# Three opinions on theorems

## 1. Think of theorem statements like an API.

Some people feel intimidated by the prospect of putting a “theorem” into their papers. They feel that their results aren’t “deep” enough to justify this. Instead, they give the derivation and result inline as part of the normal text.

Sometimes that’s best. However, the purpose of a theorem is not to shout to the world that you’ve discovered something incredible. Rather, theorems are best thought of as an “API for ideas”. There are two basic purposes:

• To separate what you are claiming from your argument for that claim.
• To provide modularity to make it easier to understand or re-use your ideas.

To decide if you should create a theorem, ask if these goals will be advanced by doing so.

A thoughtful API makes software easier to use: The goal is to allow the user as much functionality as possible with as simple an interface as possible, while isolating implementation details. If you have a long chain a mathematical argument, you should choose what parts to write as theorems/lemmas in much the same way.

Many paper intermingle definitions, assumptions, proof arguments, and the final results. Have pity on the reader, and tell them in a single place what you are claiming, and under what assumptions. The “proof” section separates your evidence for your claim from the claim itself. Most readers want to understand your result before looking at the proof.  Let them. Don’t make them hunt to figure out what your final result it.

Perhaps controversially, I suggest you should use the above reasoning even if you aren’t being fully mathematically rigorous. It’s still kinder to the reader to state your assumptions informally.

As an example of why it’s helpful to explicitly state your results, here’s an example from a seminal paper, so I’m sure the authors don’t mind. (Click on the image for a larger excerpt.)

This proof is well written. The problem is many small uncertainties that accumulate as you read it. If you try to understand exactly:

• What result is being stated?
• Under what assumptions does that result hold?

You will find that the proof “bleeds in” to the result itself. The convergence rate in 2.13 involves $Q(\theta)$ defined in 2.10, which itself involves other assumptions tracing backwards in the paper.

Of course, not every single claim needs to be written as a theorem/lemma/claim. If your result is simple to state and will only be used in a “one-off” manner, it may be clearer just to leave it in the text. That’s analogous to “inlining” a small function instead of creating another one.

## 2. Don’t fear the giant equation block.

I sometimes see a proof like this (for $0)

Take the quantity

$\frac{x-x^2}{\sqrt{x^2-2x+1}}.$

Pulling out $x$ this becomes

$x \frac{1-x}{\sqrt{x^2-2x+1}}.$

Factoring the denominator, this is

$x \frac{1-x}{\sqrt{(x-1)(x-1)}}.$

Etc.

For some proofs, the text between each line just isn’t that helpful. To a large degree it makes things more confusing– without an equality between the lines, you need to read the words to understand how each formula is supposed to be related to the previous one. Consider this alternative version of the proof:

\begin{aligned} \frac{x-x^2}{\sqrt{x^2-2x+1}} = & x \frac{1-x}{\sqrt{x^2-2x+1}} \\ = & x \frac{1-x}{\sqrt{(x-1)(x-1)}} \\ = & x \frac{1-x}{\sqrt{(x-1)^2}} \\ = & x \frac{1-x}{1-x} \\ = & x \\ \end{aligned}

In some cases, this reveals the overall structure of the proof better than a bunch of lines with interspersed text. If explanation is needed, it can be better to put it at the end, e.g. as “where line 2 follows from [blah] and line 3 follows from [blah]”.

It can also be helpful to put these explanations inline, i.e. to us a proof like

\begin{aligned} \frac{x-x^2}{\sqrt{x^2-2x+1}} = & x \frac{1-x}{\sqrt{x^2-2x+1}} & \text{ pull out x} \\ = & x \frac{1-x}{\sqrt{(x-1)(x-1)}} & \text{ factoring denominator} \\ = & x \frac{1-x}{\sqrt{(x-1)^2}} & \\ = & x \frac{1-x}{1-x} & \text{ since denominator is positive} \\ = & x & \\ \end{aligned}

Again, this is not the best solution for all (or even most) cases, but I think it should be used more often than it is.

