Estimating $\mu$ - only increasing $T$ improves estimate?

Question

Assuming an asset price $S$ follows a geometric Brownian motion (GBM), the log returns $R$ are distributed as $$ R_i := \log\left(\frac{S_i}{S_{i-1}}\right) \sim \mathcal{N}\left(\left(\mu - \frac{\sigma^2}{2}\right) \Delta t, \sigma^2 \Delta t \right), \quad i=1,\ldots,N. $$

Let $m = \left(\mu - \frac{\sigma^2}{2}\right) \Delta t$ and $s^2 = \sigma^2 \Delta t$ and consider calibrating a GBM to some returns $R_i$. We'll use the maximum likelihood estimate for $m$, and for simplicity we assume $s$ is known (as would be the case if we were generating the data through a simulation ourselves), in which case $$ \hat{m} = \frac{1}{N} \sum_{i=1}^M R_i. $$ Then the sampling distribution for the sample mean is approximately $\hat{m} \sim \mathcal{N}\left(m, \frac{s^2}{N}\right)$, and an approximate $(1-\alpha)100\%$ confidence interval for the true mean $m$ is $$ [\hat{m} - z_{\alpha/2}\frac{s}{\sqrt{N}},\: \hat{m} + z_{\alpha/2}\frac{s}{\sqrt{N}}] \qquad (1). $$ In particular, increasing the number of observations $N$ results in a smaller confidence interval. This, of course, is a standard result from elementary statistics.

On the other hand, we really need an estimate for $\mu$ in practice, and from $(1)$ we can derive a confidence interval for $\hat{\mu} = \frac{\hat{m}}{\Delta t} + \frac{\sigma^2}{2}$: \begin{align*} & \hat{m} - z_{\alpha/2}\frac{s}{\sqrt{N}} < m < \hat{m} + z_{\alpha/2}\frac{s}{\sqrt{N}} \\ & \qquad \iff \hat{m} - z_{\alpha/2}\frac{s}{\sqrt{N}} < \left(\mu - \frac{\sigma^2}{2}\right) \Delta t < \hat{m} + z_{\alpha/2}\frac{s}{\sqrt{N}} \\ & \qquad \iff \frac{\hat{m}}{\Delta t} + \frac{\sigma^2}{2} - z_{\alpha/2}\frac{s}{\Delta t\sqrt{N}} < \mu < \frac{\hat{m}}{\Delta t} + \frac{\sigma^2}{2} + z_{\alpha/2}\frac{s}{\Delta t\sqrt{N}} \\ & \qquad \iff \hat{\mu} - z_{\alpha/2}\frac{\sigma}{\sqrt{N \Delta t}} < \mu < \hat{\mu} + z_{\alpha/2}\frac{\sigma}{\sqrt{N \Delta t}}. \end{align*} Then, since $\Delta t = \frac{T}{N}$ for some final observation time $T$, a $(1-\alpha)100\%$ confidence interval for the true drift $\mu$ is $$ [\hat{\mu} - z_{\alpha/2}\frac{\sigma}{\sqrt{T}},\: \hat{\mu} + z_{\alpha/2}\frac{\sigma}{\sqrt{T}}]. $$ In particular, increasing the number of observations $N$ has no effect on the confidence interval for the drift $\mu$. Instead, we only obtain a smaller confidence interval by increasing the final time, $T$.

Indeed, for fixed $T$ we may think of obtaining higher and higher frequency data so that $N$ becomes larger and larger. But then $\Delta t$ becomes smaller and smaller by definition, such that $dt = \frac{T}{N}$. This seems quite counter intuitive: for fixed $T$, no matter if I have 1,000 or $1e16$ observations, I get no closer to my true drift $\mu$. On the other hand, if I have only 10 observations over 100 years, I get a much better estimate of $\mu$.

Have I overlooked something? Perhaps this is a well-known problem with estimating the drift that I'm not aware of?

Answer 1

Yes, you are correct. Consider the following toy example:

1. Log prices follow: $dp_t=\mu dt+\sigma dW_t$

2. Then: $r_{t+h,h}=p_{r+h,h}-p_t ~ N(\mu h, \sigma^2 h)$

3. standard ML estimators:

• $\hat{\mu}=\frac{1}{nh}\sum_{k=1} r_{kh,h}$
• $\hat{\sigma^2}=\frac{1}{nh}\sum_{k=1} (r_{kh,h}-\hat{\mu}h)^2$

Assymptotic distribution of estimators:

• $\sqrt T(\hat{\mu}-\mu) \rightarrow N(0,\sigma^2)$
• $\sqrt n (\hat{\sigma^2}-\sigma^2)\rightarrow N(0,\sigma^4)$

So when $n$ tends to infinity we get precise estimator of $\sigma^2$ , and when $T$ tends to infinity we get it for $\mu$.

This was first noted by Merton (1980).

Answer 2

It is quite intuitive as drift makes itself known over time. If you observe drift over 100 years, you can pin it with more certainty than if you observe it over small period.

Imagine two processes, one with positive drift and one with negative. To distinguish one from the other, would you want to observe over a longer period of time, or with high frequency over a day?

It isn't a problem, you just have to understand the basic of what you are estimating.