# Covariance Matrix by Multi-Factor Model

I have been trying to find literature for the derivation of the covariance matrix, following a multi-factor model. I have had no luck at all, every single article I have found on the web already gives the formula $$\Sigma_z=B\Sigma_fB'+\Sigma_{ee}$$ But not a derivation of it, could someone please point me to the right literature.

Let there be $$n$$ assets and $$k$$ factors in the market. We assume multivariate normally distributed factor returns

$$r_f\sim \mathrm{N}\left(\mu_f,\Sigma_f\right)$$

with $$k\times k$$ factor covariance matrix $$\Sigma_f$$. Conditional on the factor return, $$r_f$$, the return of an asset $$i$$, $$r_i$$, is normally distributed with mean level $$\mu_i|r_f=\beta_i^Tr_f=\beta_{i,1}r_1+\ldots+\beta_{i,k}r_k$$ and residual return variance $$\sigma_{i,\epsilon}^2$$. The residual returns between any $$i\neq j$$ are independent.

Thus, the unconditional covariance between some assets $$i$$ and $$j$$ are:

\begin{align} Cov(r_i,r_j)&=\mathrm{E}\left((\beta_{i,1}(r_1-\mu_1)+\ldots\beta_{i,k}(r_k-\mu_k)+\epsilon_i)(\beta_{j,1}(r_1-\mu_1)+\ldots\beta_{j,k}(r_k-\mu_k)+\epsilon_j)\right)\\ &=\mathrm{E}\left((\beta_i^T(r_f-\mu_f)+\epsilon_i)(\beta_j^T(r_f-\mu_f)+\epsilon_j)\right)\\ &=\mathrm{E}\left(\beta_i^T(r_f-\mu_f)(r_f-\mu_f)^T\beta_j+\beta_i^T(r_f-\mu_f)\epsilon_j+\beta_j^T(r_f-\mu_f)\epsilon_i+\epsilon_i\epsilon_j\right)\\ &=\beta_i^T\Sigma_f\beta_j+E(\epsilon_i\epsilon_j) \end{align}

If $$i=j$$, then $$E(\epsilon_i\epsilon_j)=\sigma_{i,\epsilon}^2$$, else it is zero.

Let us now collect the beta coefficient vectors for each asset into a matrix, i.e. we stack the $$\beta_i^T$$ rows into a matrix:

$$B=\begin{pmatrix} \beta_1^T\\ \beta_2^T\\ \ldots\\ \beta_n^T\\ \end{pmatrix}$$

We can now trace out all combinations of $$i,j$$:

\begin{align} Cov(r)&=\begin{pmatrix} Cov(r_1,r_1)&Cov(r_1,r_2)&\ldots&Cov(r_1,r_n)\\ Cov(r_1,r_2)&Cov(r_2,r_2)&\ldots&Cov(r_2,r_n)\\ \ldots&\ldots&\ldots&\ldots\\ Cov(r_1,r_n)&Cov(r_2,r_n)&\ldots&Cov(r_n,r_n) \end{pmatrix}\\ &=\begin{pmatrix} \beta_1^T\Sigma_f\beta_1+\sigma_{1,\epsilon}^2 &\beta_1^T\Sigma_f\beta_2&\ldots&\beta_1^T\Sigma_f\beta_n\\ \beta_1^T\Sigma_f\beta_2 &\beta_2^T\Sigma_f\beta_2+\sigma_{2,\epsilon}^2&\ldots&\beta_2^T\Sigma_f\beta_n\\ \ldots&\ldots&\ldots&\ldots\\ \beta_1^T\Sigma_f\beta_n&\beta_2^T\Sigma_f\beta_n&\ldots&\beta_n^T\Sigma_f\beta_n+\sigma_{n,\epsilon}^2 \end{pmatrix}\\ &=\begin{pmatrix} \beta_1^T\Sigma_f\beta_1 &\beta_1^T\Sigma_f\beta_2&\ldots&\beta_1^T\Sigma_f\beta_n\\ \beta_1^T\Sigma_f\beta_2 &\beta_2^T\Sigma_f\beta_2&\ldots&\beta_2^T\Sigma_f\beta_n\\ \ldots&\ldots&\ldots&\ldots\\ \beta_1^T\Sigma_f\beta_n&\beta_2^T\Sigma_f\beta_n&\ldots&\beta_n^T\Sigma_f\beta_n \end{pmatrix}+\begin{pmatrix} \sigma_{1,\epsilon}^2&0&\ldots&0\\ 0&\sigma_{2,\epsilon}^2&\ldots&0\\ \ldots&\ldots&\ldots&\ldots\\ 0&0&\ldots&\sigma_{n,\epsilon}^2 \end{pmatrix}\\ &=B\Sigma_fB^T+\Sigma_{\epsilon} \end{align}

• Thank you Sir ! May 16 at 16:21