# Statistics – the rules of the game

What is statistics about, really? It’s easy to go through a class and get the impression that it’s about manipulating intimidating formulas. But what’s the goal of them? Why did people invent them?

If you zoom out, the big picture is more conceptual than mathematical. Statistics has a crazy, grasping ambition: it wants to tell you how to best use observations to make decisions. For example, you might look at how much it rained each day in the last week, and decide if you should bring an umbrella today. Statistics converts data into ideal actions.

Here, I’ll try to explain this view. I think it’s possible to be quite precise about this while using almost no statistics background and extremely minimal math.

The two important characters that we meet are decision rules and loss functions. Informally, a decision rule is just some procedure that looks at a dataset and makes a choice. A loss function — a basic concept from decision theory– is a precise description of “how bad” a given choice is.

## Model Problem: Coinflips

Let’s say you’re confronted with a coin where the odds of heads and tails are not known ahead of time. Still, you are allowed to observe how the coin performs over a number of flips. After that, you’ll need to make a “decision” about the coin. Explicitly:

• You’ve got a coin, which comes up heads with probability $w$. You don’t know $w$.
• You flip the coin $n$ times.
• You see $k$ heads and $n-k$ tails.
• You do something, depending on $k$. (We’ll come back to this.)

Simple enough, right? Remember, $k$ is the total number of heads after $n$ flips. If you do some math, you can work out a formula for $p(k\vert w,n)$: the probability of seeing exactly $k$ heads. For our purposes, it doesn’t really matter what that formula is, just that it exists. It’s known as a Binomial distribution, and so is sometimes written $\mathrm{Binomial}(k\vert n,w)$.

Here’s an example of what this looks like with $n=21$ and $w=0.5$.

Naturally enough, if $w=.5$, with $21$ flips, you tend to see around $10-11$ heads. Here’s an example with $w=0.2$. Here, the most common value is $4$, close to $21\times.2=4.2$.

## Decisions, decisions

After observing some coin flips, what do we do next? You can imagine facing various possible situations, but we will use the following:

Our situation: After observing n coin flips, you need to guess “heads” or “tails”, for one final coin flip.

Here, you just need to “decide” what the next flip will be. You could face many other decisions, e.g. guessing the true value of w.

Now, suppose that you have a friend who seems very skilled at predicting the final coinflip. What information would you need to reproduce your friend’s skill? All you need to know is if your friend predicts heads or tails for each possible value of k. We think of this as a decision rule, which we write abstractly as

$\mathrm{Dec}(k).$

This is just a function of one integer $k$. You can think of this as just a list of what guess to make, for each possible observation, for example:

 $k$ $\mathrm{Dec}(k)$ $0$ $\mathrm{tails}$ $1$ $\mathrm{heads}$ $2$ $\mathrm{heads}$ $\vdots$ $\vdots$ $n$ $\mathrm{tails}$

One simple decision rule would be to just predict heads if you saw more heads than tails, i.e. to use

$\mathrm{Dec}(k)=\begin{cases} \mathrm{heads}, & k\geq n/2 \\ \mathrm{tails} & k

The goal of statistics is to find the best decision rule, or at least a good one. The rule above is intuitive, but not necessarily the best. And… wait a second… what does it even mean for one decision rule to be “better” than another?

## Our goal: minimize the thing that’s designed to be minimized

What happens after you make a prediction? Consider our running example. There are many possibilities, but here are two of the simplest:

• Loss A: If you predicted wrong, you lose a dollar. If you predicted correctly, nothing happens.
• Loss B: If you predict “tails” and “heads” comes up, you lose 10 dollars. If you predict “heads” and “tails” comes up, you lose 1 dollar. If you predict correctly, nothing happens.

We abstract these through a concept of a loss function. We write this as

$L(w,d)$.

The first input is the true (unknown) value $w$, while second input is the “prediction” you made. We want the loss to be small.