## 3. Use equivalence of inequalities when possible.

Many proofs involve manipulating chains of inequalities. When doing so, it should be obvious at what steps extra looseness may have been introduced. Suppose you have some positive constants $a$ and $c$ with $a^2>c$ and you need to choose $b$ so as to ensure that $b^2+c \leq a^2$.

People will often prove a result like the following:

Lemma: If $b \leq \sqrt{a^2-c}$, then $b^2+c \leq a^2$.

Proof: Under the stated condition, we have that

\begin{aligned} b^2 + c & \leq & (\sqrt{a^2-c})^2+c \\ & = & a^2-c+c \\ & = & a^2 \end{aligned}

That’s all correct, but doesn’t something feel slightly “magical” about the proof?

There are two problems: First, the proof reveals nothing anything about how you came up with the final answer. Second, the result leaves ambiguous if you have introduced additional looseness. Given the starting assumption, is it possible to prove a stronger bound?

I think the following lemma and proof are much better:

Lemma: $b^2+c \leq a^2$ if and only if $b \leq \sqrt{a^2-c}$.

Proof: The following conditions are all equivalent:

\begin{aligned} b^2+c & \leq & a^2 \\ b^2 & \leq & a^2-c \\ b & \leq & \sqrt{a^2-c}. \\ \end{aligned}

The proof shows exactly how you arrived at the final result, and shows that there is no extra looseness. It’s better not to “pull a rabbit out of a hat” in a proof if not necessary.

This is arguably one of the most basic possible proof techniques, but is bizarrely underused. I think there’s two reasons why:

1. Whatever need motivated the lemma is probably met by the first one above. The benefit of the second is mostly in providing more insight.
2. Mathematical notation doesn’t encourage it. The sentence at the beginning of the proof is essential. If you see this merely as a series of inequalities, each implied by the one before, than it will not give the “only if” part of the result. You could conceivably try to write something like $a < b \Leftrightarrow \exp a < \exp b$, but this is awkward with multiple lines.

# Exponential Families Cheatsheet

As part of the graphical models course I taught last spring, I developed a “cheatsheet” for exponential families. The basic purpose is to explain the standard moment-matching condition of maximum likelihood. The goal of the sheet was to clearly show how this property generalized to maximum likelihood in conditional exponential families, with hidden variables, or both. It’s available as an image below, or as a PDF here. Please let me know about any errors!

I use the (surprisingly controversial) convention of using a sans-serif font for random variables, rather than capital letters. I’m convinced this is the least-bad option for the machine learning literature, where many readers seem to find capital letter random variables distracting. It also allows you to distinguish matrix-valued random variables, though that isn’t used here.

# 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.

# The second and third best features of LyX you aren’t using.

LyX is a WYSIWYG editor for latex files. It’s a little bit clunky to use at first, and isn’t perfect (thank you, open source developers– I’m not ungrateful!) but after becoming familiar with it, it’s probably the single piece of software that has most improved my productivity. I like it so much I use it not just for papers but often as a scratchpad for math. Many times during random discussions I’ve used it to quickly bang out some equations and after seeing how fast it was, others immediately switched to using it.

Anyway, while macros may be the best feature you LyX aren’t using, I recently discovered another couple excellent ones I wasn’t familiar with after years of use so I thought I’d publicize. Specifically, I’ve always hated the process of including explanatory figures into LyX. Exporting plots from an experiment is tricky to improve, but when trying to create explanatory graphs, I’ve always hated the process.

Before, I thought the options were.

1) Create the graph in an external program. This is fine, of course, but is quite inconvenient when you want to go back and revise it. The external program usually saves it in some other format, so you have to open the graphic in open it again, revise it, export it to .pdf (or whatever), then open the document in LyX, compile it. Then, when you don’t like the way it looks, you have to repeat the whole process. It works, but it’s not efficient, since you can’t edit the content in place. (Which is the whole point of using a WYSIWYG editor in the first place– remove the need for thinking about anything but content.)

