Amazoncomexecobidosasin0898714540 647 so is the

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Unformatted text preview: ach position over the long run. For example, if limk→∞ p(k ) = (1/6, 1/3, 1/3, 1/6)T , then, as the game runs on indefinitely, the pea is expected to be under shell #1 for about 16.7% of the time, under shell #2 for about 33.3% of the time, etc. D E T H A Fly in the Ointment: Everything above rests on the assumption that A is convergent. But A is not convergent for the shell game because a bit of computation reveals that σ (A) = {±1, ±(1/2)}. That is, there is an eigenvalue other than 1 on the unit circle, so (7.10.33) guarantees that limk→∞ Ak does not exist. Consequently, there’s no limiting solution p to the difference equation p(k ) = Ap(k − 1), and the intuitive analysis given above does not apply. IG R Y P Ces`ro to the Rescue: However, A is summable because ρ(A) = 1, and a every eigenvalue on the unit circle is semisimple—these are the conditions in (7.10.36). So as k → ∞, I + A + · · · + Ak−1 k O C p(0) → Gp(0) = p. The job now is to interpret the meaning of this Ces`ro limit in the context of a the shell game. To do so, focus on a particular position—say the j th one—and set up “counting functions” (random variables) defined as X (0) = 1 if the pea starts under shell j , 0 otherwise, and X (i) = 1 if the pea is under shell j after the ith move, 0 otherwise, i = 1, 2, 3, . . . . Notice that X (0) + X (1) + · · · + X (k − 1) counts the number of times the pea occupies position j before the k th move, so X (0) + X (1) + · · · + X (k − 1) /k Copyright c 2000 SIAM Buy online from SIAM Buy from 7.10 Difference Equations, Limits, and Summability 637 represents the fraction of times that the pea is under shell j before the k th move. Since the expected (or mean) value of X (i) is, by definition, E [X (i)] = 1 × P X (i) = 1 + 0 × P X (i) = 0 = pj (i), It is illegal to print, duplicate, or distribute this material Please report...
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