Random walks, for quant interviews

The step process is the whole object

A random walk is a running total of independent, identical bets. Let $X_1, X_2, \dots$ be the steps — each $+1$ or $-1$ — and let $S_n = \sum_{i=1}^{n} X_i$ be your position after $n$ steps, starting from $S_0 = 0$. In the symmetric walk each step is $\pm 1$ with probability $1/2$; in the biased walk it is $+1$ with probability $p$ and $-1$ with probability $q = 1 - p$.

Before touching a specific question, say what an interviewer is really checking: whether you keep the two moments of $S_n$ straight. The step has an expectation and a variance; the sum inherits them by two different rules, and candidates who blur those rules get every downstream question — hitting times, ruin, volatility scaling — subtly wrong. State the two facts up front and the rest of the interview is bookkeeping: $\E[S_n]$ adds the step means, $\Var(S_n)$ adds the step variances because the steps are independent.

Expectation stays put; variance grows

For one symmetric step, $\E[X_i] = \tfrac12(+1) + \tfrac12(-1) = 0$, so $\E[S_n] = 0$ for every $n$: the walk has no drift and no pull. But “expected position zero” does not mean “stays near zero.” The spread is what moves. Since $\Var(X_i) = \E[X_i^2] - \E[X_i]^2 = 1 - 0 = 1$ and the steps are independent, variances add: $\Var(S_n) = n$. The standard deviation is $\sqrt{n}$.

That single line — variance linear in time, standard deviation like $\sqrt{t}$ — is the most-tested fact about random walks, so anchor it with a number. After $100$ steps the typical displacement is about $\pm 10$, not $\pm 100$ and not $\pm 50$. Run it four times as long, to $400$ steps, and the spread only doubles to $\pm 20$. Uncertainty accumulates, but it accumulates slowly, and the $\sqrt{t}$ is why.

The derivation the interviewer wants on the board

When they ask you to show it, keep it to two lines and name the assumption each line uses.

— linearity of expectation, which needs no independence. For the symmetric walk $p = q = \tfrac12$ so this is $0$.

— and this line is where independence earns its keep: the cross-covariance terms vanish only because the steps are uncorrelated. For the symmetric walk $\Var(X_i) = 1$, so $\Var(S_n) = n$ and, because the mean is zero, $\E[S_n^2] = n$ — a favorite one-line follow-up. The whole $\sqrt{t}$ law is just “independent variances add, then take a square root.”

Gambler’s ruin, with real numbers

Now the canonical trading-desk version: absorbing barriers at $0$ and $N$, and you start with $i$. In a fair game the reach-the-top probability is beautifully linear, $\Prob(\text{hit } N \text{ first}) = i/N$. Take $i = 3$, $N = 10$: you reach $10$ with probability $3/10 = 30\%$ and go broke with probability $70\%$, and the expected game length is $i(N-i) = 3 \times 7 = 21$ steps.

Then they tilt the odds and watch you. Make each step $+1$ with probability $p = 0.4$ against you, so $r = q/p = 1.5$, and the win probability becomes $\dfrac{1 - r^{\,i}}{1 - r^{\,N}}$. Plug in: $r^3 = 3.375$, $r^{10} = 57.665$, giving $\dfrac{1 - 3.375}{1 - 57.665} = \dfrac{2.375}{56.665} \approx 4.19\%$. A $10$-percentage-point edge against you did not shave $30\%$ down to $25\%$ — it collapsed it to about one in twenty-four. That geometric sensitivity, not the fair-game line, is the point of the question.

Hitting a level, recurrence, and the infinite wait

“Will the symmetric walk ever reach $+100$?” Yes — with probability $1$. The one-dimensional symmetric walk is recurrent: it returns to the origin, and reaches every level, almost surely. The catch that separates a good answer from a glib one: the expected time to get there is infinite. Certain to happen, but with no finite average wait — a distinction interviewers probe because it kills naive “it’ll mean-revert soon” reasoning.

Add drift and certainty breaks. With an upward bias, $p = 0.6$, the walk still climbs forever, but the probability it ever dips a dollar below its start is $q/p = 0.4/0.6 = 2/3$ — even a persistent tailwind leaves a two-in-three chance of being underwater at some point. Ever reaching $+1$ has probability $\min(1,\, p/q) = 1$; ever reaching $-1$ has probability $\min(1,\, q/p) = 2/3$. Recurrence is a knife-edge property of the exact symmetry, and it snaps the moment there is drift.

Why this is the market’s baseline model — and where it cracks

The reason interviewers care: the log-price of an asset is modeled as a random walk with drift, and the entire $\sqrt{t}$ machinery becomes the rule for scaling volatility across horizons. If daily returns have standard deviation $1\%$ and are independent, then annual volatility is $1\% \times \sqrt{252} \approx 15.87\%$ — you scale by $\sqrt{\text{time}}$, never by time. Every “annualize the vol” question is this fact wearing a suit.

Then name the caveats, because a candidate who recites the model without its failure modes looks junior. Real returns are not i.i.d.: volatility clusters, tails are fatter than the walk predicts, and $\sqrt{t}$ scaling breaks the instant returns are autocorrelated. And prices are bounded below by zero, so the arithmetic walk (which happily goes negative) is replaced by a walk in log price — geometric, not additive. The random walk is the right first sentence and the wrong last word.

Common mistakes

• Confusing variance with standard deviation. Variance grows linearly in $t$; the spread grows like $\sqrt{t}$. Saying the walk wanders “proportionally to time” is the single most common error and it flips every scaling question.
• Believing the symmetric walk is pulled back to zero. Mean zero is not mean reversion. There is no restoring force; the walk is recurrent but drifts arbitrarily far, and “it’s due to come back” is the gambler’s fallacy.
• Reusing fair-game intuition on a biased game. In gambler’s ruin a tiny edge compounds geometrically, not linearly. Reaching for $i/N$ when $p \neq q$ is wrong by an order of magnitude, as the $30\% \to 4\%$ example shows.
• Forgetting the wait is infinite. “Reaches the level with probability one” and “reaches it in finite expected time” are different claims for the symmetric walk; only the first is true.
• Letting modeled prices go negative. Treating the price (not log-price) as an additive random walk permits negative prices and misses drift — a tell that you memorized the model without its constraints.