Suppose we have an 512x512 image, then the value of pixel j should be $I_j = \int_{0}^{512} \int_{0}^{512} h_j(u,v)L(u,v) \,dudv$, where $h_j(u,v)$ is the filter function for pixel j, $L(u,v)$ is the corresponding radiance function.

In practice, we compute the value of pixel j by the formual $\frac{\sum_{i=1}^{N} h_j(u_i,v_i)L(u_i,v_i)}{\sum_{i=1}^{N} h_j(u_i,v_i)}$. Veach said that this estimate is slightly biased in page 308 in his thesis.

So my question is what's the expectation of this estimate?

  • $\begingroup$ I believe they want to estimate $\frac{\int_{\Omega} h_jL}{\int_{\Omega}h_j}$ the estimator above however has expectation $N\int_{\Omega^N} \frac{h_j(x_1)L(x_j)}{\sum_{k=1}^{N}h_j(x_k)}\prod_{k=1}^N p(x_k)\,dx_k$. Assuming their film coordinates are in $[0,1]^2$ then $p = 1$ and what remains is $N\int_{\Omega^N} \frac{h_j(x_1)}{\sum_{k=1}^{N}h_j(x_k)L(x_j)}\prod_{k=1}^N\,dx_k$. So the issue is really that they don't know the integral of $h_j$. $\endgroup$
    – lightxbulb
    Commented Oct 7, 2022 at 11:02
  • $\begingroup$ Typo $L(x_j)$ must be $L(x_1)$ and always in the numerator. $\endgroup$
    – lightxbulb
    Commented Oct 7, 2022 at 11:09
  • $\begingroup$ @lightxbulb : If they want to estimate $\frac{\int_{\Omega} h_j(x)L(x)\mathrm{d}x}{\int_{\Omega} h_j(x)\mathrm{d}x} $, why not calculate $H=\int_{\Omega} h_j(x)\mathrm{d}x$ first, and then use $\frac{1}{NH}\sum_{k=1}^{N} h_j(x_k)L(x_k)$ to estimate $\frac{\int_{\Omega} h_j(x)L(x)\mathrm{d}x}{\int_{\Omega} h_j(x)\mathrm{d}x} $, where $x_k\sim U[0,1)^2$? $\endgroup$
    – Andy
    Commented Oct 8, 2022 at 7:40
  • $\begingroup$ Because they wanted to have a convex linear combination with weighs that sum up to $1$: $w_k= \frac{h_j(x_k)}{\sum_i h_j(x_i)} \implies \sum_k w_k = 1$. $\endgroup$
    – lightxbulb
    Commented Oct 8, 2022 at 10:27
  • $\begingroup$ @lightxbulb : But in this case, the bias and variance is hard to analyse. $\endgroup$
    – Andy
    Commented Oct 8, 2022 at 12:50


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