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We generally estimate the covariance matrix of assets using their returns instead of prices. Why is that the case?

I can think of two possible reasons and would appreciate comments/feedback regarding them:

  • Correlation of two non-stationary time series' is spurious since they have trends embedded in them.
  • Variance of prices doesn't make sense. Consider two assets with price sequences of {100, 105, 101, 104, 102, 103} and {100, 101, 102, 103, 104, 105}. Clearly the first asset is more "variable". But the variance of prices for both is precisely the same.

In practical applications, does a covariance matrix estimated using prices actually perform worse?

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  • $\begingroup$ As far as I know, the reason is mostly the first one. It's hard to calculate a correlation between two series with non-constant means and or non-constant variance. In general, it's never even attempted although I have seen a paper or two on it. I think the second reason should be that variance is dependent on the level of the price. Specific "bad" sequences could happen with returns also. . $\endgroup$ – mark leeds Jan 31 at 11:17
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For the same reason you can't meaningfully measure covariance/correlation using price of individual assets...correlation (covariance by extension) represents the comovement in deviations from individual means. You can't represent that if the mean continues to change (ie, series considered aren't stationary). Same goes for multiple assets as is represented in a covariance matrix.

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If you assume that a financial asset price has a change that is a wiener process then you can view the future value of that asset as the initial value plus the sum of the independent daily changes (for equity or returns based then you would need log version of this):

$$ S_t = S_0 + \sum \Delta S_i $$

where $\Delta S_i = S_i - S_{i-1} $ is a wiener process.

You could also state the same for a second asset $T_t$.

If you were to evaluate the covariance of the absolute prices (rather than the changes) then you have:

$$ Cov(S,T) = \frac{1}{n}\sum_{t=1}^n (S_t - E[S])(T_t - E[T]) $$

If you recognise (under the wiener process) that $E[S] = S_0$ and same for T then you expand this out to get:

$$ \frac{1}{n} \sum_{t=1}^n (\sum_{i=1}^t \Delta S_i)(\sum_{i=1}^t \Delta T_i) $$

so the covariance result is dominated by the initial changes rather than any of the latter, which is not valid under the assumed pricing process.

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  • $\begingroup$ That's such an interesting way of looking at it and I've never seen it explained that way before. Basically, even if we make reasonable assumptions about the price change, things still don't work out correctly. Thanks for great answer. $\endgroup$ – mark leeds Feb 1 at 14:17
  • $\begingroup$ Could you please illustrate how you got the expansion to reduce to that expression? I got till $Cov(S,T)=\frac{1}{n}\Sigma_{t=1}^n(\Sigma_{i=1}^t\Delta S_i)(\Sigma_{i=1}^t\Delta T_i)$. It's not obvious how this reduces to that expression. $\endgroup$ – Amrit Prasad Feb 1 at 18:12
  • $\begingroup$ @AmritPrasad yes agreed, misstatement on the expansion but the conclusion still valid: initial terms are still much more dominant $\endgroup$ – Attack68 Feb 1 at 19:22

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