# 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?

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: $$Cov(R_i(t),R_j(t)) = (i,j)\textrm{-coeff. of} \int_0^t \underbrace {A(s)A(s)^\top}_{\Sigma_{ij}(s)} ds$$
• okay, I think 1) makes sense. Would you mind elaborating on 2), though? Or, do you know a good source explaining this change of coordinates? I'm also very interested in why covariance is considered "observed" but volatility is not. Wouldn't we have to compute (really, approximate) both of them from the observed data? E.g. $Var(R_i(t)) \approx \frac{1}{N-1}\sum_{j=1}^N (R_i(t_j) - \hat{\mu})^2$? – bcf Jun 7 '15 at 1:06
• 2) first question is an exercise you should work out yourself. Second point is a consequence: you are estimating integrated variance and this estimator only depends on the variance on the $R_i$'s not on their volatility which is not canonically defined. – AFK Jun 7 '15 at 1:41
• Thanks again, but I'm still having a bit of trouble with the change of coordinates idea. Are you saying I should work out the transformation $AdW \mapsto BAdW$? I.e., which is the "original" and "changed" coordinates? Also, when you say the law of the process is the same, are you just saying each stock at a given time is lognormally distributed? Then, under this change of coordinates, they still are, but with different volatilities? – bcf Jun 9 '15 at 18:19
• What I am saying is that if you set $V = B^{-1}A$ (assuming B invertible for simplicity) then $VdW$ is again a Brownian motion. So you go from one model to the other by replacing a Brownian motion by another so you do not change the law of the stochastic process. – AFK Jun 9 '15 at 19:11