The marginal risk contribution of asset $i$ is defined by Roncalli in his paper on ERC as follows:

$$\frac{\partial \sigma(x)}{\partial x_i} = \frac{1}{\sigma(x)} \left( w_i \sigma_i^2 + \sum_{\substack{j=1 \\ i \neq j}}^n w_j \sigma_{i,j} \right)$$

I was thinking about computing the value for a fully invested portfolio with $n=2$ assets, i.e. $x=\{w,1-w\}$.

With $n=2$, we know that:

$$\sigma(x)=\sqrt{w^2 \sigma_1^2 + (1-w)^2 \sigma_2^2 + 2 w (1-w) \sigma_{1,2}}$$

So, we can compute the marginal contribution of asset 1 (with $x_1=w$) as:

$$\frac{d \sigma(x)}{dw} = \frac{1}{2} \frac{1}{\sigma(x)} \left[ 2w \sigma_1^2 + 2(1-w)(-1) \sigma_2^2 + 2 \sigma_{1,2} (1-2w) \right] $$

$$\frac{d \sigma(x)}{dw} = \frac{1}{\sigma(x)} \left[ w \sigma_1^2 - (1-w) \sigma_2^2 + \sigma_{1,2} (1-2w) \right] $$

If I use Roncalli's definition, I get:

$$\frac{d \sigma(x)}{dw} = \frac{1}{\sigma(x)} \left[ w \sigma_1^2 + (1-w) \sigma_{1,2} \right] $$

The two results are not the same, and I'm trying to understand why. So far, I came up with the fact that Roncalli's definition is not taking into account the fact that the portfolio is fully invested. Is that correct?

  • $\begingroup$ I am not familiar with his paper but at first glance his restriction is that no short positions are allowed. Again I have not walked through all the derivations so I am not positive whether this will impact the result but fact is that such restriction (to my knowledge) is not imposed on the typical Markowitz portfolio construction. Just wanted to point this out in case. It may or may not help thus I did not formulate it as answer and just a comment. $\endgroup$
    – Matt Wolf
    Commented Jun 21, 2013 at 9:34
  • 1
    $\begingroup$ @MattWolf I don't think it forbids short positions. I just doesn't state anything about being fully invested. $\endgroup$
    – SRKX
    Commented Jun 21, 2013 at 10:04
  • $\begingroup$ this is what it says in your referenced paper: "We voluntary restrict ourselves to cases without short selling, that is 0 <= x <= 1." $\endgroup$
    – Matt Wolf
    Commented Jun 21, 2013 at 11:28
  • $\begingroup$ @MattWolf yes but this is the case for the ERC strategy because otherwise the result isn't unique. The global definitions of the risk contributions is not in the scope of this assumption. $\endgroup$
    – SRKX
    Commented Jun 21, 2013 at 12:43
  • $\begingroup$ @SRXX, again that was just an idea, I am not familiar with ERC, not even with what it stands for. But what I see from your derivation is that the contribution of asset 1 is a function of variance of asset 2 which makes little to no sense to me. And I think GusRustam is saying the same thing in different "words". $\endgroup$
    – Matt Wolf
    Commented Jun 21, 2013 at 13:25

1 Answer 1


You see, you added something new to the source formula, i.e. a dependence between weights of different assets: $w_2 = 1 - w_1$.

Let's try to forget that they are related to each other and vary them independently:

$$\sigma(x)=\sqrt{w_1^2 \sigma_1^2 + w_2^2 \sigma_2^2 + 2 w_1 w_2 \sigma_{1,2}}$$

Now $\frac{dw_2}{dw_1} = 0$ and equation becomes:

$$\frac{d \sigma(x)}{dw_1} = \frac{1}{2} \frac{1}{\sigma(x)} \left[ 2w_1 \sigma_1^2 + 2 w_2 \sigma_{1,2} \right] $$

$$\frac{d \sigma(x)}{dw_1} = \frac{1}{\sigma(x)} \left[ w_1 \sigma_1^2 + w_2 \sigma_{1,2} \right] $$

Now, let's tak $w_2 = 1 - w_1$ and you receive Roncalli's formula:

$$\frac{d \sigma(x)}{dw} = \frac{1}{\sigma(x)} \left[ w_1 \sigma_1^2 + (1-w_1) \sigma_{1,2} \right]$$ I think it's a tricky question, but I believe his way is right, because when you decompose $\sigma$ into factors, it is more correct to keep weights constant.


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