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Consider a $d$-dimensional Brownian motion $B=\left(B_1,...,B_d\right)$ whose components are independent and let $A$ be a $d\times d$ squared matrix such that $\sum_{i=1}^dA_{ii}^2=1$. Define the $W=A\,B$ the Brownian motion whose correlation matrix is $\rho=A\,A^{\dagger}$, where $\dagger$ indicate the transpose. Consider a $d$-dimensional process $\lambda_t\in\mathbb{L}_{loc}^2$ and a process $\eta_t$ that solves the stochastic differential equation

$$ d\eta_t=-\eta_t\,\lambda_t\cdot dW_t,\quad \eta_0=1, $$

where $\cdot$ indicates Euclidean product. So $\eta_t$ is a one-dimensional stochastic process. In the book by A. Pascucci it is written that the (unique) solution is

$$ \eta_t = \exp\left(-\int_0^t\lambda_s\cdot dW_s-\frac{1}{2}\,\int_0^t\left\langle\rho\,\lambda_s,\lambda_s\right\rangle\,ds\right). $$

I do not understand how exactly the co-variation $\left\langle\rho\,\lambda_s,\lambda_s\right\rangle$ is defined. It should, in principle, returns a scalar value since $\eta_t$ is scalar.

At some point in the book the co-variation between two $\mathbb{R}^d$-valued processes $X$ and $Y$ is defined as

$$ \left\langle X,Y\right\rangle_t=\left(\left\langle X^i,Y^j\right\rangle\right)_{i j=1,...,d} $$

so it seems that $\left\langle\rho\,\lambda_s,\lambda_s\right\rangle$ is a matrix. Does somebody have some clue?

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This is only a matter of notation, by noting $cov_s=\left\langle\rho\,\lambda_s,\lambda_s\right\rangle$ this can be re-ritten written in matrix notation with the d-dimensional vector $\lambda_s$ and the $d\times d $ matrix $\rho$ in the following way which might seem more amenable to you :

$$cov_s=\lambda_s^t.\rho.\lambda_s$$

which is a scalar.

Best regards

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