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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?

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    $\begingroup$ I posted a the new answer below. If you thing it is the best answer, please mark it as such. I discuss 252 instead of the Sqrt(252) as you suggested. $\endgroup$ Dec 16, 2023 at 21:29

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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).

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    $\begingroup$ 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. $\endgroup$
    – byouness
    Sep 15, 2020 at 9:07
  • $\begingroup$ Why would it also be in the denominator $\endgroup$
    – develarist
    Sep 15, 2020 at 18:01
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    $\begingroup$ 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. $\endgroup$
    – mark leeds
    Sep 15, 2020 at 18:47
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    $\begingroup$ 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. $\endgroup$ Sep 15, 2020 at 19:09
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    $\begingroup$ @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 $var(x) = cov(x, x)$). $\endgroup$
    – byouness
    Sep 16, 2020 at 8:28
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No.

By definition,

Corr[x, y] = Cov[x, y] / Sqrt( Var[x] Var[y] )

where x and y are the daily returns of two assets.

If annualizing the (variance and) covariance matrix of daily returns requires multiplication by 252, then the correlation of annual annual returns X and Y is

Corr[X, Y] = Cov[X, Y] / Sqrt( Var[X] Var[X] ) = 252 Cov[x, y] / Sqrt( 252 Var[x] 252 Var[y] )

because variances and covariances are the elements of the covariance matrix.

Canceling 252^2 inside que square root and 252 in the numerator yields makes the last member to satisfy the definition of the correlation of daily returns. Therfore

Corr[X, Y] = Corr[x, y]

Becouse 252 could be substituded by any other constant, the correlations in invariant with the period of returns.

Notes:

  1. The assumption that the covariance matrix of log returns scales with the period of such returns holds if the return of an asset is not correlated with past returns of the same or other assets, and, given an arbitrary period to measure returns, the covariances are the same whatever the start of such period.
  2. Variance is a particular case of a covariance of a variable with itself: Cov(X, X) = Var(X). Therefore, if the covariance scales with the period of the returns, so must the variance.
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  • $\begingroup$ PLZ. Check your equations, there are a few typos $\endgroup$
    – T123
    Dec 16, 2023 at 22:35

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