What is the analytic value of an asset's risk contribution, if $n=2$?

Question

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?

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.