How can an asset's variance, $\sigma_i^2$, be shown to contribute to portfolio variance, $\sigma_p^2$?

I was thinking of taking the derivative (first order conditions $\frac{\partial L_{\sigma_p^2}(w,\lambda)}{\partial \sigma_i}$) of the Lagrangean formulation of the minimum-variance portfolio's objective function, for example, but not sure if this is the right approach since the idea is to speculate before optimization, how an asset's variance will contribute to portfolio variance, based on that asset's stand-alone variance (or volatility) level. Besides, asset variances don't appear in the portfolio variance's Lagrangean.

A demonstrated derivation of the $\frac{\partial L_{\sigma_p^2}(w,\lambda)}{\partial \sigma_i}$ first order conditions might be marked best answer, but also open to alternative suggestions.


2 Answers 2


In this answer, I am assuming that you want to keep correlations constant.

To begin with, note that the $N\times N$ covariance matrix $\Sigma$ with element $\Sigma_{i,j}=Cov(x_i,x_j)$ can be written as

$$ \Sigma = \mathbf{SRS} $$

where $\mathbf{S}$ is a diagonal matrix of the simple volatilties $\sigma_i$, and $\mathbf{R}$ is the correlation matrix. Thus in a matrix sense,

$$ \frac{\partial \mathbf{\Sigma}}{\partial\sigma_i}=\frac{\partial\mathbf{SRS}}{\partial\sigma_i}=\frac{\partial\mathbf{S}}{\partial\sigma_i}\mathbf{RS}+\mathbf{SR}\frac{\partial\mathbf{S}}{\partial\sigma_i} $$

The derivative of $\mathbf{S}$ with respect to $\sigma_i$ is a diagonal matrix of zeros, whose $i$th element is $1$. It is a single-entry matrix or a selector matrix, which we shall denote $\mathbf{E}_i$. For example, for $\mathbf{E}_2$ is

$$ \mathbf{E}_2=\begin{pmatrix}0&0&0&\ldots &0\\ 0&1&0&\ldots &0\\ \ldots &\ldots &\ldots &\ldots &\ldots \\ 0&0&0&\ldots &0\\\end{pmatrix} $$


$$ \frac{\partial\mathbf{\Sigma}}{\partial\sigma_i}=\mathbf{E}_i\mathbf{RS}+\mathbf{RS}\mathbf{E}_i $$

Thus, the marginal impact of a change in (any) of the volatilities on portfolio variance $v=\mathbf{w}^T\mathbf{\Sigma} \mathbf{w}$ can be computed as (after some algebra)

$$ \frac{\partial v}{\partial \mathrm{diag(S)}}=\mathbf{w}^T\frac{\partial \mathbf{\Sigma}}{\partial \mathrm{diag(S)}}\mathbf{w}=2\mathbf{w}\otimes \left(\mathbf{RSw} \right)$$ where $\otimes$ denotes element wise multiplication, i.e. $x\otimes y = \mathrm{diag}(xy^T)$. Conveniently, this formulation returns all derivatives at once.

Example: With

$$ \mathbf{S}=\mathrm{diag}\begin{pmatrix}0.1&0.2&0.3\end{pmatrix} $$ and $$ \mathbf{R}=\begin{pmatrix}1 & 0.5 & 0.25 \\ 0.5 & 1 & 0.1 \\ 0.25 & 0.1 & 1\end{pmatrix} $$ and a weight vector

$$ \mathbf{w}=\begin{pmatrix}0.2 & 0.3 & 0.5\end{pmatrix}^T $$

we find

$$ \frac{\partial v}{\partial \mathrm{diag(S)}}=2\mathbf{w}\otimes \left(\mathbf{RSw} \right)=2\begin{pmatrix}.2\\.3\\.5\end{pmatrix}\otimes\begin{pmatrix}0.0875\\0.085\\0.161\end{pmatrix}=\begin{pmatrix}0.035\\0.051\\0.161\end{pmatrix} $$

Thus, for example, the sensitivity of the portfolio variance with respect to the first volatility is 0.035.

With a bit more of algebra, you can find the impact of standalone vols on any portfolio optimisation solution, e.g. the MVP. Using the results from above and the fact that $\sigma_{MVP}=\frac{1}{\mathbf{1}^T{\Sigma^{-1} 1}}$ and the knowledge that $\mathbf{\Sigma}^{-1}=\mathbf{S}^{-1}\mathbf{R}^{-1}\mathbf{S}^{-1}$.

