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I want to calculate a quantity $I$ which is the sum of integrals $I_k$, $$ I = \sum_k I_k\;,\quad I_k = \int_{\Omega} f_k(x)\,\textrm d x\;, $$ Every integral $I_k$ is evaluated independently by a Monte Carlo integration. So I make the approximation: $$ I_k \approx Q_k $$ where $$ Q_k = \frac V N \sum_{i=1}^N f_k(x_i)\;, \quad V = \int_{\Omega} 1 \, \textrm d x\;, $$ and $N$ is the number of Monte Carlo samples $x_i$. For every integral (for every $k$) I can now estimate the sample variance, $$ \sigma^2_k = \frac{1}{N-1}\sum_{i=1}^N \big( f_k(x_i) - \langle f_k \rangle \big)^2\;. $$ With this sample variance I can estimate the error bars, $\Delta I_k$, of the integrals, $I_k$: $$ \Delta I_k = V \frac{\sigma_k}{\sqrt N} \;. $$ So far the standard procedure.

But how can I estimate the error bars of the result $I$? Can I just use the formula for propagation of uncertainty, as we learn it in school? Like that: $$ \Delta I = \sqrt{\sum_k \left( \frac{\partial I}{\partial I_k}\Delta I_k\right)^2} = \sqrt{\sum_k \Delta I_k^2} \quad ? $$

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    $\begingroup$ in this case because you are constructing each MC integral separately and -I assume-independently, then you can use this formula. if you do something fancy like control variates to reduce the error of the intergal estimate then you might break the independence and change your error formula. $\endgroup$ – Lucas Roberts Jul 7 '17 at 10:57

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