7
$\begingroup$

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?

$\endgroup$

2 Answers 2

3
$\begingroup$

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).

$\endgroup$
1
  • $\begingroup$ Thanks for the paper link. Do you know of any more recent work on estimating the mean returns? Merton seemed to hope others would pursue this in that paper. $\endgroup$
    – bcf
    Jul 20, 2015 at 21:13
1
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ Hopefully to add to the intuition: BM is a fractal so zooming in doesn’t help. $\endgroup$
    – Bob Jansen
    Mar 27 at 19:54
  • $\begingroup$ It is a correct answer but I am not sure I agree "it isn't a problem". With some numerical examples you can show that even with 100 years of data, the estimate still has a fair amount of uncertainty. And the relevancy of the data from 98, 99 , 100 years ago is questionable on different grounds: that the economy may have changed. So there is a fundamental limit here to how well we can estimate returns $\endgroup$
    – nbbo2
    Mar 28 at 13:04
  • $\begingroup$ That's true. But OP used "problem" in the sense that the estimation was behaving at odds with what ought to be, so I corrected that. $\endgroup$
    – Arshdeep
    Mar 28 at 13:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.