) and P ( X = 12). ( 2 Points ) 4. Are two probabilities equal? Can you get the conclusion without comput- ing the probabilities? ( 2 Points ) 5. Old Etruscan die show 1 and 2, 3 and 4, 5 and 6 on opposite sides. If we use three fair Old Etruscan dies instead of morden ones, P ( X = 9) = P ( X = 12)? Why? ( 2 Points ) Solution: 1. The sample space is all possible outcomes, i.e. = { (1 , 1 , 1) , (1 , 1 , 2) (6 , 6 , 6) } . The -algebra U is the set of all subsets of . The probability measure P ( A ) = | A | / | | where A U and | | is the cardinality of the set. 2. The -algebra U ( X ) is the collection of any intersections and unions of sets, , , { (1 , 1 , 1) , i.e. the totals is 3 } , { (2 , 1 , 1) , (1 , 2 , 1) , (1 , 1 , 2) , i.e. the totals is 4 } { (6 , 6 , 6) , i.e. the totals is 18 } . In other words, U ( X ) is generated by the above sets. 3. X = X 1 + X 2 + X 3 , X 1 , X 2 , X 3 are the values of the first, second and third dies, respectively. We compute P ( X = 9) by simple counting. If X 1 = 1, then X 2 = 2 , X 3 = 6 or X 2 = 3 , X 3 = 5 or ..., totally 5 cases. X 1 number of cases 1 5 2 6 3 5 4 4 5 3 6 2 Available: Oct 09, 2007 Due: Oct 16, 2007 in class 2 TA: Roger, rdonald@acm.caltech.edu Mulin, mulinch@caltech.edu ACM116 Introduction to Stochastic Processes and Modeling Therefore, P ( X = 9) = 25 6 3 11 . 57%. For P ( X = 12), we have X 1 number of cases 1 2 2 3 3 4 4 5 5 6 6 5 So, P ( X = 12) = 25 6 3 11 . 57%....
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hw1soln - Solution 1 Mulin Cheng mulinch@caltech.edu...

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