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Suppose I have a function $f(x_1, \dots, x_N)$, where $x_i$'s are random variables. $x_i$'s have SD $\sigma_i$, and are all independent.

There are two additional assumptions, that $f$ is approximately linear within the range $x^{'}_i \pm \sigma_i$; and secondly $x_i$ and $x_j$ are uncorrelated for $i \ne j$.

How to find the Var of $f$?

sbp
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  • If $f$ is approximately $\sum a_ix_i$ the the variance is approximately $\sum a_i^{2} \sigma_i^{2}$. – Kavi Rama Murthy Jan 08 '20 at 08:01
  • True. One question: You're throwing out the cross term expectation values since they are uncorrelated right? – sbp Jan 08 '20 at 08:09
  • Yes. Actually you also assumed independence of the $x_i$'s and this automatically implies that they are uncorrelated. – Kavi Rama Murthy Jan 08 '20 at 08:11
  • So is there a way to do this for a generic function $f$? Doing a Taylor expansion around $\sigma_i$ seems the way to go? – sbp Jan 08 '20 at 08:12
  • Taylor expansion around the origin, not $\sigma_i$. – Kavi Rama Murthy Jan 08 '20 at 08:15
  • If I assume $f$ is just a function of two random variables, i.e., $f(x_1, x_2)$ and then make a Taylor expansion around $(\sigma_1, \sigma_2)$, then the terms of $\mathcal{O}(x_i^2)$ and higher drops off as the function is approximately linear around $\sigma_i$. So one can thus calculate the variance. It turns out $\mathrm{Var} f(x_1, \dots, x_N)$ = $\sum_{i=1}^{N}\sigma_i^2 f_{x_i}(\sigma_1, \dots, \sigma_N)$. Is this correct? – sbp Jan 08 '20 at 10:21

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