# Effective Time Length of Exponentially Weighted Covariance Matrix Estimate

In  Pafka, Potters and Kondor mention the following in section 2:

In contrast, if this covariance matrix estimate is used for portfolio optimization (i.e. for selecting the portfolio in a mean–variance framework, which involves the inversion of the matrix), the estimation error will be quite large for typical values of the ratio T /N (see Ref. ). In the case of exponential weighting, the results in Ref.  imply that the degree of suboptimality will depend on the ratio of the effective time length −1/ log α and the number of assets N. In particular, since the effective time corresponding to the value of the exponential decay factor α suggested by Ref.  (α = 0.94 for daily data) is shorter than the length of the time windows used in a typical standard (uniformly weighted) covariance matrix estimation, it can be expected that for the same portfolio size N the effect of noise (suboptimality of optimized portfolios) will be larger with exponential weighting than without it.

The reference  in the quoted passage links to another paper by Pafka and Kondor.

However, in neither of these papers do I find a derivation of the effective time length $$-1/\log\alpha$$, where $$\alpha$$ is the parameter of the exponentially weighted covariance matrix, nor do I find the expression "effective time length" anywhere else in the context of exponentially weighted matrices. Is there a paper that derives this result?

## 1 Answer

Usually, when one talks about exponential smoothing, they talk about it's halflife.

So, for example, suppose we exponentially smooth some quantity ( argument carries over to covariance matrix but I'd rather just rather consider the scalar quantity case ) and call the exponentially smoothed estimate $$\hat{smth_t}.$$

So, this means that we have:

$$\hat{smth_t} = \rho \times currentval_{t} + (1-\rho) \times \hat{smth_{t-1}}$$.

This can of course be re-written as

$$\hat{smth_t} = (1-\rho) \sum_{t=0}^{\infty} \rho^{t} \times currentval_{t-i}$$.

So, the half life in the exponential smoothing framework refers to the time it takes for the weight contribution of one of the past currentvals to be $$\frac{1}{2}$$ of what it was it was originally.

So, to figure that out, one sets $$\rho^{halflife} = \frac{1}{2}$$ and solve for $$halflife$$ which gives $$halflife = log(1/2)/log(\rho)$$.

In order to obtain, $$-1/log(\rho)$$, one would have to set $$\rho^{halflife} = e^{-1}$$ but I'm not clear on what the intuition would be behind doing that ? Maybe one of the papers talks about why that makes sense ?