Timeline for Properties of the image reconstruction filter in rendering
Current License: CC BY-SA 4.0
25 events
| when toggle format | what | by | license | comment | |
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| Mar 2, 2020 at 20:37 | comment | added | lightxbulb | I lost you. I have no idea what you are trying to do anymore, or why. Random new details keep popping up in your formulation without rhyme or reason (at least from my perspective). What are you trying to do? Why do you need to consider the whole path space as opposed to just the image function? I believe these two questions are related, but seeing as I do not get what your end goal is, it's hard to help even if I want to. | |
| Mar 2, 2020 at 20:00 | comment | added | 0xbadf00d | The reason why I'm considering the path space is that my actual concern is slightly more complicated. First of all: Note that (and I missed that before) the inequality (b) in the question simply follows from the fact that if $X$ is a Hilbert space (say $H$) valued random variable on a probability space $(\Omega,\mathcal A,\operatorname P)$, then the function $H\ni x\mapsto\operatorname E\left[\left\|X-x\right\|_H^2\right]$ attains its minimum at $\operatorname E[X]$. Assuming $\lambda(B)\in(0,\infty)$, we only need to apply this fact to $X=f$ and $\operatorname P:=\frac{\lambda(1_Bf)}{λ(B)}$. | |
| Mar 2, 2020 at 15:32 | comment | added | lightxbulb | Consider the image function $g(x,y) = \int_{\Omega}W(x,y, \omega)L_i(x,y,\omega)\,d\omega$. This basically gives you the intensity at every point of the film. It's a lot nicer to work with - and as far as I got it, you only care about the vertex on the film (at least the filter does). Your problem reduces to an image processing problem through that simplification. Then you simply have $I_j = \int\int h_j(x,y)g(x,y)\,dx\,dy$. | |
| Mar 2, 2020 at 15:29 | comment | added | 0xbadf00d | But $f$ is not defined on the film. We can clearly use that $I_j=\int\sigma_M^{\otimes\{0,\:1\}}({\rm d}(x_0,x_1))h_j(x_0,x_1)\int\sigma_M^{\otimes\{2,\:\ldots\:,\:k\}}({\rm d}(x_2,\ldots,x_k))f(x_0,\ldots,f_k)$ (considering paths of length $k$ only), but I don't see how this simplifies the problem. | |
| Mar 2, 2020 at 15:24 | comment | added | lightxbulb | In practice your (a) and (b) should always hold (even without all of those arguments), provided that you assume $f < \infty$, and $h_j$ is min-max preserving. I am still unsure why you are considering any dimensions beyond the film ones though. Ideally I would just work with a function $g$ defined only on the film - then you do not even need to care about any of the above, and you can just work with assumptions on $g$. | |
| Mar 2, 2020 at 15:16 | comment | added | 0xbadf00d | Finally, regarding your latest comment: Keeping the measure normalized might be problematic. Note that $\mu$ is not a finite measure (it is a $\sigma$-finite measure though); hence it cannot be normalized. However, it would be sufficient for the purpose of the question that the trace of $\mu$ on $\{f\ne0\}$ is finite (i.e. that $\mu(B)<\infty$). But I have no clue if we can assume this without loss of generality. | |
| Mar 2, 2020 at 15:12 | comment | added | 0xbadf00d | Regarding the second part of your comment: Simply note that $$\int_{B_j}\left(\frac{\int h_jf}{\mu(B_j)} \right)^2\:{\rm d}\mu=\mu(B_j)\cdot\left(\frac{\int h_jf\:{\rm d}\mu}{\mu(B_j)} \right)^2.$$ Hence the square is canceled out. Doing this with the other term as well yields the equation in my second comment. | |
| Mar 2, 2020 at 15:10 | comment | added | lightxbulb | Ok, got what you meant. Honestly, I would keep my measure normalized to avoid this, but in the alternate case I guess you would have to pop-out a constant out of the integrals. I don't think it should change much with respect to the conclusion though. Actually it should probably make the equation more symmetric. | |
