<p>To maybe give a little more intuition (and partially to increase the site's answer's per question ratio), I would add the following to @Rusan's answer.</p>

<p>If we consider estimating the parameters in a stochastic process like
$X_k = \mu \Delta + \sigma \sqrt \Delta \epsilon_k$, then we can do two things. We can either sample more points in a given interval of time (points are closer together) or we can lengthen the given interval of time (by gather data over a longer period of time, for example). When it comes to estimating $\mu$, sampling points
closer together doesn't do us much good because over shorter intervals (higher frequency), the process becomes dominated by the noise. This might help us estimate the variance of the process, but if we want a more precise estimate of the trend ($\mu$), then we need to lengthen the total interval of time ($h$ in @Rusan's answer).</p>

CC BY-SA 3.0
2014-12-31T07:36:58.030Z
2014-12-31T07:36:58.030Z
jmbejara
<p>Your question relates more broadly to modern portfolio theory, and can be illustrated via mean-variance analysis of a univariate time-series. The extension to the multi variate (normal) setting, is trivial. Below I use the term <em>accuracy</em>, in a non-formal way, relating to the variance of an estimator. In particular, when the variance of an estimator can be reduced, take this to mean the estimator can be made more <em>accurate</em>.</p>

<p>Short story: The variance of the estimator for the variance of returns can be reduced by taking more observations (finer observations - up to a limit), whereas this is not true for the variance of the estimator of the expected returns. In this sense, it is <em>more difficult</em> to <em>accurately</em> estimate expected (mean) returns compared to variance of returns. </p>

<p>To see why, consider a time-series of returns, and assume the mean (per unit time), $\mu$, and variance (per unit time), $\sigma^{2}$, are <em>constant</em> over non-overlapping time <em>intervals</em> of frequency $h$. Here, $h$ represents daily, or monthly, or quarterly periods, etc. Further assume that our time-series data are observations over time intervals of length $\Delta$, where $\Delta&lt;&lt;h$. Then, define $n=h/\Delta$, as the number of observations of returns during a time interval of $h$.</p>

<p>For the $k^{\text{th}}$ observation interval (of length $\Delta$) during a period of length $h$, without loss of generality, we can assume that the price process for an asset is given by $$S_{k+\Delta}=S_{k}\exp{(\mu\Delta+\sigma \sqrt{\Delta} \epsilon_{k})},$$ where $\epsilon_{k}$ are iid normal. The logarithmic return (over period $\Delta$) can be written, as
$$X_{k}=\mu\Delta+\sigma \sqrt{\Delta} \epsilon_{k}.$$ For mean-variance portfolio theory, we require estimates of the true values of $\mu$ and $\sigma$, since they are of course, unknown. Call these estimators $\hat{\mu}$ and $\hat{\sigma}$. To answer your question, we only need consider the properties of these estimators.</p>

<p>First
$$\hat{\mu}=\sum_{k=1}^{n}X_{k}/h, \quad \mathbb{E}(\hat{\mu})=\mu, \quad \text{Var}(\hat{\mu})=\sigma^{2}/h.$$</p>

<p>Importantly, the variance of $\hat{\mu}$ depends upon the total length of the observation period and not on the number of observations.</p>

<p>Now consider, the (biased) estimator for $\sigma$:
$$\hat{\sigma}=\sum_{k=1}^{n}X_{k}^{2}/h, \quad \mathbb{E}(\hat{\sigma})=\sigma^{2}+\mu^{2}h/n, \quad \text{Var}(\hat{\sigma})=2\sigma^{4}/n+4\mu^{2}h/n^{2}.$$</p>

<p>The bias can be neglected (see reference for details).</p>

<p>Importantly, the accuracy (variance) of the estimator for $\sigma$, $\text{Var}(\hat{\sigma})$, <strong>does</strong> depend upon the number of observations, $n$ - for fixed $h$. </p>

<p>The consequence of this is, for fixed $h$, by taking finer (smaller) observations intervals, $\Delta$, the accuracy of the variance estimator can be improved ($\text{Var}(\hat{\sigma})$ can be reduced). This is not true of the variance of the estimator of the expected (mean) return, $\text{Var}(\hat{\mu})$, whose accuracy can be improved only by increasing the observation period, $h$.</p>

<p>The reference for this is</p>

<ol>
<li>Merton, R. C. On estimating the expected return on the market. J. financ. econ. 8, 323–361 (1980).</li>
</ol>

CC BY-SA 3.0
2014-12-30T14:59:08.133Z
2014-12-31T13:10:15.340Z
2014-12-31T13:10:15.340Z
Rusan Kax
<p>In the Wikipedia article about the <a href="https://en.wikipedia.org/wiki/Black%E2%80%93Litterman_model" rel="nofollow">Black-Litterman model</a>, it states that the motivation
behind model is that "it is difficult to come up with reasonable estimates of expected returns." Why is it that expected returns are difficult to estimate, whereas the covariances are not?</p>

<econometrics><finance><financial-economics>
Black-Litterman---in what way are expected returns hard to estimate?
jmbejara
# Black-Litterman---in what way are expected returns hard to estimate?

