Point distribution model

The point distribution model is a model for representing the mean geometry of a shape and some statistical modes of geometric variation inferred from a training set of shapes.

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The average male drinks 2 liters of water when active outdoors with the standard deviation of 0.7 liters. You are planning a full day nature trip for 50 men and will bring 110 liters of water. What is the probability that you will run out of water? So let's think about what's happening here. So there's some distribution of how many liters an average man needs when they're active outdoors. And let me just draw an example. It might look something like this. So they're all going to need at least more than 0 liters, so this would be 0 liters over here. The average male, so the mean of the amount of water a man needs when active outdoors is 2 liters. So 2 liters would be right over here. So the mean is equal to 2 liters. It has a standard deviation of 0.7 liters or 0.7 liters. So the standard deviation-- maybe I'll draw it this way. So this distribution, once again, we don't know whether it's a normal distribution or not. It could just be some type of crazy distribution. So maybe some people need almost close to-- well, everyone needs a little bit of water, but maybe some people need very, very little water. Then you have a lot of people who need that, maybe some people who need more, and no one can drink more than maybe this is like 4 liters of water. So maybe this is the actual distribution. And then one standard deviation is going to be 0.7 liters away. So this is 1, 0.7 liters is-- so this would be 1 liter, 2 liters, 3 liters. So one standard deviation is going to be about that far away from the mean. If you go above it it'll be about that far, if you go below it. So let me draw. This is the standard deviation. That right there is the standard deviation to the right, that's the standard deviation to the left. And we know that the standard deviation is equal to-- I'll write the 0 in front, 0.7 liters. So that's the actual distribution of how much water the average man needs when active. Now what's interesting about this problem, we are planning a full day nature trip for 50 men and will bring 110 liters of water. What is the probability that you will run out? So the probability that you will run out-- let me write this down. The probability that I will or that you will run out is equal or is the same thing as the probability that we use more than 110 liters on our outdoor nature day, whatever we're doing. Which is the same thing as the probability, if we use more than 110 liters, that means that on average, because we had 50 men, so 110 divided by 50 is what? That's 2.-- let me get the calculator out just so we don't make any mistakes here. So this is going to be, the calculator out. So on average, if we have 110 liters that's going to be drunk by 50 men, including ourselves I guess, that means that it's the-- so we would run out if on average more than 2.2 liters is used per man. So this is the same thing as the probability of the average, or maybe we should say the sample mean-- Or let me write it this way, that the average water use per man of our 50 men is greater than, or we could say greater than or equal to, greater-- well I'll say greater than because if we're right on the money then we won't run out of water-- is greater than 2.2 liters per man. So let's think about this. We are essentially taking 50 men out of a universal sample. We got this data, who knows where we got this data from that the average man drinks 2 liters and that the standard deviation is this. Maybe there's some huge study and this was the best estimate of what the population parameters are. That this is the mean and this is the standard deviation. Now we're sampling 50 men. And what we need to do is figure out essentially what is the probability that the mean of the sample, that the sample mean, is going to be greater than 2.2 liters. And to do that we have to figure out the distribution of the sampling mean. And we know what that's called. It's the sampling distribution of the sample mean. And we know that that is going to be a normal distribution. And we know a few of the properties of that normal distribution. So this is a distribution of just all men. And then if you take samples of, say, 50 men, so this will be-- let me write this down. So down here I'm going to draw the sampling distribution of the sample mean when n, so when our sample size is equal to 50. So this is essentially telling us the likelihood of the different means when we are sampling 50 men from this population and taking their average water use. So let me draw that. So let's say that this is the frequency and then here are the different values. Now the mean value of this, the mean-- let me write it-- the mean of the sampling distribution of the sample mean, this x bar-- that's really just the sample mean right over there-- is equal to, if we were to do this millions and millions of times. If we were to plot all of the means when we keep taking samples of 50, and we were to plot them all out, we would show that this mean of distribution is actually going to be the mean of our actual population. So it's going to be the same value, I'm going to do it in that same blue. It's going to be the same value as this population over here. So that is going to be 2 liters. So we still have-- we're still centered at 2 liters. But what's neat about this is that the sampling distribution of the sample mean, so you take 50 people, find their mean, plot the frequency. This is actually going to be a normal distribution regardless of-- this one just has a well-defined standard deviation mean. It's not normal. Even though this one isn't normal, this one over here will be, and we've seen it in multiple videos already. So this is going to be a normal distribution. And the standard deviation-- and we saw this in the last video, and hopefully we've got a little bit of intuition for why this is true. The standard deviation-- actually