I am trying to implement a Microfacet BRDF shading model (similar to the Cook-Torrance model) and I am having some trouble with the Beckmann Distribution defined in this paper: https://www.cs.cornell.edu/~srm/publications/EGSR07-btdf.pdf
Where M is a microfacet normal, N is the macrofacet normal and ab is a "hardness" parameter between [0, 1].
My issue is that this distribution often returns obscenely large values, especially when ab is very small.
For instance, the Beckmann distribution is used to calculate the probability of generating a microfacet normal M per this equation :
A probability has to be between the range [0,1], so how is it possible to get a value within this range using the function above if the Beckmann distribution gives me values that are 1000000000+ in size?
So there a proper way to clamp the distribution? Or am I misunderstanding it or the probability function? I had tried simply clamping it to 1 if the value exceeded 1 but this didn't really give me the results I was looking for.
I was having the same question you did.
If you read
http://blog.selfshadow.com/publications/s2012-shading-course/hoffman/s2012_pbs_physics_math_notes.pdf
and
http://blog.selfshadow.com/publications/s2012-shading-course/hoffman/s2012_pbs_physics_math_notebook.pdf
You'll notice it's perfectly normal. To quote from the links:
"The Beckmann Αb parameter is equal to the RMS (root mean square) microfacet slope. Therefore its valid range is from 0 (non-inclusive –0 corresponds to a perfect mirror or Dirac delta and causes divide by 0 errors in the Beckmann formulation) and up to arbitrarily high values. There is no special significance to a value of 1 –this just means that the RMS slope is 1/1 or 45°.(...)"
Also another quote:
"The statistical distribution of microfacet orientations is defined via the microfacet normal distribution function D(m). Unlike F (), the value of D() is not restricted to lie between 0 and 1—although values must be non-negative, they can be arbitrarily large (indicating a very high concentration of microfacets with normals pointing in a particular direction). (...)"
You should google for Self Shadow's Physically Based Shading courses which is full of useful material (there is one blog post for each year: 2010, 2011, 2012 & 2013)
Related
I am using Keras to setup neural networks.
As input data, I use vectors in which each coordinate can be either 0 (feature not present or not measured) or a value that can range for instance between 5000 and 10000.
So my input value distribution is a kind of gaussian centered let us say around 7500 plus a very thin peak at 0.
I cannot remove the vectors with 0 in some of their coordinates because almost all of them will have some 0s at some locations.
So my question is : "how to best normalize the input vectors ?". I see two possibilities :
just substract the mean and divide by standard deviation. The problem then is that the mean is biased by the high number of meaningless 0s, and the std is overestimated, which erases the fine changes in the meaningful measurement.
compute the mean and standard deviation on the non-zeros coordinates, which is more meaningful. But then all the 0 values that correspond to non measured data will come out with high (negative) values which gives some importance to meaningless data...
Does someone have an advice on how to proceed ?
Thanks !
Instead, represent your features as 2 dimensions:
First one is normalised value of the feature if it is non zero (where normalisation is computed over non zero elements), otherwise it is 0
Second is 1 iff the feature was 0, otherwise it is 1. This makes sure that 0 from the previous feature that could either come from raw 0, or from normalised 0 can be discriminated
You can think of this as encoding extra feature saying "the other feature is missing". This way scale of each feature is normalised, and all informatino preserved
Apologies for the overlap with existing questions; mine is at a more basic skill level. I am working with very sparse occurrences spanning very large areas, so I would like to calculate probability at pixels using the density.ppp function (as opposed to relrisk.ppp, where specifying presences+absences would be computationally intractable). Is there a straightforward way to convert density (intensity) to probabilities at each point?
Maxdist=50
dtruncauchy=function(x,L=60) L/(diff(atan(c(-1,1)*Maxdist/L)) * (L^2 + x^2))
dispersfun=function(x,y) dtruncauchy(sqrt(x^2+y^2))
n=1e3; PPP=ppp(1:n,1:n, c(1,n),c(1,n), marks=rep(1,n));
density.ppp(PPP,cutoff=Maxdist,kernel=dispersfun,at="points",leaveoneout=FALSE) #convert to probabilies?
Thank you!!
I think there is a misunderstanding about fundamentals. The spatstat package is designed mainly for analysing "mapped point patterns", datasets which record the locations where events occurred or things were located. It is designed for "presence-only" data, not "presence/absence" data (with some exceptions).
The relrisk function expects input data about the presence of two different types of events, such as the mapped locations of trees belonging to two different species, and then estimates the spatially-varying probability that a tree will belong to each species.
If you have 'presence-only' data stored in a point pattern object X of class "ppp", then density(X, ....) will produce a pixel image of the spatially-varying intensity (expected number of points per unit area). For example if the spatial coordinates were expressed in metres, then the intensity values are "points per square metre". If you want to calculate the probability of presence in each pixel (i.e. for each pixel, the probability that there is at least one presence point in the pixel), you just need to multiply the intensity value by the area of one pixel, which gives the expected number of points in the pixel. If pixels are small (the usual case) then the presence probability is just equal to this value. For physically larger pixels the probability is 1 - exp(-m) where m is the expected number of points.