2) Write the graphics directly in LyX in a language like TikZ. This is more “in-place” in that you don’t have external files to find and manipulate. However, I find TikZ to be quite painful to get right with many re-compilations necessary. If the TikZ document is in place each requires a full compilation of the document. This is hilariously slow when making something like Beamer slides. Further, this totally violates the whole point of WYSIWYG since you’re looking at code, rather than the output.

There are better ways! I’ve wasted countless hours not being aware of these.

## Preview Boxes

First, LyX has a beautiful feature of “preview boxes”. Take the following very simple TikZ code, which just draws a square:

 \begin{tikzpicture} \draw[red] (0,0) -- (0,1); \draw[green] (0,1) -- (1,1); \draw[red] (1,1) -- (1,0); \draw[blue] (1,0) -- (0,0); \end{tikzpicture} 

Typically, I’d include this in LyX files by inserting a raw tex box:

And then putting the the TikZ code inside:

This is OK, but has the disadvantages from (2) above. The code can be huge, if I have a lot of graphics, I can’t tell what corresponds to what, and I have to do a (slooooooow) recompile of the whole document to see what it looks like.

However, if you just add a “preview box”:

You get something that looks like this:

So far, so pointless, right? However, when you deselect, LyX shows the graphic in-place:

You can then click on it to expand the code. This solves most of the problems: You can see what you are doing at a glance, and you don’t need to recompile the whole document to do it.

## Native SVG support

Newer versions of LyX also natively support SVG files. You first have to create the file externally using something like Inkscape, which itself saves directly to the SVG format. Then, you can include it in LyX by doing Insert->File->External Material:

And then selecting the SVG file:

Again, LyX will show it in-place and (if LyX is configured correctly…) correctly output vector graphics in the final document.

What’s even better is that LyX can automatically open the file in the external editor for you. If you right click, you can “edit externally”:

Then the external editor will automatically open the file. You can then save it with a keystroke and go back to LyX. No hunting around for the file, no cycles of exporting to other formats, and you see exactly what the final output will look like at all stages. You can really tell that LyX was created by people using it themselves.

## Bonus feature

This one is described well-enough already, but helps a lot in big documents: you can click on a point in a generated .pdf and automatically have LyX sync the editor to the corresponding point in the file.

# You deserve better than two-sided finite differences

In calc 101, the derivative is derived as $df/dx = \lim_{\epsilon \rightarrow 0} (f(x+\epsilon)-f(x))/\epsilon$. So, if you want to estimate a derivative, an easy way to do so would be to just pick some small $\epsilon$ and estimate:

$df/dx \approx \frac{1}{\epsilon} (f(x+\epsilon)-f(x))$

This can work OK. Let’s look at an example of trying to calculate the derivative of $\log (x)$, using a range of different $\epsilon$

What’s happening? Well, the result is true mathematically in the limit that $\epsilon$ is small, so it’s natural to get errors for large $\epsilon$. However, with very small $\epsilon$ you run into trouble because floating-point arithmetic can only represent finite precision. Let’s try again with a smaller value of $x=.001$

That’s somewhat concerning. We can still get a nearly correct value, but we have a limited range of steps that achieve it. This is very well-known in numerical analysis, and a common solution is to use two-sided differences, i.e. to estimate

$df/dx \approx \frac{1}{2 \epsilon} (f(x+\epsilon)-f(x-\epsilon)).$

If we try that, we indeed get better results:

What’s happening here? Basically, we are using more information about the function to make a higher-order approximation, so we can mathematically get away with using a larger value of $\epsilon$. This in turn isis helpful to avoid the numerical precision demons.

Great! But let’s go deeper. If we use $x = 10^{-5}$, we get:

Huh. 1-sided differences totally fall apart, but we seem to be running into more trouble. But never fear, you can use “four-sided differences”!

$df/dx \approx \frac{1}{12 \epsilon} (-f(x+2\epsilon)+8 f(x+\epsilon)-8 f(x+\epsilon)+ f(x-2 \epsilon)$

Then, we get what you might expect:

But what if we go deeper, with $x=10^{-7}$?