(Asked by: username_0 on 2014-12-30)

In the Wikipedia article about the [Black-Litterman model](https://en.wikipedia.org/wiki/Black%E2%80%93Litterman_model), it states that the motivation
behind model is that "it is difficult to come up with reasonable estimates of expected returns." Why is it that expected returns are difficult to estimate, whereas the covariances are not?

**Tags:** econometrics, finance, financial-economics

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## 2 answers

### Answer by username_1 on 2014-12-30. Score: 4 | Accepted answer

Your question relates more broadly to modern portfolio theory, and can be illustrated via mean-variance analysis of a univariate time-series. The extension to the multi variate (normal) setting, is trivial. Below I use the term *accuracy*, in a non-formal way, relating to the variance of an estimator. In particular, when the variance of an estimator can be reduced, take this to mean the estimator can be made more *accurate*.

Short story: The variance of the estimator for the variance of returns can be reduced by taking more observations (finer observations - up to a limit), whereas this is not true for the variance of the estimator of the expected returns. In this sense, it is *more difficult* to *accurately* estimate expected (mean) returns compared to variance of returns.

To see why, consider a time-series of returns, and assume the mean (per unit time), $\mu$, and variance (per unit time), $\sigma^{2}$, are *constant* over non-overlapping time *intervals* of frequency $h$. Here, $h$ represents daily, or monthly, or quarterly periods, etc. Further assume that our time-series data are observations over time intervals of length $\Delta$, where $\Delta<

For the $k^{\text{th}}$ observation interval (of length $\Delta$) during a period of length $h$, without loss of generality, we can assume that the price process for an asset is given by $$S\_{k+\Delta}=S\_{k}\exp{(\mu\Delta+\sigma \sqrt{\Delta} \epsilon\_{k})},$$ where $\epsilon\_{k}$ are iid normal. The logarithmic return (over period $\Delta$) can be written, as
$$X\_{k}=\mu\Delta+\sigma \sqrt{\Delta} \epsilon\_{k}.$$ For mean-variance portfolio theory, we require estimates of the true values of $\mu$ and $\sigma$, since they are of course, unknown. Call these estimators $\hat{\mu}$ and $\hat{\sigma}$. To answer your question, we only need consider the properties of these estimators.

First
$$\hat{\mu}=\sum\_{k=1}^{n}X\_{k}/h, \quad \mathbb{E}(\hat{\mu})=\mu, \quad \text{Var}(\hat{\mu})=\sigma^{2}/h.$$

Importantly, the variance of $\hat{\mu}$ depends upon the total length of the observation period and not on the number of observations.

Now consider, the (biased) estimator for $\sigma$:
$$\hat{\sigma}=\sum\_{k=1}^{n}X\_{k}^{2}/h, \quad \mathbb{E}(\hat{\sigma})=\sigma^{2}+\mu^{2}h/n, \quad \text{Var}(\hat{\sigma})=2\sigma^{4}/n+4\mu^{2}h/n^{2}.$$

The bias can be neglected (see reference for details).

Importantly, the accuracy (variance) of the estimator for $\sigma$, $\text{Var}(\hat{\sigma})$, **does** depend upon the number of observations, $n$ - for fixed $h$.

The consequence of this is, for fixed $h$, by taking finer (smaller) observations intervals, $\Delta$, the accuracy of the variance estimator can be improved ($\text{Var}(\hat{\sigma})$ can be reduced). This is not true of the variance of the estimator of the expected (mean) return, $\text{Var}(\hat{\mu})$, whose accuracy can be improved only by increasing the observation period, $h$.

The reference for this is

1. Merton, R. C. On estimating the expected return on the market. J. financ. econ. 8, 323–361 (1980).


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### Answer by username_0 on 2014-12-31. Score: 1

To maybe give a little more intuition (and partially to increase the site's answer's per question ratio), I would add the following to @Rusan's answer.

If we consider estimating the parameters in a stochastic process like
$X\_k = \mu \Delta + \sigma \sqrt \Delta \epsilon\_k$, then we can do two things. We can either sample more points in a given interval of time (points are closer together) or we can lengthen the given interval of time (by gather data over a longer period of time, for example). When it comes to estimating $\mu$, sampling points
closer together doesn't do us much good because over shorter intervals (higher frequency), the process becomes dominated by the noise. This might help us estimate the variance of the process, but if we want a more precise estimate of the trend ($\mu$), then we need to lengthen the total interval of time ($h$ in @Rusan's answer).


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