Volatility Matrices, Instantaneous Covariance, and Brownian Coordinates

# Covariance Matrix vs. Volatility Matrix


# Covariance Matrix vs. Volatility Matrix












Consider a general multidimensional market model in which each of $m$ stocks is driven by $d$ Brownian motions (as in Shreve II, p. 226), viz. $$ dS_i/S_i = \alpha_i dt + \sum_{j=1}^d \sigma_{ij}dW_j, \qquad i=1,\ldots,m, $$ where each $W_j$ is independent of the others. We could write this as a vector equation as $$ dS/S = \alpha dt + A dW $$ where $S = (S_1,\ldots,S_m)^T$, $\alpha = (\alpha_1,\ldots,\alpha_m)^T$, $dW = (dW_1,\ldots,dW_m)^T$ and $$ A = \begin{pmatrix} \sigma_{11} & \ldots & \sigma_{1d} \\ \vdots & \ddots & \\ \sigma_{m1} & \ldots & \sigma_{md} \end{pmatrix}. $$ I've seen $A$ called the "volatility matrix," for obvious reasons.

On the other hand, I've also seen (similar to eqn (1) in this article) a multidimensional market model posed in matrix-vector form as $$ dS/S = \alpha dt + B dW, $$ where again each component of $W$ is independent of the rest. The difference is the matrix $B$ is called the "square root of the covariance matrix $\Sigma$" in the article above, and they state that it is "...typical to take $B$ to be the Cholesky decomposition matrix of $\Sigma$," i.e., $BB^T = \Sigma$ where $B$ is lower triangular. Note here that the authors implicitly assume $m=d$ in the notation above.

First question: I assume by "covariance matrix" we mean $\Sigma_{i,j} = Cov(R_i, R_j)$ where $R_i = \log\left(S_i(t) / S_i(0)\right)$ are the log returns. Is this correct? If so, there seems to be a factor of $t$ missing somewhere between the two model statements. As a concrete example, let $m=d=2$ an consider any square root of the matrix $\Sigma$ (not necessarily a lower triangular one). Then in the first model $$ A = \begin{pmatrix} \sigma_{11} & \sigma_{12} \\ \sigma_{21} & \sigma_{22} \end{pmatrix} $$ and in the second $$ \Sigma = \begin{pmatrix} Var(R_1) & Cov(R_1,R_2) \\ Cov(R_2,R_1) & Var(R_2) \end{pmatrix} = t \begin{pmatrix} \sigma_{11}^2 + \sigma_{12}^2 & \sigma_{11}\sigma_{21} + \sigma_{12}\sigma_{22} \\ \sigma_{11}\sigma_{21} + \sigma_{12}\sigma_{22} & \sigma_{21}^2 + \sigma_{22}^2 \end{pmatrix}. $$ If the two model statements are consistent, then I would expect $AA^T = \Sigma$, but in fact $AA^T = \frac{1}{t}\Sigma$. So, by "covariance matrix" do we really mean $\Sigma_{ij} = \frac{1}{t}Cov(R_i,R_j)$?

Second question: In the example above, $A$ was indeed a square root of the matrix $\Sigma$ where $\Sigma_{ij} = \frac{1}{t}Cov(R_i,R_j)$, and this seemed like the obvious choice for $A$ (at least to me). But the authors of the article state $A$ could be taken to be lower triangular, which means stock $i$ is driven by only $i$ Brownian motions. This seems like an extra restriction on the general model, where each stock is driven by some linear combination of all the Brownian motions. So, is taking the "volatility" matrix to be lower triangular an extra restriction?

## Answer by AFK (score 2)

https://quant.stackexchange.com/a/18202

1) People usually consider an instantaneous covariance while you are considering a integrated covariance. In a model $$ dS_t = S_t \cdot (\alpha(t)dt + A(t)dW_t) $$ the integrated covariance of log-returns is simply the integral over time of the instantaneous covariance: $$ Cov(R_i(t),R_j(t)) = (i,j)\textrm{-coeff. of} \int_0^t \underbrace {A(s)A(s)^\top}_{\Sigma_{ij}(s)} ds $$

2) This is not a restriction since the law of the process is the same (this is just a change of coordinates of the driving brownian motion). From a modelling point of view, volatility (or correlation) is not what matters because it is never observed. Only (co)variance is.