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Fixing Random, part 37

Last time on FAIC we finally wrote a tiny handful of lines of code to correctly implement importance sampling; if we have a distribution p that we’re sampling from, and a function f that we’re running those samples through, we can compute the expected value of f even if there are “black swan” regions in p. All we need is a helper distribution q that has the same support as p, but no black swans.

Great. How are we going to find that?

A variation on this problem has come up before this series: what should the initial and proposal distributions be when using Metropolis? If we’re using Metropolis to compute a posterior from a prior then we can use the prior as the initial distribution. But it’s not at all clear in general how to choose a high-quality proposal distribution; there’s some art there.

There is also some art in choosing appropriate helper distributions when doing importance sampling. Let’s once again take a look at our “black swan” situation:

Screen Shot 2019-05-21 at 9.17.06 AM

As we’ve discussed, I contrived the “black swan” situation by ensuring that there was a region of the graph where the orange line bounded a large area, but the blue line bounded a very tiny area there.

First off: in my initial description of the problem I made the assumption that I only cared about the function on the range of 0.0 to 1.0. If you know ahead of time that there is a specific range of interest, you can always use a uniform distribution over that range as a good guess at a helper distribution. As we saw in a previous episode, doing so here gave good results. Sure, we spend a lot of time sampling a region of low area, but we could tweak that to exclude the region between 0.0 and 0.3.

What if we don’t know the region of interest ahead of time though? What if the PDF and the function we’re evaluating are defined over the whole real line and we’re not sure where to put a uniform distribution? Let’s think about some quick-and-dirty hacks for solving that problem.

What if we “stretched” the blue line a little bit around 0.75, and “squashed” it down a little bit?

Screen Shot 2019-05-23 at 2.39.51 PM.png

The light blue line is not perfect by any means, but we are now likely to get at least a few samples from the part of the orange line that has large negative area.

Since the original distribution is just a normal, we can easily make this helper distribution by increasing the standard deviation. (Code for this episode can be found here.)

var p = Normal.Distribution(0.75, 0.09);
var p2 = Normal.Distribution(0.75, 0.15);
double f(double x) => Atan(1000 * (x  .45)) * 20  31.2;
for (int i = 0; i < 10; ++i)
  Console.WriteLine(
    $”{p.ExpectedValueByImportance(f, 1.0, p2):0.###});

0.119, 0.114, 0.108, 0.117, 0.116, 0.108, 0.121, ...

Recall that the correct value is 0.113. This is not perfect, but it’s a lot better than the original.

Now, it is all well and good to say that we know how to solve the problem when the nominal distribution is normal; what if it is some arbitrary distribution and we want to “stretch” it?

That’s actually pretty straightforward. Let’s suppose we want to stretch a distribution around a particular center, and, optionally, also move it to the left or right on the real line. Here’s the code:

public sealed class Stretch : IWeightedDistribution<double>
{
  private IWeightedDistribution<double> d;
  private double shift;
  private double stretch;
  public static IWeightedDistribution<double> Distribution(
    IWeightedDistribution<double> d,
    double stretch,
    double shift = 0.0,
    double around = 0.0)
  {
    if (stretch == 1.0 && shift == 0.0) return d;
    return new Stretch(
      d, stretch, shift + around  around * stretch);
  }
  private Stretch(
    IWeightedDistribution<double> d,
    double stretch,
    double shift)
  {
    this.d = d;
    this.stretch = stretch;
    this.shift = shift;
  }
  public double Sample() =>
    d.Sample() * stretch + shift;
  // Dividing the weight by stretch preserves
  // the normalization constant 
  public double Weight(double x) =>
    d.Weight((x  shift) / stretch) / stretch;
}

And now we can get similar results:

var p = Normal.Distribution(0.75, 0.09);
var ps = Stretch.Distribution(p, 2.0, 0, 0.75);
double f(double x) => Atan(1000 * (x  .45)) * 20  31.2;
for (int i = 0; i < 10; ++i)
  Console.WriteLine(
    $”{p.ExpectedValueByImportance(f, 1.0, ps):0.###});

0.113, 0.117, 0.121, 0.124, 0.12 ...

Again, not perfect, but we’re getting at least reasonable results here.

But like I said before, these are both “more art than science” techniques; they are useful if you have a particular distribution and a particular function, and you’re looking for an expected value, and you’re willing to spend some time writing different programs and trying out different techniques and parameters to tweak it to get good results. We still have not got an algorithm where a distribution and a function go in, and an accurate estimate of the expected value comes out.