Or maybe we should go even deeper, with $x = 10^{-9}$?

Even the four-sided differences have failed us. Now, you might take a lesson here that you shouldn’t be using numerical differences (and I sort of agree). Those of certain temperament, on the other hand, would say instead that what we need is power. OK then, how do six-sided differences sound to you? Sound good?

$df/dx \approx \frac{1}{60 \epsilon}(- f(x-3 \epsilon)+9 f(x-2 \epsilon)-45 f(x-\epsilon)+45 f(x+\epsilon)-9 f(x+2\epsilon)+f(x+3 \epsilon))$

This is still tough, but there is at least some range of epsilon where you can calculate a reasonable derivative. There is, of course, a never-ended sequence of these higher-order derivative approximations. There’s even a calculator, for all your bespoke finite-difference stencil needs.

Now, you might think that the real solution here is to use automatic differentiation, and you’re mostly right. It takes more computation to sample the function at more points, and no number of samples will fundamentally stop the numerical demons from destroying your result.

However, there still remain cases where it’s worthwhile to manually compute a derivative. More importantly perhaps, when you’re implementing an automatic differentiation tool, you still need to test that it is actually correct! Here, numerically differences will probably forever remain useful. So, certainly for the cases of building a test suite for autodiff code, it certainly makes sense to use these higher-order derivatives.

(I originally intended to leave this as a comment on Tim Viera’s post on testing gradient implementations, but this kinda got out of control.)

# Sneaking up on Bayesian Inference (A fable in four acts)

Act 1: Magical Monkeys

Two monkeys, Alfred (${a}$) and Betty (${b}$) live in a parallel universe with two kinds of blocks, green (${g}$) and yellow (${y}$). Alfred likes green blocks, and Betty prefers the yellow blocks. One day, a Wizard decides to give one of the monkeys the magical power to send one block over to our universe each day.

The Wizard chooses the magical monkey by flipping a fair four-sided die. He casts a spell on Alfred if the outcome is 1, and Betty if the outcome is ${2-4}$. That is, the Wizard chooses Alfred with probability ${.25}$ and Betty with probability ${.75}$. Both Alfred and Betty send their favorite colored block with probability ${.8}$

After the Wizard has chosen, we see the first magical block, and it is green. Our problem is: What is the probability that Alfred is the magical monkey?

Intuitively speaking, we have two somewhat contradictory pieces of information. We know that the Wizard chooses Betty more often. But green is Alfred’s preferred color. Given probabilities above, is Alfred or Betty more probable?

First, we can write down all the probabilities. These are

\displaystyle \begin{aligned}Pr(M=a) & =.25\\ Pr(M=b) & =.75\\ Pr(B=g\vert M=a) & =.8\\ Pr(B=y\vert M=a) & =.2\\ Pr(B=g\vert M=b) & =.2\\ Pr(B=y\vert M=b) & =.8. \end{aligned}

Now, the quantity we are interested in is ${Pr(M=a\vert B=g).}$ We can mechanically apply Bayes’ rule to obtain

\displaystyle \begin{aligned}Pr(M=a\vert B=g) & =\frac{1}{Pr(B=g)}Pr(M=a)Pr(B=g\vert M=a)\\ & =\frac{1}{Pr(B=g)}.25\times.8. \end{aligned}

Similarly, we can calculate that

$\displaystyle Pr(M=b\vert B=g)=\frac{1}{Pr(B=g)}.75\times.2.$

But, of course, we know that one of the monkeys is magical, so these quantities must sum to one. Thus (since they both involve the quantity ${Pr(B=g)}$ that we haven’t explicitly calculated) we can normalize to obtain that

$\displaystyle Pr(M=a\vert B=g)=\frac{.25\times.8}{.25\times.8+.75\times.2}=\frac{4}{7}.$

Act 2: Magical Monkeys on Multiple Days

We wait around two more days, and two more blocks appear, both of which are yellow. Assume that the monkeys make an independent choice each day to send their favorite or less favorite block. Now, what is the probability that Alfred is the magical monkey, given that ${B=gyy}$?

