# How to show that the risk contribution function is or is not injective?

Assume a portoflio $w \in \mathbb{R}^n$, you can get the total risk contribution $\psi_i$ of asset $i$ by doing:

$$\psi_i = w_i \frac{\partial \sigma(w)}{\partial w_i}= \frac{1}{\sigma(w)} \left[ w_i^2 \sigma_i^2 + \sum_{j=1}^n w_i w_j \sigma_{i,j} \right]$$

So I can define a function $\Psi(x): \mathbb{R}^n \rightarrow \mathbb{R}^n$ which computes the contributions of a given portfolio $x$, and I will always find a unique answer.

My equation is, assume I have a set of risk contributions $\overrightarrow{\psi}=\{\psi_1, ... ,\psi_n\}$ , I'm looking to see whether $w^*=\Psi^{-1} \left( \overrightarrow{\psi} \right)$ is unique if it exists. In other words, I'm trying to see whether $\Psi(x)$ is injective.

Do you know if the proof of such statement exists, or how would you tackle the problem because the only thing I can think about now is to look at the $n$ nonlinear equations system with $n$ variable.

EDIT

I worked a bit on the problem and managed to formulate it as follows:

I have to show that $\nexists u,v \in \mathbb{R}^n$ such that $u \neq v$ and:

$$\sigma_i^2 (u_i^2-v_i^2) + \sum_{j=1, i\neq j}^n \sigma_{i,j} (u_i-vi) = 0 \quad \forall i$$

I managed to prove this for $n=2$ and $u_1+u_2=v_1+v_2=1$, but I'm struggling to prove it for $n$ assets. I tried by recurrence but the fully invested constraint prevents me from doing so...

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The $x_i$ should be a $w_i$ right? In this case $\sigma$ is a function that satisfies the Euler condition as in your other question. You could also add some constraints for your risk measure. Spontaneously, I am thinking about some kind of convexity condition and the implicit function theorem. – vanguard2k Sep 12 '13 at 12:27
Yeah thanks for the fix. I don't see how this question is related to the Euler one, although $\sigma(\cdot)$ is homogenous of degree 1 indeed. – SRKX Sep 13 '13 at 7:32
Could you clarify the relation between $\sigma$, $\sigma_i$ and $\sigma_{i,j}$? It would be useful! – Michael Grünewald Nov 12 '13 at 17:14