Next time on FAIC: I’ll make one more attempt at writing code to get a good result automatically that will fail for reasons I do not know, and sketch out a much more sophisticated algorithm but not implement it.

 

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Fixing Random, part 36

One more time! Suppose we have our nominal distribution p that possibly has “black swans” and our helper distribution q which has the same support, but no black swans.

We wish to compute the expected value of f when applied to samples from p, and we’ve seen that we can estimate it by computing the expected value of g:

x => f(x) * p.Weight(x) / q.Weight(x)

applied to samples of q.

Unfortunately, the last two times on FAIC we saw that the result will be wrong by a constant factor; the constant factor is the quotient of the normalization constants of q and p.

It seems like we’re stuck; it can be expensive or difficult to determine the normalization factor for an arbitrary distribution. We’ve created infrastructure for building weighted distributions and computing posteriors and all sorts of fun stuff, and none of it assumes that weights are normalized so that the area under the PDF is 1.0.

But… we don’t need to know the normalization factors. We never did, and every time I said we did, I lied to you. Because I am devious.

What do we really need to know? We need to know the quotient of two normalization constants. That is less information than knowing two normalization constants, and maybe there is a cheap way to compute that fraction.

Well, let’s play around with computing fractions of weights; our intuition is: maybe the quotient of the normalization constants is the average of the quotients of the weights. So let’s make a function and call it h:

x => p.Weight(x) / q.Weight(x)

What is the expected value of h when applied to samples drawn from q?

Well, we know that it could be computed by:

Area(x => h(x) * q.Weight(x)) / Area(q.Weight)

But do the algebra: that’s equal to

Area(p.Weight) / Area(q.Weight)

Which is the inverse of the quantity that we need, so we can just divide by it instead of multiplying!

Here’s our logic:

  • We can estimate the expected value of g on samples of  q by sampling.
  • We can estimate the expected value of h on samples of q by sampling.
  • The quotient of these two estimates is an estimate of the expected value of f on samples of p, which is what we’ve been after this whole time.

Whew!


Aside: I would be remiss if I did not point out that there is one additional restriction that we’ve got to put on helper distribution q : there must be no likely values of x in the support of q such that q.Weight(x) is tiny but p.Weight(x) is extremely large, because their quotient is then going to blow up huge if we happen to sample that value, and that’s going to wreck the average.


We can now actually implement some code that computes expected values using importance sampling and no quadrature. Let’s put the whole thing together, finally: (All the code can be found here.)

public static double ExpectedValueBySampling<T>(
    this IDistribution<T> d,
    Func<T, double> f,
    int samples = 1000) =>
  d.Samples().Take(samples).Select(f).Average();

public
 static double ExpectedValueByImportance(
    this IWeightedDistribution<double> p,
    Func<double, double> f,
    double qOverP,
    IWeightedDistribution<double> q,
    int n = 1000) =>
  qOverP * q.ExpectedValueBySampling(
    x => f(x) * p.Weight(x) / q.Weight(x), n);

public static double ExpectedValueByImportance(
  this IWeightedDistribution<double> p,
  Func<double, double> f,
  IWeightedDistribution<double> q,
  int n = 1000)
{
  var pOverQ = q.ExpectedValueBySampling(
    x => p.Weight(x) / q.Weight(x), n);
  return p.ExpectedValueByImportance(f, 1.0 / pOverQ, q, n);
}

Look at that; the signatures of the methods are longer than the method bodies! Basically there’s only four lines of code here. Obviously I’m omitting error handling and parameter checking and all that stuff that would be necessary in a robust implementation, but the point is: even though it took me six math-heavy episodes to justify why this is the correct code to write, actually writing the code to solve this problem is very straightforward.

Once we have that code, we can use importance sampling and never have to do any quadrature, even if we do not give the ratio of the two normalization constants:

var p = Normal.Distribution(0.75, 0.09);
double f(double x) => Atan(1000 * (x  .45)) * 20  31.2;
var u = StandardContinuousUniform.Distribution;
var expected = p.ExpectedValueByImportance(f, u);

Summing up:

  • If we have two distributions p and q with the same support…
  • and a function f that we would like to evaluate on samples of p
  • and we want to estimate the average value of f …
  • but p has “black swans” and q does not, then:
  • we can still efficiently get an estimate by sampling q
  • bonus: we can compute an estimate of the ratios of the normalization constants of p and q
  • extra bonus: if we already know one of the normalization constants, we can compute an estimate of the other from the ratio.

Super; are we done?

In the last two episodes we pointed out that there are two problems: we don’t know the correction factor, and we don’t know how to pick a good q. We’ve only solved the first of those problems.


