How Average Correlation Shrinkage Forms a Correlation Matrix

# Interpreting the average correlation figure in Pozzi et al. (2013) paper


# Interpreting the average correlation figure in Pozzi et al. (2013) paper












Has anyone read the paper "Spread of risk across financial markets:better to invest in the peripheries" by F. Pozzi, T. Di Matteo & T. Aste? If so, how do you interpret equation (1) in the dependency measure subsection of the Methods section? It looks like it produces an average correlation figure:

\begin{equation} \bar{R}_{ij}^w(t)=\frac{1}{2(\tau + 1)} \left( \sum_{s=t-\tau}^t R^w_{ij}(s) + \sum_{i=1}^{j-1} \sum_{j=2}^{N} \sum_{s=t-\tau}^t \frac{2 R_{ij}^w(s)}{N(N-1)} \right).\\ \end{equation}

That average figure can then be used (using methods from elsewhere in the paper and the cited papers, particularly citation 9, which I'm comfortable with) to form a shrunk covariance and correlation matrix. Is that right? OR, does the equation produce average values for each of the pairwise correlations, resulting, by itself, in a shrunk correlation matrix?

Not really expecting many answers, but anything is appreciated, thanks.

## Answer by Pleb (score 1)

https://quant.stackexchange.com/a/69152

## Interpretation:

The authors shrink the 6 month average of the exponentially smoothed correlations towards the average sample correlation. Thus your second formulation is correct, in the sense that equation (1) already contains shrinkage.

The basis of my statement comes from the derivations below. Here, I highlight how you can recover equation (1) using the shrinkage equation found in the paper of Ledoit and Wolf (2003).

### Basis of my statement:

I will redefine some of the statements from the paper for completeness. They define the exponentially smoothed weighted moments as:

$$\sum_{s=t-\tau}^t w_s f^w\left(r(s)\right),$$

with $w_s$ being the exponential weights, $r(s)$ being the daily returns and $f(\cdot)$ being (in our case) the summand of the empirical correlation function. If we would take the empirical average over the above function we get:

$$\bar{f}^{w}(t)=\frac{1}{\tau + 1}\sum_{s=t-\tau}^t w_s f^w\left(r(s)\right).$$

The above formulation sums over $\tau + 1$ elements, since for $\tau = 0$ the sum contains one element.

### The shrinkage equation:

We can define the shrinkage equation with $F$ being the target and $S$ being the exponentially smoothed sample correlation. Here, we set $\delta = \frac{1}{2}$ and see that:

\begin{align} \bar{R}_{ij}^w(t) &= \delta F + (1-\delta) S\\ &= \frac{1}{2} \left(F + S\right)\\ &= \frac{1}{2} \left( \frac{2}{(N-1)N}\sum_{i=1}^{j-1} \sum_{j=2}^{N} R_{ij}^w(t) + R^w_{ij}(t)\right), \end{align}

where you can find the formula for the average sample correlation in the original paper of Ledoit and Wolf (2003) (appendix A).

Averaging over the past $\tau$ months give us equation(1):

\begin{align} \bar{R}_{ij}^w(t)&=\frac{1}{2} \left( \frac{2}{(N-1)N} \frac{1}{\tau + 1}\sum_{i=1}^{j-1} \sum_{j=2}^{N}\sum_{s=t-\tau}^t R_{ij}^w(s) + \frac{1}{\tau + 1}\sum_{s=t-\tau}^t R^w_{ij}(s)\right)\\ &=\frac{1}{2(\tau + 1)} \left( \frac{2}{(N-1)N} \sum_{i=1}^{j-1} \sum_{j=2}^{N}\sum_{s=t-\tau}^t R_{ij}^w(s) + \sum_{s=t-\tau}^t R^w_{ij}(s)\right),\\ \end{align}

where the interpretation of the triple sum can be viewed as the 6 month average of the averaged pairwise sample correlations. The authors set $\tau = 125$ days which is 6 months.

As an added bonus, the authors further define the averaged weighted covariance matrix with shrinkage in the supplementary material chapter S.5. They further define $P^w$ as a diagonal matrix with averaged weighted variances over the main diagonal defined as $(\bar{s}_{kh}^w)^2=\frac{1}{\tau + 1}\sum_{k=t-\tau}^t (\hat{s}_{kh}^w)^2$.