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Why Pre-Averaged Returns Use a Difference of Price Blocks

Article Quant Q&A · Author: XY0

Summary

The document explains why a simple pre-averaging formula for noisy high-frequency log prices subtracts one block of observations from another instead of adding them as in an ordinary average. Pre-averaging is described as a method for reducing the effect of microstructure noise when estimating measures from observed prices that combine a latent price process with noise.

The derivation starts from a general weighted sum of observed log-price differences. With the triangular weighting function that rises to the midpoint and then falls, the discrete changes in the weights have opposite signs on either side of the midpoint. Rewriting the weighted returns in terms of price observations therefore yields a positive sum for the latter half of the window and a negative sum for the first half. The result assumes an even window length of at least two observations. The document gives the algebraic rationale, but does not compare alternative weights or quantify estimator performance.

Key ideas

  • Pre-averaging uses weighted observed returns to reduce microstructure noise in high-frequency price data.
  • The triangular weighting function increases to the window midpoint and decreases afterward.
  • The weight changes have opposite signs in the two halves of the window.
  • That sign change produces a difference between the later and earlier price blocks.
  • The displayed formula assumes an even window length of at least two observations.

Tags

Full text
# the pre-averaging function in Jacod et al


# the pre-averaging function in Jacod et al












In the paper of jacod et al the authors used the pre-averaging function to deal with microstructure noise. They suggest the easiest function which is $$\bar{Z_i} = \frac{1}{kn} \left( \sum_{j=kn/2}^{kn-1} Z_{i+j}-\sum_{j=0}^{kn/2-1}Z_{i+j} \right) \tag{3.13}$$ What is not really clear, why it is used a "minus" inside the parenthesis instead of a "plus" which makes more sense since we are computing an average?

## Answer by Pleb (score 1, accepted)

https://quant.stackexchange.com/a/78039

## $\bar{Z}_i$ denotes the pre-averaged returns:

The difference within the bracket, comes from the fact that $\bar{Z}_i$ is a particular instance of pre-averaged returns, when the weighting function is chosen as $g(x) = \min(x,1-x)$ and for $k_n$ being even.

### The intuition

Within High-Frequency Econometrics you often work under the assumption of observing a noisy log-price process:

$$ Z_i^n = X_i^n + \epsilon_i^n, \tag{1.1} $$ where $X_i^n$ is the true latent log-price process for $n$ intraday observations and $\epsilon_i^n$ is the associated noise-process. The primary objective of noise-robust realized measures is to mitigate the impact of noise in the observable log-price process.

One such technique, the pre-averaging estimator, achieves this goal by computing a weighted average based on the observable log-price differences (aka. noisy log-returns). This operation effectively "averages out" the noise inherent in the observable log-price process.

### Deriving $\bar{Z}_i$

The generalized formula for the pre-averaging estimator as provided by the authors, is expressed as: \begin{align*} \bar{Z}_i^n &= \sum_{j=1}^{k_n-1} g\left(\frac{j}{k_n}\right) \left(Z_{i+j} - Z_{i+j-1}\right) \\ &= -\sum_{j=0}^{k_n-1} \left(g\left(\frac{j+1}{k_n}\right) - g\left(\frac{j}{k_n}\right)\right) Z_{i+j}, \tag{3.5} \end{align*}

where, $k_n = \theta \sqrt{n}$ defines the pre-averaging window length and grows with the sample rate ($\theta$ is user-defined). Additionally, the equation numbering aligns with the conventions established in the referenced paper, although with a notation change from $V_i$ to $Z_i$.

If we employ the weighting function $g(x) = \min(x,1-x)$, it becomes apparent that:

$$ g\left(\frac{j}{k_n}\right) = \begin{cases} \frac{j}{k_n}, \qquad\qquad\:\: \text{for } \frac{j}{k_n} \leq \frac{1}{2}\\ 1- \frac{j}{k_n}, \qquad\quad \text{for } \frac{j}{k_n} > \frac{1}{2}\\ \end{cases}, $$ implying that we are in the first scenario when $j \leq \frac{1}{2}k_n$ and the latter otherwise. Substituting the weighting function into equation (3.5), bifurcates the sum into two scenarios: one where it sums over $j \leq \frac{1}{2}k_n$ and the other for $j > \frac{1}{2}k_n$. Reducing the equation yields the expected result (eq. 3.13 in the paper):

\begin{align*} \bar{Z}_i^n &= -\sum_{j=0}^{k_n/2-1} \left(\frac{j+1}{k_n} - \frac{j}{k_n}\right)Z_{i+j} - \sum_{j=k_n/2}^{k_n-1} \left(\left(1-\frac{j+1}{k_n}\right) - \left(1-\frac{j}{k_n}\right)\right)Z_{i+j}\\ &= - \frac{1}{k_n} \sum_{j=0}^{k_n/2-1} Z_{i+j} + \frac{1}{k_n} \sum_{j=k_n/2}^{k_n-1} Z_{i+j}\\ &= \frac{1}{k_n} \left(\sum_{j=k_n/2}^{k_n-1} Z_{i+j} - \sum_{j=0}^{k_n/2-1} Z_{i+j}\right) \end{align*}

As we can observe, it is necessary to assume that $k_n$ is even and $k_n \geq 2$, in order to ensure a valid sum indexation.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.