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Given the factor structure below with K factors, the return for N assets is given by (under matrix notation):

$R =\alpha + \beta F + \epsilon$

where $F$ is matrix of K factor returns and $\beta$ is matrix of NxK factor loadings and $\epsilon \sim N(0,\Omega)$. The return on a portfolio of those N stocks with weight vector $w$ can be written as:

$Ptf = wR = w\alpha + wBF + w\epsilon$

Taking the expectation and variance yields:

$E[Ptf] = w\alpha + w \beta E[F]$

$V[Ptf] = (w\beta)\Sigma (w \beta)^T + w\Omega w^T$

Where $\Sigma$ is the covariance matrix of the factors and $\Omega$ is the diagonal covariance matrix of specific risk component.

It follows that the standard deviation of the portfolio is: $\sigma_{ptf}=\sqrt{V[Ptf]}$

I am interested in the marginal risk contribution $MRC$ of the stocks, but also the factors to the portfolio. My derivation of $MRC$ which I obtained by generalizing the common case of no factor structure is as follows:

$MRC = \partial \sigma_{ptf} /\partial w = (w\beta\Sigma \beta^T + w\Omega)/\sigma_{ptf} $

My problem arises when I compute $MRC$ using the the equation above. In fact when I re-compose the risk using the weights, I do not find it is equal to the portfolio risk. Any help is appreciated. I will upload sample code shortly.

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Let $\mathbf{w}$ denote the $N\times 1$ vector of portfolio weights, $B$ the $N\times K$ factor loading matrix, $\Sigma$ the $K\times K$ matrix of factor covariances and $\Omega$ the $N\times N$ diagonal matrix of idiosyncratic risks.

Then the $N\times N$ matrix of asset variances is

$$ \Sigma_N=B\Sigma B^T+\Omega $$ and the portfolio variance is $$ \sigma_p^2=\mathbf{w}^T\Sigma_N\mathbf{w}=\mathbf{w}^TB\Sigma B^T\mathbf{w}+\mathbf{w}^T\Omega \mathbf{w} $$

The MRC vector, defined as $\mathbf{v}=\partial \sigma_p/\partial \mathbf{w}$ is:

$$ \mathbf{v}\equiv\frac{\partial \sigma_p}{\partial \mathbf{w}}=\frac{B\Sigma B^T\mathbf{w}+\Omega \mathbf{w}}{\sigma_p} $$

Clearly, if you now compute $\mathbf{w}^T\mathbf{v}$, you obtain $\sigma_p$, the portfolio risk.

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  • $\begingroup$ Thanks @Kermittfrog, I had $\Omega$ as a vector in my python implementation and the addition was still going through with no error even though dimensions were mismatched. Just wanted to make sure the math was right. $\endgroup$
    – frederico
    Commented Jun 28, 2022 at 16:16

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