Next time on FAIC: We’ll dig into the problem of finding a good helper distribution q.

Fixing Random, part 35

Last time on FAIC we deduced the idea behind the “importance sampling” technique for determining the average value of a function from double to double — call it  f — when it is applied to samples from a possibly-non-normalized weighted distribution of doubles — call it p.

Just for fun, I’m going to do the derivation all over again. I’ll again be using the technique “things equal to the same are equal to each other”, but this time we’re going to start from the other end. Let’s jump right in!

Again, for pedagogical purposes I’m going to consider only distributions with support from 0.0 to 1.0; we’ll eliminate that restriction when we can.

We discovered a while back that the expected value of f applied to samples of pis equal to the quotient of Area(x => f(x) * p.Weight(x) divided by ​​Area(x => p.Weight(x)). The latter term is the normalization constant for p, which we might not know.

Let’s take any weighted distribution q also with support from 0.0 to 1.0; it also might not be normalized.

We’re now going to do some manipulations to our expression that are obviously identities. We’ll start with the fact that

Area(x => f(x) * p.Weight(x))

obviously must be equal to:

Area(x => (f(x) * p.Weight(x) / q.Weight(x)) * q.Weight(x))

We’ve just divided and multiplied by the same quantity, so that is no change. And we’ve assumed that q has the same support as p, so the weight we’re dividing by is always non-zero.

Similarly,

Area(p.Weight)

must be equal to

(Area(q.Weight) * (Area(p.Weight) / Area(q.Weight)))

for the same reason.

So the quotient of these two quantities must also be equal to the expected value of f applied to samples from p; they’re the same quantities! Our original expression

Area(x => f(x) * p.Weight(x) /
 Area(x => p.Weight(x))

is equal to:

Area(x => (f(x) * p.Weight(x) / q.Weight(x)) * q.Weight(x)) / 
  (Area(q.Weight) * (Area(p.Weight) / Area(q.Weight)))

For any suitable q.

Let’s call that value exp_fp for “expected value of f on samples of p“. We’ve just written that value in two ways, one very straightforward, and one excessively complicated.

Unsurprisingly, my next question is: what is the expected value of function g

x => f(x) * p.Weight(x) / q.Weight(x)

over samples from q?

We know that it is estimated by this quotient of areas:

Area(x => g(x) * q.Weight(x)) /
Area(q.Weight)

The denominator is the normalization constant of q which we might not know.

Call that value exp_gq, for “expected value of g on samples of q

What is the relationship between exp_gq and exp_fp?

Well, just look at the two expressions carefully; plainly they differ by no more than a constant factor. exp_fp is equal to exp_gq * Area(q.Weight) / Area(p.Weight)

And now we are right where we ended last time. Summing up:

  • Once again, we have deduced that importance sampling works:
    • An estimate of the expected value of g applied to samples from q is proportional to the expected value of f applied to samples from p
    • the proportionality constant is exactly the quotient of the normalization constants of q and p
    • If q and p are known to be normalized, then that constant is 1.
  • Once again, we can extend this result to q and p with any support
    • All we really need for an accurate estimate is that q have support in places where f(x) * p.Weight(x) has a lot of area.
    • But it is also nice if q has low weight in where that function has small area.
  • Once again, we have two serious problems:
    • how do we find a good q?
    • we are required to know the normalization constants of both p and q, either of which might be hard to compute in general
  • Once again, the previous statement contains a subtle error.
    • I’m so devious.
    • I asked what the error was in the previous episode as well, and there is already an answer in the comments to that episode, so beware of spoilers if you want to try to figure it out for yourself.

We are so close to success, but it seems to be just out of our grasp. It’s vexing!


Next time on FAIC: Amazingly, we’ll implement a version of this algorithm entirely based on sampling; we do not actually need to do the quadrature to compute the normalization factors!

Fixing Random, part 34

Last time on FAIC we implemented a better technique for estimating the expected value of a function f applied to samples from a distribution p:

  • Compute the total area (including negative areas) under the function x => f(x) * p.Weight(x)
  • Compute the total area under x => p.Weight(x)
    • This is 1.0 for a normalized PDF, or the normalizing constant of a non-normalized PDF; if we already know it, we don’t have to compute it.
  • The quotient of these areas is the expected value

Essentially our technique was to use quadrature to get an approximate numerical solution to an integral calculus problem.

However, we also noted that it seems like there might still be room for improvement, in two main areas:

  • This technique only works when we have a good bound on the support of the distribution; for my contrived example I chose a “profit function” and a distribution where I said that I was only interested in the region from 0.0 to 1.0.
  • Our initial intuition that implementing an estimate of “average of many samples” by, you know, averaging many samples, seems like it was on the right track; can we get back there?