Now, one point might be confusing. We defined our situation as predicting the next coinflip, but now $L$ is defined comparing $d$ to $w$, not to a new coinflip. We do this because comparing to $w$ gives the most generality. To deal with our situation, just use the average amount of money you’d lose if the true value of the coin were $w$. Take loss A. If you predict “tails”, you’ll be wrong with probability $w$, while if you predict “heads”, you’ll be wrong with probability $1-w$, and so lose $1-w$ dollars on average. This leads to the loss

$L_{A}(w,d)=\begin{cases} w & d=\mathrm{tails}\\ 1-w & d=\mathrm{heads} \end{cases}.$

For loss B, the situation is slightly different, in that you lose 10 times as much in the first case. Thus, the loss is

$L_{B}(w,d)=\begin{cases} 10w & d=\mathrm{tails}\\ 1-w & d=\mathrm{heads} \end{cases}.$

The definition of a loss function might feel circular– we minimize the loss because we defined the loss as the thing that we want to minimize. What’s going on? Well, a statistical problem has two separate parts: a model of the data generating process, and a loss function describing your goals. Neither of these things is determined by the other.

So, the loss function is part of the problem. Statistics wants to give you what you want. But you need to tell statistics what that is.

Despite the name, a “loss” can be negative– you still just want to minimize it. Machine learning, always optimistic, favors “reward” functions that are to be maximized. Plus ça change.

## Model + Loss = Risk

OK! So, we’ve got a model of our data generating process, and we specified some loss function. For a given w, we know the distribution over k, so… I guess… we want to minimize it?

Let’s define the risk to be the average loss that a decision rule gives for a particular value of w. That is,

$R(w,\mathrm{Dec})=\sum_{k}p(k\vert w,n)L(w,\mathrm{Dec}(k)).$

Here, the second input to $R$ is a decision rule– a precise recipe of what decision to make in each possible situation.

Let’s visualize this. As a set of possible decision rules, I will just consider rules that predict “heads” if they’ve seen at least m heads, and “tails” otherwise:

$\mathrm{Dec}_{m}(k)=\begin{cases} \mathrm{heads} & k\geq m\\ \mathrm{tails} & k

With $n=21$ there are $22$ such decision rules, corresponding to $m=0$, (always predict heads), $m=1$ (predict heads if you see at least one heads), up to $m=22$ (always predict tails). These are shown here:

These rules are intuitive: if you’d predict heads after observing 16 heads out of 21, it would be odd to predict tails after seeing 17 instead! It’s true that for losses $L_{A}$ and $L_{B}$, you don’t lose anything by restricting to this kind of decision rule. However, there are losses for which these decision rules are not enough. (Imagine you lose more when your guess is correct.)

With those decision rules in place, we can visualize what risk looks like. Here, I fix $w=0.2$, and I sweep through all the decision rules (by changing $m$) with loss $L_{A}$:

The value $R_A$ in the bottom plot is the total area of the green bars in the middle. You can do the same sweep for $w=0.4$, which you can is pictured here:

We can visualize the risk in one figure with various $w$ and $m$. Notice that the curves for $w=0.2$ and $w=0.4$ are exactly the same as we saw above.

Of course, we get a different risk depending on what loss function we use. If we repeat the whole process using loss $L_{B}$ we get the following:

## Dealing with risk

What’s the point of risk? It tells us how good a decision rule is. We want a decision rule where risk is as low as possible. So you might ask, why not just choose the decision rule that minimizes $R(w,\mathrm{Dec})$?

The answer is: because we don’t know $w$! How do we deal with that? Believe it or not, there isn’t a single well-agreed upon “right” thing to do, and so we meet two different schools of thought.

### Option 1 : All probability all the time

Bayesian statistics (don’t ask about the name) defines a “prior” distribution $p(w)$ over $w$. This says which values of $w$ we think are more and less likely. Then, we define the Bayesian risk as the average of $R$ over the prior:

$R_{\mathrm{Bayes}}(\mathrm{Dec})=\int_{w=0}^{1}p(w)R(w,\mathrm{Dec})dw.$

This just amounts to “averaging” over all the risk curves, weighted by how “probable” we think $w$ is. Here’s the Bayes risk corresponding to $L_{A}$ with a uniform prior $p(w)=1$:

For reference, the risk curves $R(w,\mathrm{Dec}_m)$ are shown in light grey. Naturally enough, for each value of $m$, the Bayes risk is just the average of the regular risks for each $w$.

Here’s the risk corresponding to $L_{B}$:

That’s all quite natural. But we haven’t really searched through all the decision rules, only the simple ones $\mathrm{Dec}_m$. For other losses, these simple ones might not be enough, and there are a lot of decision rules. (Even for this toy problem there are $2^{22}$, since you can output heads or tails for each of $k=0$, $k=1$, …, $k=21$.)