Intuitively speaking, what do we expect? Betty is more likely to be chosen, and 2/3 of the blocks we’ve seen are suggestive of Betty (since she prefers yellow).

Now, we can mechanically calculate the probabilities. This is just like before, except we use the fact that the blocks seen on each day are conditionally independent given a particular monkey.

\displaystyle \begin{aligned}Pr(M=a\vert B=gyy) & =\frac{1}{Pr(B=gyy)}Pr(M=a)Pr(B=gyy\vert M=a)\\ & =\frac{1}{Pr(B=gyy)}Pr(M=a)Pr(B=g\vert M=a)Pr(B=y\vert M=a)Pr(B=y\vert M=a)\\ & =\frac{1}{Pr(B=gyy)}.25\times.8\times.2\times.2\\ Pr(M=b\vert B=gyy) & =\frac{1}{Pr(B=gyy)}Pr(M=b)Pr(B=g\vert M=b)Pr(B=y\vert M=b)Pr(B=y\vert M=b)\\ & =\frac{1}{Pr(B=gyy)}.75\times.2\times.8\times.8 \end{aligned}

Again, we know that these two probabilities sum to one. Thus, we can normalize and get that

$\displaystyle Pr(M=a\vert B=gyy)=\frac{.25\times.8\times.2\times.2}{.25\times.8\times.2\times.2+.75\times.2\times.8\times.8}=\frac{1}{13}.$

So now, Betty looks more likely by far.

Act 3: Magical Monkeys and the Weather

Now, suppose the Wizard gives us some additional information. The magical monkey (whilst relaxing over in the other universe) is able to perceive our weather, which is either clear ${(c)}$ or rainy ${(r)}$. Both the monkeys prefer clear weather. When the weather is clear, they send their preferred block with probability ${.8}$, while if the weather is rainy, they angrily send their preferred block with probability ${.2.}$ That is, we have

\displaystyle \begin{aligned}Pr(B=g\vert M=a,W=c) & =.8\\ Pr(B=g\vert M=a,W=r) & =.2\\ Pr(B=g\vert M=b,W=c) & =.2\\ Pr(B=g\vert M=b,W=r) & =.8. \end{aligned}

Along with seeing the previous sequence of blocks ${B=gyy}$, we observed the weather sequence ${W=rrc}$. Now, what is the probability that Alfred is the magical monkey?

We ask the Wizard what the distribution over rainy and clear weather is. The wizard haughtily responds that this is irrelevant, but does confirm that the weather is independent of which monkey was made magical.

Can we do anything without knowing the distribution over the weather? Can we calculate the probability that Alfred is the magical monkey?

Let’s give it a try. We can apply Bayes’ equation to get that

\displaystyle \begin{aligned}Pr(M=a\vert B=gyy,W=rrc) & =\frac{1}{Pr(B=gyy,W=rrc)}Pr(M=a)Pr(B=gyy,W=rrc\vert M=a).\end{aligned}

Now, since the weather is independent of the monkey, we know that

\displaystyle \begin{aligned}Pr(B=gyy,W=rrc\vert M=a) & =Pr(W=rrc\vert M=a)Pr(B=gyy\vert M=a,W=rrc)\\ & =Pr(W=rrc)Pr(B=gyy\vert M=a,W=rrc). \end{aligned}

Thus, we have that

\displaystyle \begin{aligned}Pr(M=a\vert B=gyy,W=rrc) =&\frac{Pr(W=rrc)}{Pr(B=gyy,W=rrc)}Pr(M=a)Pr(B=gyy\vert M=a,W=rrc)\\ =&\frac{Pr(W=rrc)}{Pr(B=gyy,W=rrc)}Pr(M=a) \\ & \times Pr(B=g\vert M=a,W=r)Pr(B=y\vert M=a,W=r)Pr(B=y\vert M=a,W=c)\\ =&\frac{Pr(W=rrc)}{Pr(B=gyy,W=rrc)}.25\times.2\times.8\times.8. \end{aligned}

