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Consider a random variable $r_t$ which represents the return of an asset at time t. In the univariate case, we just consider $r_t$ to be the return of a single security at time t. Generally, we assume $\mathbb{E}(r_t)=0$ for high frequency returns over a short time period (~1 day). Thus, we can 'standardize' this and define $r_t=\sigma_tz_t$ where $z_t$ has mean zero and variance 1.

In the general case consider the vector of returns $r_t=(r_{1, t}, ..., r_{N, t})^T$. My professor defines $r_t=\Sigma_t^{1/2}z_t$, where $z_t$ is analogously $(z_{1,t}, ..., z_{N,t})^T$ and $\Sigma_t$ is the variance-covariance matrix.

My question is about the covariance matrix. What does it mean to take the square root of a matrix? Why is it applicable in this sense?

Thanks.

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Every covariance matrix $\Sigma$ is positive semi-definite (PSD). Because $\Sigma$ is PSD, then there exists a unique PSD and symmetric matrix, let's call it $\Sigma^{1/2}$, such that $\Sigma = \Sigma^{1/2}\Sigma^{1/2}$ (see here for a proof). We often call $\Sigma^{1/2}$ the square root of $\Sigma$, although, strictly speaking, there are no actual square root operations involved. Note that we could have referred to the "square root" of $\Sigma$ as $B$, such that $\Sigma = BB$.

Regarding the question "why is it applicable in this sense", it is not immediately clear, since you would have to define $r_t$ as $r_t=\left(\Sigma_t^{1/2}\right)^{-1}z_t$ for $r_t$ to be the "whitened" version of $z_t$. That is, if $z_t$ has covariance matrix $\Sigma_t$, then $r_t=\left(\Sigma_t^{1/2}\right)^{-1}z_t$ has covariance matrix $I$. In the 1-dimensional case, if $r$ and $z$ are random variables, and if $z$ has variance $\sigma^2$, then $r = \left(\left(\sigma^{2}\right)^{1/2}\right)^{-1}z = z/\sigma$ has variance $1$. Note the similarity here between $\sigma^2$ and the covariance matrix $\Sigma$.

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  • $\begingroup$ Given that the matrix sigma is symmetric, is sigma^(1/2) also symmetric? $\endgroup$
    – johnf42
    Commented Nov 29, 2023 at 15:01
  • $\begingroup$ Yes, $\Sigma^{1/2}$ is also symmetric. $\endgroup$
    – mhdadk
    Commented Nov 29, 2023 at 15:02
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    $\begingroup$ There are usually many square roots: the assertion of uniqueness is incorrect. (Simple counterexample: let $\Sigma=\pmatrix{2&1\\1&1}=T^\prime T$ where $T=\pmatrix{1&0\\1&1}.$) However, there is a unique symmetric PSD square root. $\endgroup$
    – whuber
    Commented Nov 29, 2023 at 15:54
  • $\begingroup$ @whuber I think more commonly, the square root matrix $T$ is such that $TT = \Sigma$, instead of $T'T = \Sigma$? $\endgroup$
    – Zhanxiong
    Commented Nov 30, 2023 at 1:14
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    $\begingroup$ @Zhanxiong I think the challenge here is whether positive semidefiniteness is defined for non-symmetric matrices or not, since a matrix $A$ and its symmetrized version $\frac{1}{2}(A + A^T)$ define the same quadratic form. Normally, we restrict this definition to symmetric matrices as this implies that their eigenvalues are real (and therefore have a well-defined ordering). However, in the case of the $T$ mentioned by whuber, it is not symmetric, yet its eigenvalues are positive and real, and so it is technically PSD. We can also let $S = T'$ such that $\Sigma = SS'$. $\endgroup$
    – mhdadk
    Commented Dec 1, 2023 at 14:56
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The operator theorists sent me to talk to yall about Matrix Functions.

Given some analytic function $f$ defined on $\mathbb{R}$, we can extend its definition to square matrices by plugging a matrix into the series expansion of the function, where raising a matrix to a power is interpreted as iterated multiplication. It's easy to see that this is equivalent to simply applying $f$ to each of the eigenvalues of the matrix and leaving the eigenvectors alone (or at least, this is the case for diagonalizable matrices, like covariance matrices).

This gives us the technology to make sense of $\Sigma^{\frac{1}{2}}$ in a unique manner. We think of $f(x) = \sqrt{x}$, and so $\Sigma^{\frac{1}{2}}$ as simply the matrix which has the same eigenvectors as $\Sigma$, except the associated eigenvalues are each square rooted.

I'd like to point out that this isn't contrary to what other commenters have said about the matrix square root being non-unique; these are simply different conventions as to what one means by matrix square root. And indeed, we have implicitly been using the principal branch of the square root function, which is not written in stone. (Also, some people refer to the Cholesky decomposition as a matrix square root, further muddying the waters).

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    $\begingroup$ (+1) Nice insight. For anyone interested in learning more about matrix functions, I recommend "Chapter V: Operator Monotone and Operator Convex Functions" in Bhatia's Matrix Analysis book (1997). $\endgroup$
    – mhdadk
    Commented Dec 1, 2023 at 15:53
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    $\begingroup$ +1. But to obtain a positive definite root you are limited to that principal branch. $\endgroup$
    – whuber
    Commented Dec 1, 2023 at 22:10

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