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BBy Bot
Jun 09'24
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In the gambler's ruin problem, assume that the gambler initial stake is 1 dollar, and assume that her probability of success on any one game is [math]p[/math]. Let [math]T[/math] be the

number of games until 0 is reached (the gambler is ruined). Show that the generating function for [math]T[/math] is

[[math]] h(z) = \frac{1 - \sqrt{1 - 4pqz^2}}{2pz}\ , [[/math]]

and that

[[math]] h(1) = \left \{ \begin{array}{ll} q/p, & \mbox{if $q \leq p$}, \\ 1, & \mbox{if $q \geq p,$} \end{array} \right. [[/math]]

and

[[math]] h'(1) = \left \{ \begin{array}{ll} 1/(q - p), & \mbox{if $q \gt p$}, \\ \infty, & \mbox{if $q = p.$} \end{array} \right. [[/math]]

Interpret your results in terms of the time [math]T[/math] to reach 0. (See also Example.)

BBy Bot
Jun 09'24
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Show that the Taylor series expansion for [math]\sqrt{1 - x}[/math] is

[[math]] \sqrt{1 - x} = \sum_{n = 0}^\infty {{1/2} \choose n} x^n\ , [[/math]]

where the binomial coefficient [math]{1/2} \choose n[/math] is

[[math]] {{1/2} \choose n} = \frac{(1/2)(1/2 - 1) \cdots (1/2 - n + 1)}{n!}\ . [[/math]]

Using this and the result of Exercise, show that the probability that the gambler is ruined on the [math]n[/math]th step is

[[math]] p_T(n) = \left \{ \begin{array}{ll} \frac{(-1)^{k - 1}}{2p} {{1/2} \choose k} (4pq)^k, & \mbox{if $n = 2k - 1$,} \\ 0, & \mbox{if $n = 2k$.} \end{array} \right. [[/math]]

BBy Bot
Jun 09'24
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For the gambler's ruin problem, assume that the

gambler starts with [math]k[/math] dollars. Let [math]T_k[/math] be the time to reach 0 for the first time.

  • Show that the generating function [math]h_k(t)[/math] for [math]T_k[/math] is the [math]k[/math]th power of the generating function for the time [math]T[/math] to ruin starting at 1. Hint: Let [math]T_k = U_1 + U_2 +\cdots+ U_k[/math], where [math]U_j[/math] is the time for the walk starting at [math]j[/math] to reach [math]j - 1[/math] for the first time.
  • Find [math]h_k(1)[/math] and [math]h_k'(1)[/math] and interpret your results.
BBy Bot
Jun 09'24
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(The next three problems come from Feller.[Notes 1])

As in the text, assume that [math]M[/math] is a fixed positive integer.

  • Show that if a gambler starts with an stake of 0 (and is allowed to have a negative amount of money), then the probability that her stake reaches the value of [math]M[/math] before it returns to 0 equals [math]p(1 - q_1)[/math].
  • Show that if the gambler starts with a stake of [math]M[/math] then the probability that her stake reaches 0 before it returns to [math]M[/math] equals [math]qq_{M-1}[/math].

Notes

  1. W. Feller, op. cit., pg. 367.
BBy Bot
Jun 09'24
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Suppose that a gambler starts with a stake of 0 dollars.

  • Show that the probability that her stake never reaches [math]M[/math] before returning to 0 equals [math]1 - p(1 - q_1)[/math].
  • Show that the probability that her stake reaches the value [math]M[/math] exactly [math]k[/math] times before returning to 0 equals [math]p(1-q_1)(1 - qq_{M-1})^{k-1}(qq_{M-1})[/math]. Hint: Use Exercise.
BBy Bot
Jun 09'24
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In the text, it was shown that if [math]q \lt p[/math], there is a positive probability that a gambler, starting with a stake of 0 dollars, will never return to the origin. Thus, we will now assume that [math]q \ge p[/math]. Using Exercise, show that if a gambler starts with a stake of 0 dollars, then the expected number of times her stake equals [math]M[/math] before returning to 0 equals [math](p/q)^M[/math], if [math]q \gt p[/math] and 1, if [math]q = p[/math]. (We quote from Feller: “The truly amazing implications of this result appear best in the language of fair games. A perfect coin is tossed until the first equalization of the accumulated numbers of heads and tails. The gambler receives one penny for every time that the accumulated number of heads exceeds the accumulated number of tails by [math]m[/math]. The ‘fair entrance fee’ equals 1 independent of [math]m[/math].”)

BBy Bot
Jun 09'24
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In the game in Exercise, let [math]p = q = 1/2[/math] and [math]M = 10[/math]. What is the probability that the gambler's stake equals [math]M[/math] at least 20 times before it returns to 0?

BBy Bot
Jun 09'24
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Write a computer program which simulates the game in Exercise for the case [math]p = q = 1/2[/math], and [math]M = 10[/math].

BBy Bot
Jun 09'24
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In de Moivre's description of the game, we can modify the definition of player A's fortune in such a way that the game is still a martingale (and the calculations are simpler). We do this by assigning nominal values to the counters in the same way as de Moivre, but each player's current fortune is defined to be just the value of the counter which is being wagered on the next game. So, if player A has [math]a[/math] counters, then his current fortune is [math](q/p)^a[/math] (we stipulate this to be true even if [math]a = 0[/math]). Show that under this definition, player A's expected fortune after one play equals his fortune before the play, if [math]p \ne q[/math]. Then, as de Moivre does, write an equation which expresses the fact that player A's expected final fortune equals his initial fortune. Use this equation to find the probability of ruin of player A.

BBy Bot
Jun 09'24
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Assume in the gambler's ruin problem that [math]p = q = 1/2[/math].

  • Using Equation, together with the facts that [math]q_0 = 1[/math] and [math]q_M = 0[/math], show that for [math]0 \le z \le M[/math],
    [[math]] q_z = {{M - z}\over M}\ . [[/math]]
  • In Equation, let [math]p \rightarrow 1/2[/math] (and since [math]q = 1 - p[/math], [math]q \rightarrow 1/2[/math] as well). Show that in the limit,
    [[math]] q_z = {{M - z}\over M}\ . [[/math]]
    Hint: Replace [math]q[/math] by [math]1-p[/math], and use L'Hopital's rule.