# Principal Components Analysis, Eigenfaces, and Karhunen-Loève Expansions

Principal Components Analysis

Instead of photographs, let’s suppose we’re just looking at vectors, i.e. points in space, randomly drawn from some unknown distribution.

Part of the crabs dataset in R’s MASS package.

In the image above, the points clearly vary most along some diagonal line. There are a few things we might want to do with this:

1. We might want to plot the data so that the axes represent key features. In this case, we’d like to rotate the plot, so that this diagonal line is along the horizontal axis.

2. We might want to plot the data so that a point’s position on one axis is linearly independent of its position along the other ones. In this way, we could generate new points from the distribution without having to overly worry about covariance issues, because we can generate the position along each axis separately.

Happily, PCA does both of these.

The new perpendicular axes. Note one appears to go along some sort of central line in the data.

The data plotted with respect to the new axes. The data’s variance clearly increases from left to right, but the covariance is zero.

I might get round to looking at the mathematics behind how PCA does this in a later post. For now, let’s see how PCA works for our photographs.

Eigenface Decomposition

The first step is to find the mean face, which we did last time. If we now subtract the red, blue and green intensities for pixels of this mean face from those of the originals, we get this shifty group.

The “difference faces”, with thresholding.

These faces look very dark. The reason for this is fairly simple: when we subtract the mean face, a large number of the colour intensities will end up being negative, rather than in [0,1]. R refuses to plot anything when this happens, so I’ve simply told it to count negative values as being zero — pitch black. Later on, I’ll also count values greater than one as being one — pitch white. Jeremy Kun notices the same tendency to darkness, but doesn’t really explain why: I suspect Mathematica, the software Kun uses in his post, does this “thresholding” automatically.

As an alternative to thresholding, I’ll often be linearly “rescaling” images, so that the smallest intensity goes to zero, and the largest goes to one. For the “difference faces”, since subtracting the mean face is also a linear transformation, rescaling means we get an image that’s pretty similar to the original.

The difference faces with rescaling.

So far, so good. Now for the eigenvectors, or “eigenfaces”. Remember, these represent the directions in which we expect variation to be independent, with highest-variance directions first. The photographs are $66$ by $58$ pixels in three colours, so these vectors have $66\times58\times3=11484$ elements each, and are each normalised so that the sum of the squares of their intensities is equal to one, so it shouldn’t be too surprising that their intensities are all close to zero. This makes thresholding useless, so we just show the rescaled version.

The eigenfaces, rescaled. The rescaling was done ignoring the last face, as this is essentially random noise, with a far larger intensity range, but no informative value.

These are certainly less sinister than those in Kun’s article. I assume this is the consequence of looking at such a homogenous set of people, and Türk doing a good job of positioning everyone in the same place in the frame. Such a small and cooperative data set makes this blogger a happy little theorist.

We wouldn’t really expect these eigenfaces to exactly align with introducing or removing different face features, but looking at them shows some features a lot more obviously than others. For example: the first eigenface appears to be determining asymmetry in lighting on the two sides of the face; the second one partially accounts for glasses, and how far down the hair comes over the forehead; eigenface nine accounts for asymmetry in how far the hair comes down on each side of the forehead, and also looks rather similar to the original face nine, since that is the only face with such a drastic asymmetry; and so on.

So, what can we do with these eigenfaces? Well, we can easily decompose a face into the mean face, plus a linear sum of the eigenfaces — twenty-five of them, here. If we want to reduce the size of the data, we can start throwing out some of the eigenfaces. In particular, since we’ve calculated how much variability there is along each eigenface, we can throw out the least variable ones first. This way, we minimise how much of the variance between the faces is thrown out, and so keep the faces as distinct as possible.

To illustrate this, we can introduce the eigenfaces for a face one at a time, to observe the effect on the image:

Progression for face one.

Progression for the more distinctive face nine.

