Does using box or gaussian filter introduce bias to the image when reconstructing the pixel?

Bias does not seem to be talked in the Chapter 7.8 of PBRT

  • $\begingroup$ It says "box filter allows high-frequency sample data to leak into the reconstructed values". How do you define bias besides the one implied here ? $\endgroup$
    – Kaan E.
    Jan 15, 2021 at 13:46
  • $\begingroup$ I am trying to wrap my head around what is bias in path tracers, when a path tracer is no longer unbiased. E.g. Indigo renderer describes it as error manifesting as noise, higher sample rate always converging to the right solution. In PBRT it mentions that gaussian filter will introduce blur, which is good for antialiasing purposes. I am wondering that which is unbiased, blurred image or aliased image? Or are neither actually introducing bias, and will converge to correct solution eventually? $\endgroup$
    – Toothery
    Jan 29, 2021 at 11:01
  • $\begingroup$ You have to be careful about what is meant by "correct solution". With a gaussian blur for instance, you could say the "correct solution" is a gaussian-blurred version of the image, and it will converge to that; that is still unbiased. A classic example of bias would be photon mapping. After creating a photon map with N photons, raytracing does not converge to the true lighting in the scene no matter how long you run it for, because it has the sampling error of the finite photon map baked in. $\endgroup$ Feb 28, 2021 at 19:37
  • $\begingroup$ ^ this. Box and Gaussian filters are settings which will produce (slightly) different images, but neither is biased. Think of it like you are switching between using two different cameras to produce your image. $\endgroup$
    – Peter
    Mar 2, 2021 at 18:44

1 Answer 1


This is an answer to your comment rather than the title of your question. It can be considered as a partial answer.

The bias in path tracers are introduced by the estimators. For example, think of the following two equations: $$ I = \int_{\Omega} f(\omega) d\omega $$ $$ F = \frac{1}{N} \sum_{i}^{N} \frac{f(\omega)}{p(\omega)}$$ where $p: \Omega \to R$. It should be evident that as the number of samples N increases, the function $F$, ie the estimator, outputs values that are closer to $I$.

The unbiased estimator satisfies the following $E[F_N] \approx I$ or $\lim_{N\to\infty}F_N=I$, that is the expected value of the estimator matches the desired outcome in $I$ as the number of samples goes to infinity. The biased estimator does not offer such a guarantee. You can read more from D. Van Antwerpen's master thesis, I am giving the full citation in case the link goes away:

Van Antwerpen, Dietger. “Unbiased Physically Based Rendering on the GPU.” Master of Science in Computer Science, Delft University of Technology, 2010.


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