Hypothesis Test for Drift in a Brownian Motion

Worked Solution

How to Think About It: This is the bread-and-butter problem of detecting a signal (drift) buried in noise (diffusion). In practice, you observe daily returns and ask: is there statistically significant alpha, or is the drift indistinguishable from zero? The key structural fact is that increments of Brownian motion are i.i.d. Normal, so classical MLE and testing theory apply directly. The punchline -- and the reason this matters on a trading desk -- is that detecting drift is incredibly hard because the signal-to-noise ratio grows only as $\sqrt{T}$, where $T$ is the total observation window.

Quick Estimate: Suppose $\mu = 0.10$ (10% annualized drift) and $\sigma = 0.20$ (20% annualized vol). With daily data ($\Delta = 1/252$) over one year ($n = 252$), the per-step drift is $\mu \Delta = 0.000397$ and per-step std is $\sigma \sqrt{\Delta} \approx 0.0126$. The t-statistic is roughly $\hat{\mu} \sqrt{n\Delta} / \hat{\sigma} = 0.10 \cdot \sqrt{1}/0.20 = 0.50$. That is nowhere near the 1.96 threshold at $\alpha = 0.05$. You need about 15 years of data to get a t-stat of 2. This is the fundamental difficulty of detecting drift.

Formal Solution:

Part (i): MLE for the drift.

Define the increments $Y_i = X_{i\Delta} - X_{(i-1)\Delta}$ for $i = 1, \ldots, n$. By the properties of Brownian motion with drift, these are i.i.d.:

$Y_i \sim N(\mu \Delta, \sigma^2 \Delta)$

The log-likelihood is:

$\ell(\mu, \sigma^2) = -\frac{n}{2} \ln(2\pi \sigma^2 \Delta) - \frac{1}{2\sigma^2 \Delta} \sum_{i=1}^{n}(Y_i - \mu \Delta)^2$

Taking $\partial \ell / \partial \mu = 0$:

$\hat{\mu} = \frac{1}{n\Delta} \sum_{i=1}^{n} Y_i = \frac{X_{n\Delta} - X_0}{n\Delta}$

So the MLE for the drift is just the total displacement divided by total time -- the slope of the line from start to finish. This makes intuitive sense: you are fitting a linear trend.

Since $\hat{\mu} = \bar{Y}/\Delta$ and $\bar{Y} \sim N(\mu \Delta, \sigma^2 \Delta / n)$:

$\hat{\mu} \sim N\!\left(\mu, \frac{\sigma^2}{n\Delta}\right)$

The standard error is:

$\text{SE}(\hat{\mu}) = \frac{\sigma}{\sqrt{n\Delta}}$

In practice, replace $\sigma$ with its MLE $\hat{\sigma}$, where:

$\hat{\sigma}^2 = \frac{1}{n\Delta} \sum_{i=1}^{n}(Y_i - \hat{\mu}\Delta)^2$

Part (ii): Two-sided test for $H_0: \mu = 0$.

Under $H_0$, $\hat{\mu} \sim N(0, \sigma^2 / (n\Delta))$. The test statistic is:

$T = \frac{\hat{\mu}}{\hat{\sigma} / \sqrt{n\Delta}} = \frac{\hat{\mu} \sqrt{n\Delta}}{\hat{\sigma}}$

Under $H_0$, $T$ is asymptotically $N(0,1)$ (and in finite samples has a $t_{n-1}$ distribution, though for large $n$ this is essentially Normal).

The level-$\alpha$ two-sided test rejects $H_0$ when:

$|T| > z_{\alpha/2}$

where $z_{\alpha/2}$ is the upper $\alpha/2$ quantile of the standard Normal. For $\alpha = 0.05$, reject when $|T| > 1.96$.

Part (iii): Distribution of the annualized Sharpe ratio under $H_0$.

The annualized Sharpe ratio is:

$S = \frac{\hat{\mu}}{\hat{\sigma}}$

Notice the relationship to the test statistic:

$T = S \cdot \sqrt{n\Delta} = S \cdot \sqrt{T_{\text{total}}}$

where $T_{\text{total}} = n\Delta$ is the total observation period (in years). Since $T \sim N(0,1)$ under $H_0$:

$S = \frac{T}{\sqrt{n\Delta}} \sim N\!\left(0, \frac{1}{n\Delta}\right)$

Equivalently, $S \sqrt{n\Delta} \sim N(0,1)$ under the null. The standard deviation of $S$ is