Fortunately, we can get a formula for the best decision rule for any loss. First, re-write the Bayes risk as

$R_{\mathrm{Bayes}}(\mathrm{Dec})=\sum_{k} \left( \int_{w=0}^{1}p(w)p(k\vert n,w)L(w,\mathrm{Dec}(k))dw \right).$

This is a sum over $k$ where each term only depends on a single value $\mathrm{Dec}(k)$. So, we just need to make the best decision for each individual value of $k$ separately. This leads to the Bayes-optimal decision rule of

$\mathrm{Dec}_{\text{Bayes}}(k)=\arg\min_{d}\int_{w=0}^{1}p(w)p(k\vert w,n)L(w,d)dw.$

With a uniform prior $p(w)=1$, here’s the optimal Bayesian decision rules with loss $L_{A}$:

And here it is for loss $L_B$:

Look at that! Just mechanically plugging the loss function into the Bayes-optimal decision rule naturally gives us the behavior we expected– for $L_{B}$, the rule is very hesitant to predict tails, since the loss is so high if you’re wrong. (Again, these happen to fit in the parameterized family $\mathrm{Dec}_{m}$ defined above, but we didn’t use this assumption in deriving the rules.)

The nice thing about the Bayesian approach is that it’s so systematic. No creativity or cleverness is required. If you specify the data generating process ($p(k\vert w,n)$) the loss function ($L(w,d)$) and the prior distribution ($p(w$)) then the optimal Bayesian decision rule is determined.

There are some disadvantages as well:

• Firstly, you need to make up the prior, and if you do a terrible job, you’ll get a poor decision rule. If you have little prior knowledge, this can feel incredibly arbitrary. (Imagine you’re trying to estimate Big G.) Different people can have different priors, and then get different results.
• Actually computing the decision rule requires doing an integral over w, which can be tricky in practice.
• Even if your prior is good, the decision rule is only optimal when averaged over the prior. Suppose, for every day for the next 10,000 years, a random coin is created with $w$ drawn from $p(w)$. Then, no decision rule will incur less loss than $\mathrm{Dec}_{\text{Bayes}}$. However, on any particular day, some other decision rule could certainly be better.

So, if you have little idea of your prior, and/or you’re only making a single decision, you might not find much comfort in the Bayesian guarantee.

Some argue that these aren’t really disadvantages. Prediction is impossible without some assumptions, and priors are upfront and explicit. And no method can be optimal for every single day. If you just can’t handle that the risk isn’t optimal for each individual trial, then… maybe go for a walk or something?

### Option 2 : Be pessimistic

Frequentist statistics (Why “frequentist”? Don’t think about it!) often takes a different path. Instead of defining a prior over w, let’s take a worst-case view. Let’s define the worst-case risk as

$R_{\mathrm{worst}}(\mathrm{Dec})=\max_{w}R(w,\mathrm{Dec}).$

Then, we’d like to choose an estimator to minimize the worst-case risk. We call this a “minimax” estimator since we minimize the max (worst-case) risk.

Let’s visualize this with our running example and $L_{A}$:

As you can see, for each individual decision rule, it searches over the space of parameters $w$ to find the worst case. We can visualize the risk with $L_{B}$ as:

What’s the corresponding minimax decision rule? This is a little tricky to deal with– to see why, let’s expand the worst-case risk a bit more:

$R_{\mathrm{Worst}}(\mathrm{Dec})=\max_{w}\sum_{k}p(k\vert n,w)L(w,\mathrm{Dec}(k)).$

Unfortunately, we can’t interchange the max and the sum, like we did with the integral and the sum for Bayesian decision rules. This makes it more difficult to write down a closed-form solution. At least in this case, we can still find the best decision rule by searching over our simple rules $\mathrm{Dec}_m$. But be very mindful that this doesn’t work in general!

For $L_{A}$ we end up with the same decision rule as when minimizing Bayesian risk:

For $L_{B}$, meanwhile, we get something slightly different:

This is even more conservative than the Bayesian decision rule. $\mathrm{Dec}_{B-\mathrm{Bayes}}(2)=\mathrm{tails}$, while $\mathrm{Dec}_{B-\mathrm{minimax}}(2)=\mathrm{heads}$. That is, the Bayesian method predicts heads when it observes 2 or more, while the minimax rule predicts heads if it observes even one. This makes sense intuitively: The minimax decision rule proceeds as if the “worst” w (a small number) is fixed, whereas the Bayesian decision rule less pessimistically averages over all w.

