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?

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    $\begingroup$ yes, I edited my post $\endgroup$
    – XY0
    Jan 17 at 19:03

1 Answer 1


$\bar{Z}_i$ 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.


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