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During my studies years ago I came over a formula that states something like if $x_i$ are independent normally distributed variables with variances $\sigma^2_i$ and $f(x_i)$ is differentiable (and probably otherwise "good" I don't remember), then the variance of $f$ would be something along the lines of $$\sum_i\left(\frac{\partial{f}}{\partial{x_i}}\right)^2\sigma^2_i$$

but I'm not sure neither about the powers not the multipliers , the only thing I remember that it involved partial derivatives.

Does anyone know the correct formula? Thank you in advance.

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    $\begingroup$ Well this looks similar en.wikipedia.org/wiki/… $\endgroup$
    – Thomas
    Jul 7 '14 at 9:58
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    $\begingroup$ Your formula seems correct to me. You can check that it appears in both the Gaussian concentration inequality and the Poincaré inequality for Gaussian measures. $\endgroup$
    – Siméon
    Jul 7 '14 at 10:19
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    $\begingroup$ It doesn't make sense. Consider a simple example: $f(x)=x^2$ with rv $Y$ having mean 0 and variance $\sigma^2.$ Then the variance of $f(Y)$ is $E(Y^4)-(E(Y^2) )^2= 2\sigma^4.$ The given expression is $4x^2\sigma^2.$ What is $x$? $\endgroup$
    – Mr.Spot
    Jul 12 '14 at 5:27

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