In Thierry Roncalli's book Introduction to Risk Parity and Budgeting (2013), he gives an example of particular solutions to the Risk Budgeting portfolio such as for the $n=2$ asset case.

The risk contributions are:

$$ \frac{1}{\sigma(x)} \cdot \begin{bmatrix} w^2\sigma_1^2 + \rho w(1-w) \sigma_1 \sigma_2 \\ (1-w)^2\sigma_2^2 + \rho w(1-w) \sigma_1 \sigma_2 \\ \end{bmatrix} $$

The vector $[b,1-b]$ are the risk budgets.

He presents the optimal weight $w$ as:

$$ w^* = \frac {(b - \frac{1}{2}) \rho \sigma_1\sigma_2 - b\sigma_2^2 +\sigma_1\sigma_2\sqrt{(b - \frac{1}{2})^2\rho^2 + b(1-b)}} {(1-b)\sigma_1^2 - b\sigma_2^2 + 2(b - \frac{1}{2})\rho\sigma_1\sigma_2} $$

How are these weights derived ? I don't need a full derivation (it would be helpful though), I just don't know how it is derived.

Is it done by setting the risk contributions equal to the budgets?

$$ \begin{bmatrix} b \\ 1-b \\ \end{bmatrix} = \frac{1}{\sigma(x)} \cdot \begin{bmatrix} w^2\sigma_1^2 + \rho w(1-w) \sigma_1 \sigma_2 \\ (1-w)^2\sigma_2^2 + \rho w(1-w) \sigma_1 \sigma_2 \\ \end{bmatrix} $$


1 Answer 1


You are correct in your assumption, this is specified at the start of section 2.2.1 Definition of a risk budgeting portfolio.

We consider a set of given risk budgets $\{B_1,\dots,B_n\}$. Here $B_i$ is an amount of risk measured in dollars. We denote $\mathcal{RC}_i(x_1,\dots,x_n)$ the risk contribution of asset $i$ with respect to portfolio $x=(x_1,\dots,x_n)$. The risk budgeting portfolio is then defined by the following constraints:

$$ \mathcal{RC}_1(x_1,\dots,x_2)=B_1 \\ \vdots \\ \mathcal{RC}_i(x_1,\dots,x_2)=B_i \\ \vdots \\ \mathcal{RC}_n(x_1,\dots,x_2)=B_n \\ $$

The two asset case

We can rewrite the two equations into a singular formula by solving for $\sigma(x)$ for both of them and subsequently eliminating $\sigma(x)$ by setting the two formulations of $\sigma(x)$ equal to each other:

$$ \frac{w^2\sigma_1^2+w(1-w)\rho\sigma_1\sigma_2}{b}= \frac{(1-w)^2\sigma_2^2+w(1-w)\rho\sigma_1\sigma_2}{1-b} $$

After rearranging this becomes a (complicated) quadratic equation in $w$, which can be solved via the quadratic formula. By cancelling terms in the resulting fraction and observing that $0\leq w\leq 1$ (and thus eliminating one of the solutions of the quadratic formula) you should arrive at the optimal $w^*$.

  • $\begingroup$ Thank you. Is there some trick required to arrive at the final weight ? I get as far as solving for $w$ in one equation after rearranging as a quadratic in $w$ and substituting the result in the other, but it’s quite messy. $\endgroup$
    – FISR
    Commented Mar 29, 2023 at 12:16
  • $\begingroup$ If you have know of a reference that contains at least a partial derivation that would be very helpful. $\endgroup$
    – FISR
    Commented Mar 29, 2023 at 21:01
  • 1
    $\begingroup$ Edited the question, let me know if you need more details. $\endgroup$ Commented Mar 30, 2023 at 6:07

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