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If I have a complicated function of multivariables $f(x_1,x_2,x_3,\ldots,x_n)$, and I were to find the variance approximation through the delta method, say $\sigma^2_{approx}$, would the 95% confidence interval directly follow as:

$$ f(x_1,x_2,x_3,\ldots,x_n)\pm 1.96\cdot \sqrt{\sigma^2_{approx}} $$

Or would there be something else I need to do on $f(x_1,x_2,x_3,\ldots,x_n)$?

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you are missing a $\sqrt{n}$, ie

$$ f(x_1,\ldots,x_n) \pm 1.96 \sqrt{\sigma^2_{approx}/n} $$ Assuming by $\sigma^2_{approx}$ you mean $\sigma^2_{approx} = f'(x_1,\ldots,x_n)^2\sigma^2$.

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  • $\begingroup$ I don't think those approximations came from the delta method. $\endgroup$
    – bdeonovic
    Commented Jan 7, 2022 at 20:36
  • $\begingroup$ I read the paper and it looks like they use taylor series approximations. In such a case, what would be the valid confidence interval for $\frac{X_1}{X_2}$? Would it be $\frac{X_1}{X_2}\pm 1.96 \sqrt{Var\left(\frac{X_1}{X_2}\right)}$ or $\frac{X_1}{X_2}\pm 1.96 \sqrt{Var\left(\frac{X_1}{X_2}\right)/n}$? thanks $\endgroup$
    – user321627
    Commented Jan 7, 2022 at 20:44

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