I'm currently studying Monte Carlo sampling, referencing Veach's "Robust Monte Carlo Methods for Light Transport Simulation".

On page 63, he writes:

The idea of Monte Carlo integration is to evaluate the integral

$I = \int_{\Omega}f(x)d\mu(x) $

using random sampling. In its basic form, this is done by independently sampling $N$ points $X_1, ..., X_N$ according to some convenient density function $p$, and then computing the estimate

$F_N = \frac{1}{N}\sum_{i=1}^{N}\frac{f(X_i)}{p(X_i)}$

I've been playing around with this, and I understand the the technique when using uniform sampling on arbitrary domains $\Omega=[a, b]$, where $p(X_i) = \frac{1}{b-a}$. I've written a small python test program and it seems to work well.

However, I'm confused by his statement regarding "independently sampling $N$ points $X_1, ..., X_N$ according to some convenient density function $p$".

I'm assuming this means I can choose any arbitrary probability density function I want for $p(X_i)$?

As a simple test, I chose the Gaussian distribution $N(0.5, 0.15)$ to get a PDF centered at $0.5$ and roughly fitted to the interval $[0,1]$. To me, this seems "convenient".

I'm trying to apply the formula by drawing samples $X_i$ using this PDF, and for each iteration of the summation I can evaluate the integrand at each sample as $f(X_i)$, and divide by the probability $p(X_i)$ of each chosen sample.

For simplicity, I'm attempting to integrate trivial functions such as $f(x) = 1$, and $f(x) = x$ etc.

However, this does not seem to work at all, and I get values significantly different from the true value of the integral (eg. we expect $\int_{0}^{1}1 = 1$, $\int_{0}^{1}x = 0.5$), even with large samples sizes of $N = 100,000$ etc. The values I get are off by ~1, and vary noticeably between runs.

I suspect 1 of 2 things: either MC integration of this form requires uniform sampling, or I'm misunderstanding something... I'd appreciate any insights you may have!

  • $\begingroup$ You're using a different measure $\mu$. $\endgroup$
    – saulspatz
    Commented May 17, 2020 at 17:29
  • $\begingroup$ Hmm, how so? $\mu = 0.5$ is the center of the interval $[0,1]$ $\endgroup$
    – b1skit
    Commented May 17, 2020 at 17:37
  • $\begingroup$ Do you understand what the $\mathrm{d}\mu(x)$ means in the integral? Why are you expecting to calculate $\int f(x)\,\mathrm{d}x$ if you don't use a uniform distribution? $\endgroup$
    – saulspatz
    Commented May 17, 2020 at 17:42
  • $\begingroup$ Thank you - Now that you point it out, no, I don't think I understand that. I read $d\mu(x)$ as "change in $\mu$ as a function of x", but I'm not really sure how to interpret that. I glossed over it, since I'm trying to implement the approximation $F_N$. I'd appreciate any insight you have on this. Are you saying that I can only use a uniform distribution? $\endgroup$
    – b1skit
    Commented May 17, 2020 at 17:48
  • $\begingroup$ Yes, that's what I'm saying. $\mathrm{d}x$ means a uniform distribution is the only appropriate one. You can read about probability distributions to understand more about what $\mathrm{d}\mu$ means. $\endgroup$
    – saulspatz
    Commented May 17, 2020 at 17:52

1 Answer 1


For example, to compute $\int_0^2f(x)\,\mathrm dx$ you might compute $\int_0^1f(x)\,\mathrm dx$ with $1000$ samples and spearately compute $\int_1^2f(x)\,\mathrm dx$ with $2000$ samples (because the function is "wilder" over that interval and needs more sample for an equally reliable result) and add the parts. This is equivalent to making $3000$ samples over the original interval $[0,2]$ but weighting the samples differently, namely assign twice the weight to those samples in the subinterval with half the probability density (which also means that the weighted number of samples becomes $\sim 4000$). Taking this further, you can use any other nice probability density and the induced weighting by the reciprocal of the density.

  • 1
    $\begingroup$ Hmm, I don't think this is the problem. For example, I'm trying to integrate the function $f(x) = 1$, which is a perfectly flat line, yet the technique doesn't seem to work. $\endgroup$
    – b1skit
    Commented May 17, 2020 at 17:33

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