Measuring share contribution of each var/cov term to the standard deviation of a sum of variables

Question

Say, for a simple example, I have a random variable $X = \alpha_1 X_1 + \alpha_2 X_2$, where $X_i$ are random variables and $\alpha_i$ are weights. I then calculate the standard deviation of $X$ as $\sigma_X = \sqrt{(\alpha_1)^2 \mathrm{Var}(X_1) + (\alpha_2)^2 \mathrm{Var}(X_2) + 2\alpha_1 \alpha_2 \mathrm{Cov}(X_1,X_2)}$.

What I'm wondering is: what is a good way to calculate the "share" of each of the three individual variance/covariance terms in the total standard deviation of $X$?

It seems like, because of the concavity of the square root function, e.g. $(\alpha_1)^2 \mathrm{Var}(X_1) / \mathrm{Var}(X)$ would not be a good, representative measure of the variance of $\alpha_1 X_1$'s "share contribution" to $X$'s standard deviation; and neither would the share of the square root of each variance/covariance term in the sum of these individual square roots, not the least because I would run into issues trying to take the square root of a negative covariance term.

The motivation for my question is as follows: I'm measuring sampling error in survey-based economic indexes (e.g. the Institute for Supply Management's PMI). For composite indexes (i.e. indexes that are calculated from responses to multiple questions, as weighted sums--the PMI is an example of this), both variance and covariance terms enter into the sampling error formula, as in the example above. I want to look at the time series of the "contributions" to the sampling error of the composite index of each of these individual variance and covariance terms. I want to do this to examine how both dispersion in responses to different questions and co-variation in responses to different questions (a) vary with economic conditions and (b) affect uncertainty in the composite index.

Does this question make sense? If so, do you have any ideas or suggestions?

Answer 1

Here's my current best guess, in case someone later has this same question.

Make the problem more general in the following way: say you are looking at the expression $x = \sqrt{x_1 + x_2 + \ldots + x_n}$ and want to figure out the "share contribution" of $x_i$ in $x$ (the $x_i$'s are simply the variance and covariance terms in the expression in the question). Suppose we have reason to believe that $\sum x_i \geq 0$ (as in variance expression in the question).

The concavity of the square root makes this kind of difficult. So too does the fact that some $x_i$ could be negative. But, let's look at how this might work for a simple example with just two terms: $x = \sqrt{x_1 + x_2}$. Say $x_1 = 36$ and $x_2 = 64$, so that $x = \sqrt{100} = 10$. We might say that $x_1$ contributes $(36/100)\cdot 10 = 3.6$ and $x_2$ contributes $(64/100)\cdot 10 = 6.4$ to this figure. Or, we might instead say $x_1$ contributes $(\sqrt{36}/(\sqrt{36}+\sqrt{64}))\cdot 10 \approx 4.29$ and $x_2$ contributes $\approx 5.71$ to this figure.

One problem with the first method is, of course, that it understates the contributions of smaller terms. By itself, $\sqrt{x_1}$ is $6$, only $25\%$ lower than $\sqrt{x_2} = 8$; while its calculated contribution is almost $45\%$ lower than that of $x_2$. This doesn't seem right. On the other hand, at least the denominator ($x_1+x_2$) has some natural connection to $x = \sqrt{x_1+x_2}$; while the denominator in the second method, $\sqrt{x_1}+\sqrt{x_2}$, is hardly natural.

I'm going to propose a third method and claim that it is better than these first two, though I'm having trouble supporting the claim; it just seems right to me.

Here's the third method, with the simple example from above. Look at $\min\{x_1,x_2\} = x_1$. Both $x_1$ and $x_2$ contribute the amount $x_1$ to the sum inside the square root; so, we should figure out how much of $x$ is contributed by the first $x_1$ amount of both terms, together, and then divide it evenly between $x_1$ and $x_2$; and this amount will eventually be included in the total contributed by each term (in fact, it will represent all of the contribution from $x_1$. To do this, we simply compute $s_1 = \frac{1}{2}\cdot \sqrt{x_1+x_1} = sqrt{72}/2 \approx 4.24$. So, $x_1$'s share contribution to $x$ is $.424$; while $x_2$'s is $.424$ plus something else. Here's how we compute that "something else": we figure out the additional amount contributed by all terms of size at least $x_2$ (in this case, just $x_2$), i.e. compute $\sqrt{x_1 + x_2} - sqrt{x_1 + x_1}$, and add this amount to the $.424$ contributed by the first $x_1$ amount of $x_2$. In numbers, this is $\sqrt{100}-sqrt{72} \approx 10 - 8.49 = 1.51$. So, $x_2$ contributes $\approx 4.24 + 1.51 = 5.75$ to $x$, so that its share contribution is $\approx .575$.

Now I'll write out general formulas for the case with $x = \sqrt{x_1 + \ldots + x_n}$. Reorder the $x_i$ so that $x_1 \leq x_2 \leq \ldots \leq x_n$. Then the contribution of $x_1$ is $s_1 = \frac{1}{n}\cdot\sqrt{n x_1}$; the contribution of $x_2$ is $s_2 = s_1 + \frac{1}{n-1}\cdot(\sqrt{(n-1)x_2 + x_1}-\sqrt{n x_1})$; of $x_3$ is $s_3 = s_1 + s_2 + \frac{1}{n-2}\cdot(\sqrt{(n-2) x_3 + x_2 + x_1}-\sqrt{(n-1)x_2 + x_1})$; and so on. The shares are obtained simply by dividing $s_i$ by $x$ (that the shares add up to $1$ is straightforward to verify).

As long as $\sum x_i \geq 0$, we can account for negative $x_i$ simply by taking the negative of the square root of the absolute value of the expression, as long as the expression inside the square root is negative. In computer code, for example, this would be achieved by replacing sqrt(...) with (real(sqrt(...))-imag(sqrt(...))).

If anyone reads this and has any thoughts (it sucks, it makes sense, have a way to explain why it makes sense because I'm kind of failing at that), please share; I'd really appreciate it.