I'm not really following what's going on in this slide. why is $p(r, \theta) \approx r$? If anything wouldn't it be $(1 / r) \cdot (1 / \theta)$ because $r$ and $\theta$ are chosen independently from each other and uniformly amongst their distributions?

BryceSummers

Let R be the radius of the circle we are sampling from. (Constant)
Let r be a fixed and arbitrary target radius.
Let $\theta$ be a fixed and arbitrary target angle.

$p(r, \theta) = p(r) * p(\theta)$

$ = (1/R) * p(\theta)$

$ = (1/R) * 1/(2pir)$

$= (1/(R2pi)) * (1/r)$

For direct proportionality, we can eliminate the constant terms.

$p(r, \theta) \approx (1/r)$

mchoquet

@dvernet: let me given an alternative explanation: we want $p(r) \sim r$ (not $\approx$, $\sim$ means "proportional to"), so the probability of our sample being at some radius $r$ increases linearly as that radius increases. This is reasonable because the amount of circle to sample points from at radius r increases linearly with r.

mchoquet

@doodooloo: the points everywhere in the circle should be equally likely to be chosen; having p(r) scale linearly with r achieves that effect.

BryceSummers

@doodooloo

A naive method will indeed lead to there being more density in the center.
Image

A correct uniform method like those on the slides after this one will have equal density everywhere.

kayvonf

I updated the slide title to make it more clear. I also fixed the subtitle to say: "points farther from the center are less likely to be chosen." thanks everyone.

dvernet

@kayvonf FYI it doesn't like your update took effect.

I'm not really following what's going on in this slide. why is $p(r, \theta) \approx r$? If anything wouldn't it be $(1 / r) \cdot (1 / \theta)$ because $r$ and $\theta$ are chosen independently from each other and uniformly amongst their distributions?

Let R be the radius of the circle we are sampling from. (Constant) Let r be a fixed and arbitrary target radius. Let $\theta$ be a fixed and arbitrary target angle.

$p(r, \theta) = p(r) * p(\theta)$

$ = (1/R) * p(\theta)$

$ = (1/R) * 1/(2

pir)$$= (1/(R

2pi)) * (1/r)$For direct proportionality, we can eliminate the constant terms.

$p(r, \theta) \approx (1/r)$

@dvernet: let me given an alternative explanation: we want $p(r) \sim r$ (not $\approx$, $\sim$ means "proportional to"), so the probability of our sample being at some radius $r$ increases linearly as that radius increases. This is reasonable because the amount of circle to sample points from at radius r increases linearly with r.

@doodooloo: the points everywhere in the circle should be equally likely to be chosen; having p(r) scale linearly with r achieves that effect.

@doodooloo

A naive method will indeed lead to there being more density in the center. Image

A correct uniform method like those on the slides after this one will have equal density everywhere.

I updated the slide title to make it more clear. I also fixed the subtitle to say: "points farther from the center are

less likelyto be chosen." thanks everyone.@kayvonf FYI it doesn't like your update took effect.