I find the slide a little bit confusing, especially with the different occurrence of f(x). It was helpful for me to think of it this way:

our goal is to calculate
$$
I=\int f(x)dx
$$
We can rewrite this as
$$
I=\int {f(x)\over p(x)}p(x)dx
$$
which is $E[{f(x)\over p(x)}]$.

By the derivation of the next slide, we can estimate this with ${1\over N}\sum {f(x_i)\over p(x_i)}$, where $x_i$~$p(x)$

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mmp

That's good feedback. In retrospect I think that what you're suggesting--basically starting with the derivation we used later in the lecture, would make the ideas clearer. (The students next year can benefit from this!)

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2

Tianye

Sorry I forgot what V(p,p') was... Is it BRDF?

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mmp

V is just a function that is 1 if the two points are mutually visible and 0 if there is another object occluding the line segment between them.

The BRDF comes in starting in next lecture.

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1

atrytko

The phrase "decreases linearly with sample size" was confusing to me. After checking with the teaching staff, I thought I'd post here to clarify that this means 'is inversely proportional'.

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2

brianjo

Small technical note: the right hand side of the first line of the proof should read E[f(X_i)/p(X_i)].

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mmp

Good catch--thanks. Will fix in the master slides.

I find the slide a little bit confusing, especially with the different occurrence of f(x). It was helpful for me to think of it this way:

our goal is to calculate $$ I=\int f(x)dx $$ We can rewrite this as $$ I=\int {f(x)\over p(x)}p(x)dx $$ which is $E[{f(x)\over p(x)}]$.

By the derivation of the next slide, we can estimate this with ${1\over N}\sum {f(x_i)\over p(x_i)}$, where $x_i$~$p(x)$

This comment was marked helpful 0 times.

That's good feedback. In retrospect I think that what you're suggesting--basically starting with the derivation we used later in the lecture, would make the ideas clearer. (The students next year can benefit from this!)

This comment was marked helpful 0 times.