Solution: Expected Exit Time from a Bounded Interval

Exercise: Expected Exit Time from a Bounded Interval

Part 1

$Y_{n+1} = S_{n+1}^2 - (n+1) = (S_n + X_{n+1})^2 - n - 1 = S_n^2 + 2S_n X_{n+1} + X_{n+1}^2 - n - 1$ .

Taking conditional expectation, using independence and $\mathbb{E}[X_{n+1}] = 0$ , $\mathbb{E}[X_{n+1}^2] = 1$ :

\mathbb{E}[Y_{n+1} \mid \mathcal{F}_n] = S_n^2 + 0 + 1 - n - 1 = S_n^2 - n = Y_n.

Martingale. ✓

Part 2

Increments: $|Y_{n+1} - Y_n| = |2S_n X_{n+1} + X_{n+1}^2 - 1|$ . On $\{\tau > n\}$ , $S_n \in [0, N]$ , so $|Y_{n+1} - Y_n| \le 2N + 2$ — bounded.

For $\mathbb{E}[\tau] < \infty$ : in any block of $N$ consecutive $+1$ increments the walk would be forced out. The probability of such a block is $2^{-N}$ , and the blocks are disjoint, so:

\mathbb{P}(\tau > kN) \le (1 - 2^{-N})^k,

which is summable, so $\mathbb{E}[\tau] < \infty$ by a standard expectation-via-tail-sum argument. OST condition (3) is satisfied.

Part 3

OST: $\mathbb{E}[Y_\tau] = Y_0 = k^2$ .

Also, $\mathbb{E}[Y_\tau] = \mathbb{E}[S_\tau^2] - \mathbb{E}[\tau]$ . On $\{S_\tau = 0\}$ : $S_\tau^2 = 0$ . On $\{S_\tau = N\}$ : $S_\tau^2 = N^2$ . So:

\mathbb{E}[S_\tau^2] = N^2\cdot \mathbb{P}(S_\tau = N) = N^2\cdot k/N = kN.

Substituting: $k^2 = kN - \mathbb{E}[\tau]$ , so $\mathbb{E}[\tau] = kN - k^2 = k(N - k)$ . ✓

Part 4

(k, N) = (50, 100)

gives

\mathbb{E}[\tau] = 50\cdot 50 = 2500

steps, i.e. 2500 seconds ≈ 42 minutes on average.

import numpy as np
import matplotlib.pyplot as plt

k_vals = np.arange(1, 100)
E_tau = k_vals * (100 - k_vals)

plt.plot(k_vals, E_tau)
plt.xlabel('starting position k')
plt.ylabel('E[tau]')
plt.title('E[tau] = k(N-k), N=100')
# Parabola peaking at k=50 with E[tau]=2500 and zero at k=0, 100 (instant exit).

The parabola peaks at the midpoint — farthest from both barriers, longest expected time — and drops to zero at each boundary (instant exit).

Takeaways

Two-martingale OST identity. Apply OST to $S_n$ to get the exit probabilities; apply OST to $S_n^2 - n$ to get the exit time. Both are one-line computations once the right martingale is chosen.
$\mathbb{E}[\tau] = k(N - k)$ — the canonical quadratic exit-time formula. Memorise it; it appears in every trading-rule mean-holding-time argument.
Continuous-time analogue. For Brownian motion started at $x \in (a, b)$ , the expected exit time from $(a, b)$ is $(x - a)(b - x)/\sigma^2$ — exact same formula, scaled by diffusion variance. Same martingale ( $W_t^2 - t$ ) drives the derivation.
$Y_n = S_n^2 - n$ is the discrete quadratic-variation martingale. It is to the discrete walk what $W_t^2 - t$ is to Brownian motion — the foundational $L^2$ -martingale of the process.