Some faces will become clear faster than others, but it seems like both faces become close to the originals by, say, eighteen components. Indeed, if we only use eighteen components for each face, we get the following:

Class Average I, reconstructed using only eighteen of twenty-five components.

Faces like number ten are a bit vague, but that’s mostly pretty good.

So, what else can we do? Well, since we know how faces vary along each eigenface, we can try generating sets of new faces. The results can be, well, rather mixed. Sometimes the results look OK, sometimes they don’t look convincing at all.

This set doesn’t look too convincing to me.

This one looks better.

The one on the left’s been in the wars. The one on the right looks shocked to be here.

This is probability mostly due to my sampling the value of the eigenface components as independent normal distributions, which makes no sense in the context of the problem.

That’s about it for now. There are a few diagnostic plots I can show off once I find them again, allowing you to do things like assigning a quantity to how distinctive each face is from the rest of the set (nine and twenty-one stand out the most), and a more quantitive assessment of how many eigenfaces to throw out while keeping the faces distinguishable.

# Principal Components Analysis, Eigenfaces, and Karhunen-Loève Expansions: Part One

Last time, we were talking about interpolation and orthogonal polynomials. For this series, we’re also going to end up talking about orthogonal vectors and orthogonal functions. But first, we’ll look at some old photos from the Seventies.

Principal Components Analysis
The Budapest National Gallery has its floors arranged in chronological order, so the first floor is given over to an overwhelming sea of medieval paintings of Bible stories and prophets. Well-painted as they are, when the founder of a country is a King Saint, you know you’re in for the long haul on this floor. However, the visitor who hasn’t run out of patience before they reach the floors for the 20th Century will find a cute pair of photographs by Péter Türk.

One is a collation of shots of Türk’s class. For the second, he chopped each of the shots into a grid of squares. The top-left squares from each shot have then been placed together in one large square. The squares of the next grid place along have then been placed alongside, and so on. The result is something that, from a distance, looks like a new face.

Class Average I, Péter Türk, 1979.

Class Average II, Péter Türk, 1979.

Today, this would be done by computer — a mechanical Türk, so to speak — and the squares would be blended together instead of being placed in blocks, but what we have here is some idea of a “mean face”, the average of all the individual photos. It’s not exactly like any of the individual photos, but clearly shows a lot of their shared features. Especially the hair. If we let the computer take the mean over each pixel position, rather than placing them next to each other, we obtain something that looks surprisingly human.

The true mean face, or as close as we can get with low-quality online scans of the originals.

Class Average II rescaled, for comparison.

It’s worth mentioning — not for the last time — how good a job Türk did at keeping his photographs consistent, with regards to face positioning. This is a small group of people, yes, but if we take the eyes as an example, the eye positions are so similar that the mean face still has pupils!

This tinkering is well and good, but consider what we might want to do with a large collection of such photos.
1. Say we wanted to make a digital database with thousands of these photos. Maybe we want to set up some sort of photographic identification system, where we compare a new photo, of someone requesting access to something, against the photos we have on file. How would we decide if the new photo was similar enough to one on the database to grant them access?
2. Along similar lines, suppose we’re not interested specifically in the people in the photos, but we are interested in finding some rules of thumb for telling between different people in the same population. How would we do so?
3. Now suppose we’d also like to compress the photos somehow, to save on memory. How should we do so while keeping the photos as distinguishable as possible?

One method we can use for this is Principal Components Analysis, which I’ll be talking about over the next few posts. However, here’s a brief taste of what it allows us to do, statistically rather than by guesswork:
1. PCA gives us a way to take a new photo, and make some measure of its “distance” from our originals. We can then decide that it’s a photo of the person in the closest original, or that it’s a new person if all the “distances” are too large.
2. The most important features for distinguishing between people in the above set of photos are the side of their face the light source is on, and how far down their fringe comes.
3. We can compress the photos in such a way that we know how much of the original variance, or distinctiveness between the photos, that we keep. If we don’t mind compressing by different amounts for different photos, we could also keep a set amount of distinctiveness for each photo, rather than across the whole group.
4. We can try — with variable levels of success — to generate new faces from scratch.