maybe put it a better way. The variance of the sample mean is going to be the variance. So remember, it's going to be-- this is standard deviation, so it's going to be the variance of the population divided by n. And if you wanted the standard deviation of this distribution right here, you just take the square root of both sides. If you take the square root of both sides of that we have the standard deviation of the sample mean is going to be equal to the square root of this side over here, is going to be equal to the standard deviation of the population divided by the square root of n. And what's this going to be in our case? We know what the standard deviation of the population is. It is 0.7. And what is n? We have 50 men. So 0.7 over the square root of 50. Now let's figure out what that is with the calculator. So we have 0.7 divided by the square root of 50. And we have 0.09-- well I'll say 0.098-- well it's pretty close the 0.99. So I'll just write that down. So this is equal to 0.099. That's going to be the standard deviation of this. It's going to have a lower standard deviation. So the distribution is going to be normal, it's going to look something like this. So this is 3 liters over here, this is 1 liter. The standard deviation is almost a tenth, so it's going to be a much narrower distribution. It's going to look something-- I'm trying my best to draw it-- it's going to look something like this. You get the idea. Where the standard deviation right now is almost 0.1, so it's 0.09, almost a tenth. So it's going to be something-- one standard deviation away is going to look something like that. So we have our distribution. It's a normal distribution. And now let's go back to our question that we're asking. We want to know the probability that our sample will have an average greater than 2.2. So this is the distribution of all of the possible samples. The means of all of the possible samples. Now to be greater than 2.2, 2.2 is going to be right around here. So we essentially are asking we will run out if our sample mean falls into this bucket over here. So we essentially need to figure out what is-- you can even view it as what's this area under this curve there? And to figure that out we just have to figure out how many standard deviations above the mean we are, which is going to be our Z-score. And then we could use a Z-table to figure out what this area right over here is. So we want to know when we're above 2.2 liters, so 2.2 liters-- we could even do it in our head-- 2.2 liters is what we care about. That's right over here. Our mean is 2, so we are 0.2 above the mean. And if we want that in terms of standard deviations, we just divide this by the standard deviation of this distribution over here. And we figured out what that is. The standard deviation of this distribution is 0.099. So if we take-- and you'll see a formula where you take this value minus the mean and divide it by the standard deviation-- that's all we're doing. We're just figuring how many standard deviations above the mean we are. So you just take this number right over here divided by the standard deviation, so 0.099 or 0.099, and then we get-- let's get our calculator. And actually we had the exact number over here. So we can just take 0.2-- we could just take this 0.2 divided by this value over here. On this calculator when I press second answer it just means the last answer. So I'm taking 0.2 divided by this value over there and I get 2.020. So that means that this value, or I should write this probability is the same probability of being 2.02 standard deviations-- or maybe I should write it this way-- more than-- Let me write it down here where I have more space. So this all boils down to the probability of running out of water is the probability that the sample mean will be more than-- just the 50 that we happened to select-- remember, if we take a bunch of samples of 50 and plot all of them we'll get this whole distribution. But the one 50, the group of 50 that we happened to select, the probability of running out of water is the same thing as the probability of the mean of those people, will be more than 2.020 standard deviations above the mean of this distribution, which they're actually the same distribution. So what is that going to be? And here we just have to look up our Z-table. Remember, this 2.02 is just this value right here. 0.2 divided by 0.09. I just had to pause the video because there's some type of fighter jet outside or something. So anyway, hopefully they won't come back. But anyway, so we need to figure out the probability that the sample mean will be more than 2.02 standard deviations above the mean. And to figure that out we go to a Z-table, and you could find this pretty much anywhere. Usually it's in any stat book or on the internet, wherever. And so essentially we want to know the probability-- the Z-table will tell you how much area is below this value. So if you go to z of 2.02-- that was the value that we were dealing with, right. You have 2.02, it was-- so you go for the first digit. We go to 2.0, and it was 2.02. 2.02 is right over there. So we have 2.0, and then in the next digit you go up here. So 2.02 is right over there. So this 0.9783-- let me write it down over here-- this 0.9783-- I want to be very careful. 0.9783, that Z-table, that's not this value over here. This 0.9783 on the Z-table, that is giving us this whole area over here. It's giving us the probability that we are below that value. That we are less than 2.02 standard deviations above the mean. So it's giving us that value over here. So to answer our question, to answer this probability, we just have to subtract this from 1 because these will all add up to 1. So we just take our calculator back out and we just take 1 minus 0.9783 is equal to 0.0217. So this right here is 0.0217. Or another way you could say it, it is a 2.17% probability that we will run out of water. And we are done. Let me make sure I got that number right. So that number it was, yeah, 0.0217, right. So it's a 2.17% chance we run out of water.