Example:
X <- redwood
D <- density(X, 0.2)
pixarea <- with(D, xstep * ystep)
M <- pixarea * D
p <- 1 - exp(-M)
then M and p are images which should be almost equal, and can both be interpreted as probability of presence.
For more information see Chapter 6 of the spatstat book.
If, instead, you had a pixel image of presence/absence data, with pixel values equal to 1 or 0 for presence or absence respectively, then you can just use the function blur in the spatstat package to perform kernel smoothing of the image, and the resulting pixel values are presence probabilities.
As a beginner to differential privacy, I would like to why the variance for noise mechanisms needs to be calibrated with sensitivity? What is the purpose of that? What happens if we don't calibrate it and add a random variance?
Example scenario here In Laplacian noise, why scale parameter is calibrated?
One way you can understand this intuitively is by imagining a function that returns either of two values, say 0 and a for some real a.
Suppose further that we have an additive noise mechanism, so that we end up with two probability distributions on the real line, as in the image from your attached link (this is an example of the setup above, with a=1):
In pure DP, we are interested in computing the maximum of the ratio of these distributions over the entire real line. As the calculation in your link shows, this ratio is bounded everywhere by e to the power of epsilon.
Now, imagine moving the centers of these distributions further apart, say by shifting the red distribution further to the right (IE, increasing a). Clearly this will place less probability mass from the red distribution on the value 0, which is where the maximum of this ratio will be achieved. Therefore the ratio between these distributions at 0 will be increased--a constant (the mass the blue distribution places on 0) is divided by a smaller number.
One way we could move the ratio back down would be to "fatten" the distributions out. This would correspond pictorally to moving the peaks of the distributions lower, and spreading the mass out over a wider area (since they have to integrate to 1, these two things are necessarily coupled for a distribution like the Laplace). Mathematically we would accomplish this by increasing the variance in the Laplace distribution (increasing b in the parameterization here), which has the effect of lowering the peak of the blue distribution at 0 and raising the mass the red distribution places at 0, thereby reducing the ratio between them back down (a smaller numerator and a larger denominator).
If you perform the calculations, you will find that the relationship between the variance parameter b and the sensitivity of the function f is in fact linear; that is, setting b to be
fixes the maximum of this ratio, to
which is precisely the definition of pure differential privacy.
If you add arbitrary amounts of random noise, you simply end up with random data. Sure, it preserves privacy, but at the same time as destroying any real value in the data. The noise you add needs to match your existing distribution so that it preserves privacy without destroying the value of the data. That’s what the calibration step does.
Hi everyone I am reading a book "Element of Statistical Learning) and came across the below paragraph which i dont I understand. (explains how the training data was generated)
We generated 10 means mk from a bivariate Gaussian distribution N((0,1)T,I) and labeled this class as blue. Similraly, 10 more were drawn from from N((0,1)T,I) and labeled class Orange. Then for each class we generated 100 observations as follows: for each observation, we picked an mk at random with probability 1/10, and then generated a N(mk, I/5), thus leading to a mixture of Gaussian cluster for each class.
I would appreciate if you could explain the above paragraph and especially N((0,1)T,I)
by the way- (0,1) to the power of T for Transpose.
Is this notation mathmatically common or related to a specific computer language.
In the paragraph N stands for the Normal distribution; more specifically, in this case it stands for the Multivariate normal distribution. It is not specific to any programming languages. It comes from statistics and probability theory, but due to numerous appealing properties and important applications of this probability distribution it is also widely used in programming, so you should be able to perform the described procedure in any language.
The part (0,1)^T is a vector of means. That is, we have in mind a random vector of length two, where the first element on average is 0, and the second one on average is 1.
"I" stands for the 2x2 identity matrix whose role is the variance-covariance matrix. That is, the variance of both random vector components is 1 (i.e., the diagonal terms), while off-diagonal points are 0 and correspond to the covariance between the two random variables.
we have skew normal distribution with location=0, scale =1 and shape =0 then it is same as standard normal distribution with mean 0 and variance 1.but if we change the shape parameter say shape=5 then mean and variance also changes.how can we fix mean and variance with different values of shape parameter
Just look after how the mean and variance of a skew normal distribution can be computed and you got the answer! Knowing that the mean looks like:
and
You can see, that with a xi=0 (location), omega=1 (scale) and alpha=0 (shape) you really get a standard normal distribution (with mean=0, standard deviation=1):
If you only change the alpha (shape) to 5, you can except the mean will differ a lot, and will be positive. If you want to hold the mean around zero with a higher alpha (shape), you will have to decrease other parameters, e.g.: the omega (scale). The most obvious solution could be to set it to zero instead of 1. See:
Mean is set, we have to get a variance equal to zero with a omega set to zero and shape set to 5. The formula is known:
With our known parameters:
Which is insane :) That cannot be done this way. You may also go back and alter the value of xi instead of omega to get a mean equal to zero. But that way you might first compute the only possible value of omega with the formula of variance given.
Then the omega should be around 1.605681 (negative or positive).
Getting back to mean:
So, with the following parameters you should get a distribution you was intended to:
location = 1.256269 (negative or positive), scale = 1.605681 (negative or positive) and shape = 5.
Please, someone test it, as I might miscalculated somewhere with the given example.