Which decision rule will work better? Well, if w happens to be near the worst-case value, the minimax rule will be better. If you repeat the whole experiment many times with w drawn from the prior, the Bayesian decision rule will be.

If you do the experiment at some w far from the worst-case value, or you repeat the experiment many times with w drawn from a distribution different from your prior, then you have no guarantees.

Neither approach is “better” than the other, they just provide different guarantees. You need to choose what guarantee you want. (You can kind of think of this as a “meta” loss.)

## So what about all those formulas, then?

For real problems, the data generating process is usually much more complex than a Binomial. The “decision” is usually more complex than predicting a coinflip– the most common decision is making a guess for the value of $w$. Even calculating $R(w,\mathrm{Dec})$ for fixed $w$ and $\mathrm{Dec}$ is often computationally hard, since you need to integrate over all possible observations. In general, finding exact Bayes or minimax optimal decision rules is a huge computational challenge, and at least some degree of approximation is required. That’s the game, that’s why statistics is hard. Still, even for complex situations the rules are the same– you win by finding a decision rule with low risk.

# Linear Classifiers and Loss Functions

A linear classifier is probably the simplest machine learning technique. Given some input vector $\bf x$, predict some output $y$. One trains a vector of “weights” $\bf w$, that determine the behavior of the classifier. Given some new input $\bf x$, the prediction $f$ will be:

$f({\bf x}) = I[{\bf w}^T {\bf x}>0]$

Here, $I[\text{expr}]$ is the indicator function– $1$ if $\text{expr}$ is true, and -1 otherwise. So, $y$ is one of two classes.

Often the classifier will be written like $I[{\bf w}^T {\bf x}+b>0]$, i.e. including a “bias” term $b$. Here I’ll ignore this– if you want a bias term, just append a $1$ to every vector $\bf x$.

Now, it seems like nothing could be simpler than a linear classifier, but there is considerable subtlety in how to train a linear classifier. Given some training data $\{({\hat {\bf x}},{\hat y})\}$, how do we find the vector of weights $\bf w$?

There are various popular ways of doing this. Three popular techniques are:

• The Perceptron Algorithm
• Logistic Regression
• Support Vector Machines (Hinge loss)

The perceptron algorithm works, roughly speaking as follows: Go through the training data, element by element. If one datum is misclassified, slightly nudge the weights so that it is closer to being correct. This algorithm has the remarkable property that, if the training data is linearly separable, it will find a $\bf w$ that correctly classifies it in a finite number of iterations. (The number of iterations that it takes depends on the “how separable” the data is, making fewer mistakes on data with a larger margin.)

Unfortunately, the assumption that the training data is linearly separable is basically never true in practice. There is still a bound in terms of how much points would need to be moved in order to be linearly separable, but it isn’t as strong as we might like. (I don’t know what the folklore is for how this tends to work in practice.)

Logistic regression and support vector machines take a different approach. There, a “loss function” is defined in terms of the weights, and one just optimizes the loss to find the best weights. (I’m ignoring regularization here for simplicity.) For logistic regression, the loss is

$L = -\sum_{({\hat {\bf x}},{\hat y})} \log p({\hat y}|{\hat {\bf x}})$

where $p(y=1|{\bf x})=\sigma({\bf w}^T {\bf x})$, for a sigmoid function $\sigma$, and $p(y=-1|{\bf x})=1-p(y=1|{\bf x})$.

For a support vector machine, the loss is

$L = \sum_{({\hat {\bf x}},{\hat y})} (1-{\hat y} \cdot {\bf w}^T {\hat {\bf x}})_+$

where $(a)_+$ is $a$ if $a>0$ and 0 otherwise. This is a hinge loss. Notice that it will be zero if ${\hat y} \cdot {\bf w}^T {\hat {\bf x}} > 1$, or if that particular training element is comfortably on the correct side of the decision boundary. Otherwise, the “pain” is proportional to “how wrong” the classification is.

Both the hinge-loss and logistic regression are convex loss functions. This means that if we apply a nonlinear search procedure to find a local minima, that will also be the global minima.

These different loss functions are compared here.

WHY NOT DO THE OBVIOUS THING?

Now, the critical point about these three above methods is that none of them do the seemingly obvious thing: find the vector of weights ${\bf w}$ that has the lowest classification error on the training data. Why not? Some would defend logistic regression or SVM for theoretical reasons (namely a meaningful probabilistic interpretation, and the reasonableness and theoretical guarantees of max-margin methods, respectively).