Also worth noting is that, apart from 4., all of this can be done with only a covariance estimate: we make no assumptions about the distribution the photos are drawn from.

We’ll come back to these photos, and these applications, later in the series. Next time, we’ll look at something a bit simpler first.

It’s been about five months since I said this post was in draft, so it’s about time I reined in my perfectionism and published the damn thing.

Since this is a graduate student blog at the moment, it seems reasonable I should write a bit more about what I’m learning at any particular time. Late last year, our department had a talk by Erik van Doorn, from the University of Twente, which looked at birth-death processes, and asymptotic results for how quickly they can be expected to converge to their equilibrium distribution. It was an overview of his paper Representations for the decay parameter of a birth-death process based on the Courant-Fischer theorem, to appear in the Journal of Applied Probability.

A good deal of the talk introduced and used basic results on orthogonal polynomials, so I went to see if any of my books mentioned the subject. It turned out there was a chapter on them in Approximation Theory and Methods by Michael Powell – a book that’s been on my bookshelf for about five years but hardly been read – regarding their use in Gaussian quadratures. The following is mostly spliced together from that chapter, and my undergraduate notes on Numerical Analysis.

Interpolation

Before we talk about quadratures, it’s best if we start with interpolation. Say we have some function over some interval, where we can take a few sample values, with no measurement error, but we have no explicit formula and can’t afford to sample it everywhere. We thus would like to use our sample values to fit an approximating function to the whole interval. One simple way to do this is to try to fit a polynomial through the sample points. We can do this by assigning each sample point a Lagrange polynomial

$l_k(x) = \prod_{n \neq k} \frac{x-x_n} {x_k-x_n} ,$

with value one at that sample point and zero at all the others. For example, if we take our sample points at -1,-0.5,0,0.5, and 1, then the Lagrange polynomials are those in the plot below. There’s a light grey line at one to help check they are equal to one or zero at the sample points.

Our fitted curve will then just be a sum of these Lagrange polynomials, multiplied by their corresponding sample value, so we get a polynomial passing through all the sample points, and estimate the function $f(x)$ as

$\hat{f}(x) = \sum_k f(x_k) l_k(x) .$

This gives a curve that passes through all the interpolation points with the smallest-order polynomial possible. It works well for estimating functions that are, indeed, polynomials, but for other functions it can run into problems. In particular, there are cases where the difference between the interpolation curve and the true function at certain points increases when we increase the number of sample points, so we can’t necessarily improve the approximation by adding points. There’s also the question of where to sample the original function, if we have control over that. I’ll pass over these issues, and move on to integration.

Now say that, instead of approximating a function with some samples, we want to approximate a function’s integral by sampling its value at a few points, or

$\int_a^b f(x) \, \textrm{d} x \simeq \omega_0f(x_0) +\omega_1f(x_1) +\ldots +\omega_nf(x_n) .$

If we want to focus on making the integration accurate when $f$ is a low-order polynomial, the quadrature with $n+1$ sample points is exact for polynomials up to order $n$ if we set the weights as

$\omega_k = \int_a^b l_k(x) \, \textrm{d} x.$

In other words, a quadrature is equivalent to fitting an interpolation curve, and integrating over it. For example, if we’re integrating a function over the interval $[-1,1] ,$ we could simply take one sample, with weight one. This would give a quadrature of $2f(x_0) ,$ which is exact for any zero-order, constant function, regardless of the position of $x_0.$

We could take samples at the endpoints, to get the quadrature $\frac{1} {2} f(-1) +\frac{1} {2} f(1) ,$ and we can set $f(x)$ to be constant, or proportional to $x,$ to see the result for first-order polynomials is exact.

