Solution: Variance and Drift of a Biased Random Walk

Exercise: Variance and Drift of a Biased Random Walk

Part 1 — Single-tick moments

Each $X_i$ takes value $+1$ with probability $p = 0.51$ and $-1$ with probability $q = 0.49$:

For the variance:

Part 2 — $n$-step moments

Since $S_n = \sum_{i=1}^{n} X_i$ with $X_i$ i.i.d., linearity of expectation gives:

Independence of the $X_i$ makes the variance additive:

Part 3 — One day of ticks ($n = 10{,}000$)

The expected gain is $200, but the standard deviation is $100. One standard deviation below the expected path is still $100 profit; one standard deviation above is $300. The distribution of daily P&L is wide relative to its mean.

Part 4 — Signal-to-noise ratio

Setting this equal to 1:

So after 2500 ticks (roughly a quarter-day), the expected move is one standard deviation above zero. More usefully, the signal-to-noise ratio grows only as $\sqrt{n}$ — to get a ratio of 2 (a meaningful statistical signal) requires $n = 10{,}000$, and a ratio of 3 requires $n = 22{,}500$.

Takeaways

• Drift scales as $n$, noise scales as $\sqrt{n}$. Their ratio — the Sharpe-like signal-to-noise — grows only as $\sqrt{n}$. This is why statistical significance for small drifts requires huge samples.
• $\mathbb{E}[X_i^2] = 1$ whenever $X_i \in \{\pm 1\}$, regardless of $p$. The variance reduces to $1 - \mu^2 = 4pq$, which is maximised at $p = 1/2$ and degenerate at the extremes.
• Practical consequence. When fitting a random-walk model to intraday data, tiny drift estimates come with large standard errors. A "statistically significant" daily drift of 2 bps needs thousands of days to detect reliably — a classic reason why expected-return estimation is harder than volatility estimation.