However, probably the more significant hurdle is computational considerations. Namely, the problem of finding the weights with lowest classification error is (in order of increasing horribleness) non-convex, non-differentiable, and NP-hard.

In fact it is NP-hard even to find an approximate solution, with worst-case guarantees. (See citation 2 below, which, interestingly, gives an approximate algorithm in terms of a property of the data.)

Nevertheless, if classification error is what we want, I don’t see how it makes sense to minimize some alternative loss function. As such, I decided today to try the following algorithm.

1. Apply an SVM to get an initial solution.
2. Apply heuristic search to minimize classification error, initialized to the solution of step 1.

Now, I am well aware of the problems with non-convex optimization. However, simply the fact that logistic regression or hinge loss is convex is not an argument in their favor. If theoretical considerations dictate that we minimize classification error, just substituting a different loss, and then refusing to look at the classification rate is highly questionable. Sure, that can lead to a convex optimization problem, but at the that’s because a different problem is being solved! The goal is accurate prediction of future data, not accurate minimization of a loss on the training data. If we use our best convex approximation to initialize the heuristic optimization of the loss we really want, we will never do worse.

EXPERIMENT

There are some heuristics available for doing this (Reference 3), but they seem a little expensive. I decided to try something really simple.

1. Fit a classifier (by hinge loss or logistic regression).
2. Throw out a few of the most badly misclassified points.
3. Repeat until all remaining points are correctly classified.

The idea is that, by “giving up” on badly misclassified points, the boundary might be movable into a position where it correctly classifies other points. There is no guarantee in general that this will find a linear classifier with a lower misclassification rate, but it should not do worse.

To test this out, I got some data from the MNIST handwritten digit database. To make the problem harder, I took one class to be random images from either the set of 1’s or 2’s, while the other class was 3’s or 4’s. The images were downsampled to 7×7, and a constant of one added. So we are looking for a linear classifier in a 50 dimensional space.

The data was trained on a database of 10,000 examples, with a test set of 10,000 examples.

Here are the results where we throw out the worst 100 training points in step 2:

Here are the results when we throw out 20 at a time:

Here are the results when we throw out 5 at a time:

There seems to be a trade-off in step 2: Fewer models need to be fit if we throw out more points at a time. However, this seems to come with a small penalty in terms of terms of the classification error on the final model.

So, in summary– a drop in classification error on test data from .941 to .078. Thats a 17% drop. (Or a 21% drop, depending upon which rate you use as a base.) This from a method that you can implement in basically zero extra work if you already have a linear classifier. Seems worth a try.

References:

[2] Efficient Learning of Linear Perceptrons by Shai Ben-david, Hans Ulrich Simo

[3] Optimizing 0/1 Loss for Perceptrons by Random Coordinate Descent by L. Li and H.-T. Lin

# Marginal Beliefs of Random MRFs

A pairwise Markov Random Field is a way of defining a probability distribution over some vector ${\bf x}$. One way to write one is

$p({\bf x}) \propto \exp( \sum_i \phi(x_i) + \sum_{(i,j)} \psi(x_i,x_j) )$.

Where the first sum is over all the variables, and the second sum is over neighboring pairs. Here, I generated some random distributions over binary valued variables. For each $i$, I set $\phi(x_i=0)=0$, and $\phi(x_i=1)=r_i$ where $r_i$ is some value randomly chosen from a standard Gaussian. For the pairwise terms, I used $\psi(x_i,x_j) = .75 \cdot I(x_i=x_j)$. (i.e. $\psi(x_i,x_j)$ is .75 when the arguments are the same, and zero otherwise.) This is an “attractive network”, where neighboring variables want to have the same value.

Computing marginals $p(x_i)$ is hard in graphs that are not treelike. Here, I approximate them using a nonlinear minimization of a “free energy” similar to that used in loopy belief propagation.

Here, I show the random single-variate biases $r_i$ and the resulting beliefs.  What we see is constant valued regions (encouraged by $\psi$) interrupted where the $\phi$ is very strong.

Now, with more variables.

Now, a “repellent” network. I repeated the procedure above, but changed the pairwise interactions to $\psi(x_i,x_j) = -.75 \cdot I(x_i\not=x_j)$. Neighboring variables want to have different values.  Notice this is the opposite of the above behavior– regions of “checkerboard” interrupted where the $\phi$ outvotes $\psi$.

Now, the repellent network with more variables.