# How to annualize the correlation matrix?

If asset returns are daily, and the asset return covariance matrix, $$\Sigma$$, is annualized by $$\Sigma \times 252$$, do I also multiply the correlation matrix by 252 to annualize it?

## 1 Answer

No, because correlation is a unitless quantity. As you use volatilities to do the scaling, the $$\sqrt{252}$$ factor should already be taken into account in them.

If you take a correlation of 1 between two assets, multiplying your correlation matrix by a factor $$C \neq 1$$ risks either to underestimate correlations (by hiding perfect (anti)correlations) or have your matrix not making any sense (correlation greater than 1).

• To make it easy to understand, recall that correlation is $\frac{cov(x,y)}{\sqrt{var(x) var(y)}}$. So, the $252$ in the numerator and denominator will simplify. Sep 15 '20 at 9:07
• Why would it also be in the denominator Sep 15 '20 at 18:01
• Hi: It's in the denominator because the variances are in the diagonals of the covariance matrix so they need to be multiplied by the same factor. But I think it's better to think of it the following way: covariance is the analogue of variance so it increases with time. On the other hand, correlation doesn't increase with time so there's no need to do anything to it. Whether one measures it over a year or a day shouldn't change it. Sep 15 '20 at 18:47
• The above is technically correct, but correlation of daily returns will not necessarily be equal to the correlation of annual returns. The role of random variation (noise) in returns diminishes over longer periods as the signal accumulates, and correlations get stronger. Sep 15 '20 at 19:09
• @develarist, Some important things to keep in mind: (1) correlations are between $-1$ and $1$, if you multiply by $\sqrt{252}$ it won't be a correlation anymore. (2) the rescaling by this $\sqrt{252}$ doesn't work if there is autocorrelation or mean-reversion, works only when iid. (3) To answer your question why it should be applied to the denominator as well, if the increments are iid, then it applies to the covariance but also the variance (particular case as $var(x) = cov(x, x)$). Sep 16 '20 at 8:28