Through the same logic we can calculate that

$\displaystyle Pr(M=b\vert B=gyy,W=rrc)=\frac{Pr(W=rrc)}{Pr(B=gyy,W=rrc)}.75\times.8\times.2\times.2.$

Again, we know that these probabilities need to sum to one. Since the factor ${Pr(W=rrc)/Pr(B=gyy,W=rrc)}$ is constant between the two, we can normalize it out. Thus, we get that

$\displaystyle Pr(M=a\vert B=gyy,W=rrc)=\frac{.25\times.2\times.8\times.8.}{.25\times.2\times.8\times.8.+.75\times.8\times.2\times.2.}=\frac{4}{7}.$

Again, it looks like Alfred was the more likely monkey. And– oh wait– we somehow got away with not knowing the distribution over the weather…

Act 4: Predicting the next block

Now, suppose that after seeing the previous sequences (namely ${B=gyy}$ and ${W=rrc}$) , on the fourth day, we find that it has rained. What is the probability that we will get a green block on the fourth day? Mathematically, we want to calculate

$\displaystyle Pr(B'=g\vert W'=r,B=ggy,W=rrc).$

How can we go about this? Well, if we knew that the magical monkey was Alfred (which we do not!), it would be easy to calculate. We would just have

$\displaystyle Pr(B'=g\vert W'=r,M=a),$

and similarly if the magical monkey were Betty. Now, we don’t know which monkey is magical, but we know the probabilities that each monkeys is magical given the available information– we just calculated them in the previous section! So, we can factorize the distribution of interest as

\displaystyle \begin{aligned}Pr(B'=g\vert W'=r,B=ggy,W=rrc) & =Pr(B'=g\vert W'=r,M=a)Pr(M=a\vert B=ggy,W=rrc)\\ & +Pr(B'=g\vert W'=r,M=b)Pr(M=b\vert B=ggy,W=rrc)\\ & =.2\times\frac{4}{7}+.8\times\frac{3}{7}\\ & =\frac{33}{70} \end{aligned}

So we are slightly less likely than even to see a green block on the next day.

Epilogue: That’s Bayesian Inference

This is how Bayesian inference works (in the simplest possible setting). In general Bayesian inference, you have:

• A set of possible inputs (for us these are weather patterns)
• A set of possible outputs (for us these are colored blocks)
• A set of possible models, that map a given input to a distribution over the outputs (for us these are monkeys)
• A “prior” distribution over which models are more likely (for us, this is the wizard and the four-sided die)
• A dataset of inputs and outputs (for us, this is ${W=rrc}$ and ${B=ggy}$)
• A new input (for us this is ${W'=r}$)
• An unknown output that we’d like to predict (for us this is ${B'}$)

Bayesian inference essentially proceeds in two steps:

• In the first step, one uses the prior over which models are more likely along with the dataset of inputs and outputs to build a “posterior” distribution over which models are more likely. We did this by using

$\displaystyle Pr(M\vert W,B)\propto Pr(M)\prod_{i}Pr(B_{i}\vert W_{i},M),$

where ${i}$ indexes the different days of observation.

• In the second step, one “integrates over the posterior”. That is, each model gives us a distribution over ${B'}$ given the new input ${W'}$. So, we simply sum over the different models

$\displaystyle Pr(B'\vert W')=\sum_{m}Pr(M=m\vert W,B)Pr(B'\vert W',M=m).$

In the general case, the set of possible models is much larger (typically infinite) and more complex computational methods need to be used to integrate over it. (Commonly Markov chain Monte Carlo or variational methods). Also, of course, we don’t typically have a Wizard telling us exactly what the prior distribution over the models is, meaning one must make assumptions, or try to “learn” the prior as in, say, empirical Bayesian methods.