Say I have a random variable $Z$ that is itself a function of two independent random variables $g(X,Y)$, where

$$Z = g(X,Y) = \frac{X}{Y}$$

I know that the propagation of uncertainty gives a good approximation of $Var[g(X,Y)]$ using Taylor expansions around the means $\mu_X,\mu_Y$, such that, when $X, Y$ are independent,

$$Var[Z] \approx \Bigg(\frac{\delta g(x,y)}{\delta x}\Bigg)^2 \sigma_x^2 + \Bigg(\frac{\delta g(x,y)}{\delta y}\Bigg)^2 \sigma_y^2 = \frac{\sigma_x^2}{\mu_Y^2} + \frac{\sigma_y^2 \mu_X^2}{\mu_Y^4} $$

Suppose now that $Z$ is a function of two series of independent random variables, $g(X_1,...,X_k,Y_1,...Y_k)$, such that

$$Z = g(X_1,...,X_k,Y_1,...Y_k) = \frac{X_1}{Y_1} + \dotsb + \frac{X_k}{Y_k}$$

If each of the ratios are independent, I am wondering if the variance of $Z$ can be calculated by taking the sum of propagation of uncertainty for each of the fractions, i.e.,

$$Var[Z] = \sum_{i=1}^k Var\Bigg[\frac{X_i}{Y_i}\Bigg] \\ \approx \sum_{i=1}^k \Bigg[\Bigg(\frac{\delta g(x_i,y_i)}{\delta x_i}\Bigg)^2 \sigma_{x_i}^2 + \Bigg(\frac{\delta g(x_i,y_i)}{\delta y_i}\Bigg)^2 \sigma_{y_i}^2 \Bigg]$$

where for the $j^{th}$ term, I calculate partial derivatives with respect to $x_j$ and $y_j$ as in the basic case presented above and ignore the other $i \neq j$ terms? Or do I need to do out each partial derivative with respect to the complete series of $\{x_i\},\{y_i\}$ ?


1 Answer 1


Yes, the variance of the sum is the sum of the variances, in this case.

By the law of the unconscious statistician, for $i\neq j$: \begin{align*} E\left[\frac{X_i}{Y_i}\frac{X_j}{Y_j} \right] &= \iiiint \frac{X_i}{Y_i}\frac{X_j}{Y_j} f(x_i)f(x_j)f(y_i)f(y_j)dx_i dx_j dy_i dy_j \\ &= E\left[\frac{X_i}{Y_i}\right] E\left[\frac{X_j}{Y_j} \right]. \end{align*} So $\text{Cov}\left(\frac{X_i}{Y_i},\frac{X_j}{Y_j}\right)=0$.

Edit: Responding to your comment: \begin{align*} \text{Var}\left(\sum_i \frac{X_i}{Y_i} \right) &= \text{Cov}\left(\sum_i \frac{X_i}{Y_i},\sum_i \frac{X_i}{Y_i} \right) \tag{defn}\\ &= \sum_i \sum_j \text{Cov}\left(\frac{X_i}{Y_i} ,\frac{X_j}{Y_j} \right) \tag{bilinearity of cov}\\ &= \sum_{i=j}\text{Cov}\left(\frac{X_i}{Y_i} ,\frac{X_j}{Y_j} \right) \tag{previous}\\ &= \sum_i \text{Var}\left( \frac{X_i}{Y_i} \right) \tag{defn} \end{align*}

  • $\begingroup$ Thanks! I don't totally follow how we get to variance still -- I'm just trying to tighten my intuition on this and perhaps you could provide a bit more detail in your answer or in a comment? In the above example, is it true that my variance estimate using propagation of uncertainty can just be swapped in for $Var(X_j/Y_j), j \in \{i\}$ in the more complex example (even though its not a proper analytic solution but an estimate)? You seem to be suggesting that I've just got a generalization equivalent to $Var(\sum_i X_j) = \sum_i Var(X_j)$ in other words? (Sorry for sloppy or unorthodox usage!) $\endgroup$ Commented Jun 14, 2017 at 1:42
  • $\begingroup$ @isosceleswheel I haven't checked your math for the approximation to the variance of a single ratio. However, I just added some stuff to my answer to show you why you can interchange the order of the sum and the variance operator. Let me know if this helps. $\endgroup$
    – Taylor
    Commented Jun 14, 2017 at 1:59
  • $\begingroup$ Thanks, I think I can see why I'm allowed to separate the Variances in this case, given the conditions. I will do some additional editing for a fuller version of my exact problem, since there are some other conditions that might break down the independence (but I don't think so because the details are just additional constants for the most part...). Short of a typographical error, the math should be OK for the propagation of uncertainty, as it is just straight from a book/online examples. Anyway, if you happen to stop by in a day or so maybe you can weigh in again. ;) $\endgroup$ Commented Jun 14, 2017 at 3:11

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