  • $\begingroup$ While this gives the derivative of portfolio variance with respect to individual variance, the results can sometimes be negative. When an investor asks "what is the contribution of asset A's risk to your total risk?", if you reply with a negative number, you're fired. The derivative here gives the marginal contribution to risk, just like the Lagrangian 'solution' to this problem. $\endgroup$ Commented Nov 12, 2020 at 20:13
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    $\begingroup$ I would argue that negative portfolio weights may result in negative derivatives and are perfectly normal, think risk reducing exposure or hedges. In that regard, I think the metric‘s goal needs to be defined properly. Also, I think the Original question was about the impact of $\sigma_i$, not the $i$th asset‘s contribution to overall risk, no? $\endgroup$ Commented Nov 12, 2020 at 20:33
  • $\begingroup$ are the elements in the vector $2\mathbf{w}\otimes \left(\mathbf{RSw} \right)$ unitless? what does a 0.035 sensitivity of the portfolio variance really mean, besides being numerically relative to the other elements in the same vector? also, does this vector's sum, norm, re-scaling, etc have any meaning or usefulness $\endgroup$
    – develarist
    Commented Nov 12, 2020 at 20:49
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    $\begingroup$ @steveo'america, yes: If an assets (more exactly: a position's) contribution to overall variance is sought, and if the contributions are to be positive and are to be summing to total variance, then your answer is the correct one. In that case, I suggest you could replace the matrix square root equation with (maybe) an Eigen decomposition and show that the squared elements of the resulting vector sum to total variance. If, on the other hand, the sensitivity of the portfolio volatility with respect to an individual asset volatility is sought after, my response would be correct. $\endgroup$ Commented Nov 13, 2020 at 7:10
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    $\begingroup$ @Kermittfrog yes, any square root of $\Sigma$ will yield the decomposition of risk. Many of them, however, lack the 'equivariance' property. For example, if you use the Cholesky square root of $\Sigma$, then re-arrange the order of the assets in your vector, the results are not reordered in the same way. In this case it is hard to identify the elements of $\Sigma^{1/2}\vec{w}$ with the assets themselves. $\endgroup$ Commented Nov 13, 2020 at 16:45

The Lagrangian 'solution' can yield negative contributions to portfolio risk, which is a bad look. An alternative definition is via the symmetric square root of the covariance, $\Sigma^{1/2}$. For portfolio $\vec{w}$ define $$ \vec{r} = \Sigma^{1/2}\vec{w}. $$ The norm of $\vec{r}$ is the volatility of the portfolio. Moreover, this definition is equivariant with respect to rotations of the asset space (using the Cholesky square root would not yield this property), and thus elementwise $\vec{r}$ can be identified with the individual assets. See also no parity like risk parity.

  • $\begingroup$ could you start higher up because I don't know where $\Sigma^{1/2} w$ even comes from. what is that. or what is $\Sigma w$ $\endgroup$
    – develarist
    Commented Nov 12, 2020 at 20:11
  • $\begingroup$ $\Sigma^{1/2}$ is the symmetric square root of $\Sigma.$ It is a symmetric matrix $A$ such that $AA = \Sigma$. Computed by sqrtm in matlab, for example. In contrast to the Cholesky square root, which is a lower triangular $A$ such that $AA^{\top} = \Sigma$. $\endgroup$ Commented Nov 12, 2020 at 20:15
  • $\begingroup$ but why would you multiply a covariance matrix by the portfolio weight vector to compute returns? what is $r$ $\endgroup$
    – develarist
    Commented Nov 12, 2020 at 20:16
  • $\begingroup$ The vector $\vec{r}$ has the property that $\vec{r}^{\top}\vec{r} = \sigma^2_p = \vec{w}^{\top}\Sigma\vec{w}$. (I should also mention you want to take the absolute value of $\vec{r}$ to avoid negative values. Doing so still gives this decomposition of total variance.) $\endgroup$ Commented Nov 12, 2020 at 20:21
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    $\begingroup$ Elements of $|\vec{r}|$ are the risk contribution of each asset to total portfolio risk. It is like an 'elementwise square root' of the portfolio variance. It is the quantity that you seek to compute. The answer. $\endgroup$ Commented Nov 12, 2020 at 20:26

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