We could also take the endpoints and the midpoint. Then we have $\frac{1} {3} f(-1) +\frac{4} {3} f(0) +\frac{1} {3} f(1) ,$ which is exact for polynomials up to order two.

However, occasionally we stumble onto a quadrature that does a little better than expected. For the first quadrature above, since our interval is symmetric around zero, if we let $x_0=0$ any first-order term will be antisymmetric around this midpoint, so this quadrature is exact for first-order polynomials too. Similarly, the second quadrature is exact for quadratic terms, but the third quadrature can still only deal with quadratics, and can’t handle cubics.

Considering what happened when we placed the sample points for the first quadrature at zero, we might guess this is something to do with where we place our sample points. If so, how should we place our sample points, and what’s the highest-order function we can exactly integrate with any set number of samples? To answer this, we can use orthogonal polynomials.

Orthogonal polynomials

We say two vectors are orthogonal when their inner product is equal to zero. For example, if the inner product is simply the dot product, then

$\langle x,y \rangle = \sum_{k=1}^n x_ky_k =0,$

and so vectors are orthogonal if they are perpendicular to each other.

We have a similar definition and example for orthogonal polynomials, but now we choose an inner product that integrates over an interval instead of summing over two vectors’ dimensions:

$\langle f,g \rangle = \int_a^b f(x)g(x)\,\textrm{d} x =0.$

We can then choose a sequence of polynomials with increasing order that are all orthogonal to each other. For example, we can start the sequence with $f_0(x)=1,$ or some multiple of it. We then seek a first-order polynomial $f_1(x)=Ax+B$ such that

$\int_a^b 1\times(Ax+B) \,\textrm{d} x =\frac{A} {2} (b^2-a^2) +B(b-a) =0.$

This can be any multiple of $x-(b+a) /2 .$ In many cases we wish the orthogonal polynomials to have be orthonormal, i.e. $\langle f_k,f_k\rangle =1,$ so for the above we require

$\int_a^b C_0^2 \,\textrm{d}x = C_0^2 (b-a) = 1,$

\begin{aligned} \int_a^b C_1^2(x-(b+a)/2)^2 \,\textrm{d}x &= C_1^2 \frac{1}{3} \left[\left(b-\frac{b+a}{2}\right)^3 -\left(a-\frac{b+a}{2}\right)^3\right] \\&= C_1^2 \frac{2} {3} \left(\frac{b-a} {2}\right)^3 \\&=1,\end{aligned}

and so on, giving a condition for the value of each scaling factor $C_k.$ We can then find the next term by looking for a second-order polynomial that is orthogonal to $1$ and $x,$ and so on. In the case where $a=-1$ and $b=1$ this gives a simple sequence of polynomials that begins with

$f_0(x)=\frac{1}{\sqrt{2}} ,\, f_1(x)=\sqrt{\frac{3}{2}} x,$

$f_2(x)=\sqrt{\frac{5}{2}} (\frac{3}{2} x^2-\frac{1}{2}) ,\, f_3(x) =\sqrt{\frac{7}{2}} (\frac{5}{2}x^3-\frac{3}{2}x),\ldots$

This is an orthonormal version of the Legendre polynomials.

Since any polynomial can then be expressed as a linear combination of members of this sequence, each polynomial in the sequence is also orthogonal to any polynomial with lower order. So, for example, $f_3$ is orthogonal to all constant, linear, and quadratic polynomials.

To be continued
The next post will explain why these orthogonal polynomials help us decide on interpolation points.

After a few more posts, I’m planning to return to quadratures to talk about something I’ve mentioned on other topics: since the above procedure for quadratures gives a point estimate, we have no idea of how much uncertainty we have in our estimate. I’m therefore going to talk a bit about Bayesian quadratures. In particular, I’m going to start with a 1988 paper by Persi Diaconis called “Bayesian Data Analysis”, and fill in the gaps for those of us whose knowledge of integrated Brownian motion isn’t quite up to speed.