# 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?

• yes, I edited my post
– XY0
Jan 17 at 19:03

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