| Mar 2, 2020 at 15:07 | comment | added | 0xbadf00d | Sure. You're using Jensen's inequality. Hölder's inequality is the same in this particular case. You cannot use it cause you would need to apply it for the second function to be the constant function $1$. Now you only need to note that the $p$-norm of the constant $1$ is equal to the measure of the entire space. So you we need to divide this factor out (hence normalize the measure) again. | |
| Mar 2, 2020 at 15:04 | comment | added | lightxbulb | You mentioned that $(\int u)^2 \leq \int u^2$ doesn't have to hold. Do you have a reference for this? As far as I know Holder's inequality holds regardless of the normalization of the measure. I am not sure what the issue with $\mu^2$ is - or at least I cannot find the mistake. I haven't pulled anything out of the integral - I just took the common denominator so I could merge the terms. | |
| Mar 2, 2020 at 13:11 | comment | added | 0xbadf00d | You've made a mistake in your third equation. The $\mu^2$ in the denominator should be a $\mu$. I guess you've pulled the corresponding term out of the integral and forgot to multiply it with the measure over which the integral is taken. | |
| Mar 2, 2020 at 13:07 | comment | added | lightxbulb | The ugly fractions. | |
| Mar 2, 2020 at 13:06 | comment | added | 0xbadf00d | Substitute variables for what? | |
| Mar 2, 2020 at 13:06 | comment | added | lightxbulb | The equations in the second edit are equivalent - it's just algebraic transformations. I have a meeting in a few minutes, so I cannot exactly go in details, but try writing it down ($a,b$ are just substitute variables). | |
| Mar 2, 2020 at 13:06 | comment | added | 0xbadf00d | And note that your claim "$\left(\int u\right)^2 \leq \int u^2$" is not correct, unless you normalize the measure over which you're taking the integrals. | |
| Mar 2, 2020 at 13:04 | comment | added | 0xbadf00d | How do I need to read your equation lines? Are they meant to be equivalent to each other or are they meant to be implications (eq. 1 implies eq.2 and so on)? I begin to struggle what you did from (and including) the third equation on. Do you agree that your first equation is equivalent to the one in my second comment? | |
| Mar 2, 2020 at 13:00 | comment | added | lightxbulb | If you set $a=b=1$, then $h_j \leq \sqrt{\alpha}$ is your requirement. But I believe you allow for $\alpha$ to be any, thus you just require for $h_j$ to never go to infinity (which holds for filters). Since it's also an integral, I would say that it's even ok for it to explode on sets of measure zero. | |
| Mar 2, 2020 at 12:57 | comment | added | 0xbadf00d | And let me note that the aspect of (b) I'm looking for is that we can bound the left-hand side in terms of the variance of $f$. So, the factor $\frac1{\mu(B)}$ before the integral in the definition of $f_0$ is not important for me. | |
| Mar 2, 2020 at 12:55 | history | edited | lightxbulb | CC BY-SA 4.0 |
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| Mar 2, 2020 at 12:47 | comment | added | 0xbadf00d | Regarding your "whole term equation": Note that it simplifies to $$\mu|h_jf|^2-\frac{|\mu(h_jf)|^2}{\mu(B_j)}\le\alpha\left[\mu|f|^2-\frac{|\mu f|^2}{\mu(B)}\right],$$ where $B:=\{f\ne0\}$, $B_j:=B\cap\{h_j>0\}$ and $\mu g:=\int g\:{\rm d}\mu$. And it might be useful to note that, if each $h_j$ is a box filter and hence takes only the values $0$ or $1$, we've got $|h_j|^2=h_j$. | |
| Mar 2, 2020 at 12:32 | history | edited | lightxbulb | CC BY-SA 4.0 |
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| Mar 2, 2020 at 12:30 | comment | added | 0xbadf00d | Thank you for your answer. It would be enough for me to show (b) for the box filter with radius 1/2. | |
| Mar 2, 2020 at 12:29 | history | edited | lightxbulb | CC BY-SA 4.0 |
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| Mar 2, 2020 at 12:07 | history | edited | lightxbulb | CC BY-SA 4.0 |
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| Mar 2, 2020 at 12:00 | history | answered | lightxbulb | CC BY-SA 4.0 |