Effective Time Length of Exponentially Weighted Covariance Matrix Estimate

Question

In [1] 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. [10]). In the case of exponential weighting, the results in Ref. [10] 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. [12] (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 [10] 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] https://arxiv.org/abs/cond-mat/0402573

[10] https://arxiv.org/abs/cond-mat/0205119

Answer 1

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 ?