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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.

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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 $i,j$:

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

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  • $\begingroup$ Thank you Sir ! $\endgroup$ May 16 at 16:21

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