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Unformatted text preview: 6.262 Discrete Stochastic Processes, Spring 2010 Problem Set 5 — Solutions due: Friday, March 12, 2010 Problem 1 (Exercise 3.4) To show equality between events, we proceed as follows. Given events A and B , A = B if and only if for any given ω ∈ A we can show that ω also belongs to B and for any given ω ∈ B , ω also belongs to A . a) Comparing { N ( t ) < n } to { S n > t } , we see that if there are fewer than n arrivals in (0 ,t ], then the n th arrival must have taken longer than t . Thus, for any sample path ω ∈ { N ( t ) < n } , we also have ω ∈ { S n > t } . Conversely, if the n th arrival took longer than t , then there must be fewer than n arrivals in (0 ,t ]. Thus, for any sample path ω ∈ { S n > t } , we also have ω ∈ { N ( t ) < n } . Thus, { S n > t } = { N ( t ) < n } . b) If ω ∈ { N ( t ) ≤ n } , then it need not be true that S n ( ω ) ≥ t — in fact, if ω is such that N ( t )( ω ) = n , then S n ( ω ) ≤ t . So, { N ( t ) ≤ n } 6 = { S n ≥ t } c) If ω ∈ { S n < t } , then the process must have registered at least n arrivals in (0 ,t ]. We can therefore say that ω ∈ { N ( t ) ≥ n } . However, to show that ω need not belong to { N ( t ) > n } (which is the event we are actually interested in) let ω be such that S n ( ω ) < t , but S n +1 ( ω ) > t . Then, ω belongs to { S n < t } . However, N ( t )( ω ) = n , so ω 6∈ { N ( t ) > n } . Therefore, { S n < t } 6 = { N ( t ) > n } Problem 2 (Exercise 3.11) We certainly don’t need Wald’s equality to compute the number of Bernoulli trials up to and including the k th success. However, the point is to get you to understand Wald’s equality on a simple example before we start applying it to more complex situations. a) Let P( X i = 1) = p and P( X i = 0) = 1 p , where the event { X i = 1 } corresponds to a “success” on the i th trial. Let J = min { n  S n ≥ k } and notice that J is a stopping rule, as the event { J = n } = { k 1 sucesses in first n 1 trials and a success on the n th trial } only depends on X 1 ,...,X n . 1 By Wald’s equality, we have that E( S J ) = E( J )E( X 1 ). Note that E ( S J ) = k , since S n changes in increments of 1. Also, E ( X 1 ) = p . It follows that E( J ) = E( S J ) E ( X 1 ) = k p . 1 To show that J is a stopping rule, we also need to show that J is finite with probability 1. However, in this case, the easiest way to show that is by computing E( J ) directly, without Wald’s equality, and showing it’s finite. (Why would that imply that J is finite with probability 1?) Again, keep in mind that the point of this exercise is to illustrate the mechanics of Wald’s equality, not to convince you of its power....
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 Spring '08
 Moon,J
 Probability, Probability theory, Sn, Wald

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