Starting at value 0, the fortune of an investor increases per week by 200 with probability 3/8, remains constant with probability 3/8 and decreases by 200 with probability 2/8. The weekly increments of the investor's fortune are assumed to be independent. The investor stops the 'game' as soon as he has made a total fortune of 2000 or a loss of 1000, whichever occurs first. By using suitable martingales and applying the optional stopping theorem, determine:

The probability $p_{2000}$ that the investor finishes the 'game' with a profit of 2000, the probability $p_{-1000}$ that the investor finishes the 'game' with a loss of 1000, and the mean duration of the 'game', $E(N)$.

Good evening, I have tried to solve this problem, but I don't know how to solve it. In my class we have seen the theory, and have not resulted trouble applying that theory. I am stuck, could you guide me to solve the exercise? Please. I really want to understand the link between theory and application


By assumption, the weekly increment of the investor's fortune has the distribution

$$\nu := \frac{3}{8} \delta_{200} + \frac{3}{8} \delta_0 + \frac{2}{8} \delta_{-200}$$

where $\delta_x$ denotes the Dirac measure at $x \in \mathbb{R}$. Since the increments are assumed to independent, this means that the fortune $S_n$ after the end of the $n$-th week is given by

$$S_n = \sum_{j=1}^n X_j$$

where the $X_j$ are independent random variables such that $X_j \sim \nu$. Through the remaining part, we assume that

$$X_j \sim \tilde{\nu} := \frac{3}{8} \delta_1 + \frac{3}{8} \delta_0 + \frac{2}{8} \delta_{-1}.$$

This doesn't do any harm; we can obain the original process by multiplying with the factor $200$. (It simplifies the calculations in the last part a lot.)

It is not difficult to see that $\mu := \mathbb{E}X_j = \frac{1}{8}> 0$ and this means that $(S_n)_n$ is not a martingale (with respect to the canonical filtration). Therefore, we set

$$\tilde{S}_n := \sum_{j=1}^n X_j - n \cdot \mu = S_n - n \mu.$$

It follows from the independence of the random variables $X_j$ that this process is a martingale. Now define a stopping time

$$\tau := \inf\{n \geq 0; S_n \geq 10\} \wedge \inf\{n \geq 0; S_n \leq -5\}.$$

(Here $a \wedge b = \min\{a,b\}$.) Note that

$$\tau = \inf\{n \geq 0; S_n = 10\} \wedge \inf\{n \geq 0; S_n = -5\}.$$

Applying the optional stopping theorem to the (bounded) stopping time $\tau \wedge n$ yields $\mathbb{E}\tilde{S}_{n \wedge \tau}=0$, i.e. $$\mu \mathbb{E}(\tau \wedge n) = \mathbb{E}(S_{n \wedge \tau}).$$ Since $\mathbb{E}(|S_{n \wedge \tau}|) \leq 10$, the monotone convergence theorem shows $\tau \in L^1$. Therefore, the dominated convergence theorem yields

$$\mathbb{E}\tau = \frac{1}{\mu} \mathbb{E}(S_{\tau}) = \frac{1}{\mu} (10 p_1 - 5 p_2) \tag{1}$$ where $p_1 := \mathbb{P}(S_{\tau}=10)$ and $p_2 := \mathbb{P}(S_{\tau}=-5)$. Obviously, $$p_1+p_2 = 1 \tag{2}$$

So far we have 2 equalities, but 3 unknown variables ($p_1,p_2$ and $\mathbb{E}\tau$); this means that we need some more information. To this end, we note that

$$Z_n := \left( \frac{2}{3} \right)^{S_n}$$

defines a martingale. Therefore, it follows again from the optional stopping theorem that $\mathbb{E}Z_{\tau} = \mathbb{E}Z_0=1$. This yields

$$1 = \mathbb{E}Z_{\tau} = \left( \frac{2}{3} \right)^{10} p_1 + \left( \frac{2}{3} \right)^{-5} p_2. \tag{3}$$

Summarizing, we have

$$\begin{align*} p_1+p_2 &= 1 \\ \left( \frac{2}{3} \right)^{10} p_1 + \left( \frac{2}{3} \right)^{-5} p_2 &= 1. \end{align*}$$

Solving this linear system yields $p_1$, $p_2$. $\mathbb{E}\tau$ can be calculated using $(1)$.

Remark Note that it is essential that $\mu = \mathbb{E}X_1 \neq 0$. For example if $$X_j \sim \frac{1}{4} \delta_{-1} + \frac{1}{2} \delta_0 + \frac{1}{4} \delta_1$$ then the above reasoning does not apply.

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