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Exercise 3. Show that Equation (22) is indeed a solution of Equation (20). Verify that, taking po 1, this result gives the values of P(T₁) = p₁ and previously. P(T2) = P2 we obtained

Exercise 3. Show that Equation (22) is indeed a solution of Equation (20). Verify that, taking po 1, this result gives the values of P(T₁) = p₁ and previously. P(T2) = P2 we obtained

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Exercise 3
Page 6 Reflection This is a favorite problem of mine for an intermediate level probability course. Let's think about the solution a bit. The solution for the last part seems like something of a cheat. How does one know that the limit exists in the first place? And it seems strange that the limiting probability doesn't depend on the probability that the deck is initially together by suit. Is this right? Is it possible to get a closed form expression for P(Tk)? For simplicity, let's write pk = P(T), then Equation (17) is 6 of 7
Transcribed Image Text:Page 6 Reflection This is a favorite problem of mine for an intermediate level probability course. Let's think about the solution a bit. The solution for the last part seems like something of a cheat. How does one know that the limit exists in the first place? And it seems strange that the limiting probability doesn't depend on the probability that the deck is initially together by suit. Is this right? Is it possible to get a closed form expression for P(Tk)? For simplicity, let's write pk = P(T), then Equation (17) is 6 of 7
4 1 Pk+1 Pki Po 15 15 (20) We could use transform techniques to solve Equation (20); on the other hand, our experience with those techniques gives us an idea of what a solution should look like. In particular, our intuition is that a solution of Equation (20) will have the form: 1 Pk = + Cak; k ≥0 (21) 4 for appropriate constants C and a. Indeed, one finds 1 Pk = +0 - + (Po − 1)()*; ; k > 0 22 4 Note that this result shows that the limiting probability really doesn't depend on po, the probability the cards in the deck are initially together by suit. Exercise 3. Show that Equation (22) is indeed a solution of Equation (20). Verify that, taking po = 1, this result gives the values of P(T₁) = P₁ and P(T2) p2 we obtained previously.
Transcribed Image Text:4 1 Pk+1 Pki Po 15 15 (20) We could use transform techniques to solve Equation (20); on the other hand, our experience with those techniques gives us an idea of what a solution should look like. In particular, our intuition is that a solution of Equation (20) will have the form: 1 Pk = + Cak; k ≥0 (21) 4 for appropriate constants C and a. Indeed, one finds 1 Pk = +0 - + (Po − 1)()*; ; k > 0 22 4 Note that this result shows that the limiting probability really doesn't depend on po, the probability the cards in the deck are initially together by suit. Exercise 3. Show that Equation (22) is indeed a solution of Equation (20). Verify that, taking po = 1, this result gives the values of P(T₁) = P₁ and P(T2) p2 we obtained previously.
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