Background

The point distribution model concept has been developed by Cootes,^[1] Taylor et al.^[2] and became a standard in computer vision for the statistical study of shape^[3] and for segmentation of medical images^[2] where shape priors really help interpretation of noisy and low-contrasted pixels/voxels. The latter point leads to active shape models (ASM) and active appearance models (AAM).

Point distribution models rely on landmark points. A landmark is an annotating point posed by an anatomist onto a given locus for every shape instance across the training set population. For instance, the same landmark will designate the tip of the index finger in a training set of 2D hands outlines. Principal component analysis (PCA), for instance, is a relevant tool for studying correlations of movement between groups of landmarks among the training set population. Typically, it might detect that all the landmarks located along the same finger move exactly together across the training set examples showing different finger spacing for a flat-posed hands collection.

Details

First, a set of training images are manually landmarked with enough corresponding landmarks to sufficiently approximate the geometry of the original shapes. These landmarks are aligned using the generalized procrustes analysis, which minimizes the least squared error between the points.

$k$ aligned landmarks in two dimensions are given as

\mathbf {X} =(x_{1},y_{1},\ldots ,x_{k},y_{k})

It's important to note that each landmark $i\in \lbrace 1,\ldots k\rbrace$ should represent the same anatomical location. For example, landmark #3, $(x_{3},y_{3})$ might represent the tip of the ring finger across all training images.

Now the shape outlines are reduced to sequences of $k$ landmarks, so that a given training shape is defined as the vector $\mathbf {X} \in \mathbb {R} ^{2k}$ . Assuming the scattering is gaussian in this space, PCA is used to compute normalized eigenvectors and eigenvalues of the covariance matrix across all training shapes. The matrix of the top $d$ eigenvectors is given as $\mathbf {P} \in \mathbb {R} ^{2k\times d}$ , and each eigenvector describes a principal mode of variation along the set.

Finally, a linear combination of the eigenvectors is used to define a new shape $\mathbf {X} '$ , mathematically defined as:

\mathbf {X} '={\overline {\mathbf {X} }}+\mathbf {P} \mathbf {b}

where ${\overline {\mathbf {X} }}$ is defined as the mean shape across all training images, and $\mathbf {b}$ is a vector of scaling values for each principal component. Therefore, by modifying the variable $\mathbf {b}$ an infinite number of shapes can be defined. To ensure that the new shapes are all within the variation seen in the training set, it is common to only allow each element of $\mathbf {b}$ to be within $\pm$ 3 standard deviations, where the standard deviation of a given principal component is defined as the square root of its corresponding eigenvalue.

PDM's can be extended to any arbitrary number of dimensions, but are typically used in 2D image and 3D volume applications (where each landmark point is $\mathbb {R} ^{2}$ or $\mathbb {R} ^{3}$ ).

Discussion

An eigenvector, interpreted in euclidean space, can be seen as a sequence of $k$ euclidean vectors associated to corresponding landmark and designating a compound move for the whole shape. Global nonlinear variation is usually well handled provided nonlinear variation is kept to a reasonable level. Typically, a twisting nematode worm is used as an example in the teaching of kernel PCA-based methods.

Due to the PCA properties: eigenvectors are mutually orthogonal, form a basis of the training set cloud in the shape space, and cross at the 0 in this space, which represents the mean shape. Also, PCA is a traditional way of fitting a closed ellipsoid to a Gaussian cloud of points (whatever their dimension): this suggests the concept of bounded variation.

The idea behind PDMs is that eigenvectors can be linearly combined to create an infinity of new shape instances that will 'look like' the one in the training set. The coefficients are bounded alike the values of the corresponding eigenvalues, so as to ensure the generated 2n/3n-dimensional dot will remain into the hyper-ellipsoidal allowed domain—allowable shape domain (ASD).^[2]

References

^ T. F. Cootes (May 2004), Statistical models of appearance for computer vision (PDF)
^ ^a ^b ^c D.H. Cooper; T.F. Cootes; C.J. Taylor; J. Graham (1995), "Active shape models—their training and application", Computer Vision and Image Understanding (61): 38–59
^ Rhodri H. Davies and Carole J. Twining and P. Daniel Allen and Tim F. Cootes and Chris J. Taylor (2003). Shape discrimination in the Hippocampus using an MDL Model. IMPI. Archived from the original on 2008-10-08. Retrieved 2007-07-27.

External links

Flexible Models for Computer Vision, Tim Cootes, Manchester University.
A practical introduction to PDM and ASMs.

This page was last edited on 11 January 2022, at 15:53

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