In this episode I’m going to stick to the restriction to distributions with support over 0.0 to 1.0 for pedagogic reasons, but our aim is to find a technique that gets us back to sampling over arbitrary distributions.

The argument that I’m going to make here (several times!) is: two things that are both equal to the same third thing are also equal to each other.

Recall that we arrived at our quadrature implementation by estimating that our continuous distribution’s expected value is close to the expected value of a very similar discrete distribution. I’m going to make my argument a little bit more general here by removing the assumption that p is a normalized distribution. That means that we’ll need to know the normalizing factor np, which as we’ve noted is Area(p.Weight).

We said that we could estimate the expected value like this:

  • Imagine that we create a 1000 sided “unfair die” discrete distribution.
  • Each side corresponds to a 0.001 wide slice from the range 0.0 to 1.0; let’s say that we have a variable x that takes on values 0.000, 0.001, 0.002, and so on, corresponding to the 1000 sides.
  • The weight of each side is the probability of choosing this slice: p.Weight(x) / 1000 / np
  • The value of each side is the “profit function” f(x)
  • The expected value of “rolling this die” is the sum of (value times weight): the sum of f(x) * (p.Weight(x) / 1000 / np)over our thousand values of x

Here’s the trick:

  • Consider the standard continuous uniform distribution u. That’s a perfectly good distribution with support 0.0 to 1.0.
  • Consider the function w(x) which is x => f(x) * p.Weight(x) / np.  That’s a perfectly good function from double to double.
  • Question: What is an estimate of the expected value of w over samples from u?

We can use the same technique:

  • Imagine we create a 1000-sided “unfair die” discrete distribution
  • x is as before
  • The weight of each side is the probability of choosing that slice, but since this is a uniform distribution, every weight is the same — so, turns out, it is not an unfair die! The weight of each side is 0.001.
  •  The value of each side is our function w(x)
  • The expected value of rolling this die is the sum of (value times weight): the sum of (f(x) * p.Weight(x) / np) * 0.001 over our thousand values of x

But compare those two expressions; we are computing exactly the same sum both times. These two expected values must be the same value.

Things equal to the same are equal to each other, which implies this conclusion:

If we can compute an estimate of the expected value of w applied to samples from u by any technique then we have also computed an estimate of the expected value of f applied to samples from p.

Why is this important?

The problem with the naïve algorithm in our original case was that there was a “black swan” — a region of large (negative) area that is sampled only one time in 1400 samples. But that region is sampled one time in about 14 samples when we sample from a uniform distribution, so we will get a much better and more consistent estimate of the expected value if we use the naïve technique over the uniform distribution.

In order to get 100x times as many samples in the black swan region, we do not have to do 100x times as many samples overall. We can just sample from a helper distribution that targets that important region more often.

Let’s try it! (Code can be found here.)

In order to not get confused here, I’m going to rename some of our methods so that they’re not all called ExpectedValue. The one that just takes any distribution and averages a bunch of samples is now ExpectedValueBySampling and the one that computes two areas and takes their quotient is ExpectedValueByQuadrature.

var p = Normal.Distribution(0.75, 0.09);
double f(double x) => Atan(1000 * (x  .45)) * 20  31.2;
var u = StandardContinuousUniform.Distribution;
double np = 1.0; // p is normalized
double w(double x) => f(x) * p.Weight(x) / np;
for (int i = 0; i < 100; ++i)
  Console.WriteLine($”{u.ExpectedValueBySampling(w):0.###});

Remember, the correct answer that we computed by quadrature is 0.113. When sampling p directly we got values ranging from 0.7 to 0.14. But now we get:

0.114, 0.109, 0.109, 0.118, 0.111, 0.107, 0.113, 0.112, ...

So much better!

This is awesome, but wait, it gets more awesome. What is so special about the uniform distribution? Nothing, that’s what. I’m going to do this argument one more time:

Suppose I have distribution q, any distribution whatsoever, so long as its support is the same as p — in this case, 0.0 to 1.0. In particular, suppose that q is not necessarily a normalized distribution, but that again, we know its normalization factor. Call it nq.  Recall that the normalization factor can be computed by nq = Area(q.Weight).

Our special function g(x) is this oddity:

x => (f(x) * (p.Weight(x) / q.Weight(x)) * (nq / np)

What is the expected value of g over distribution q?  One more time:

  • Create a 1000-sided unfair die, x as before.
  • The weight of each side is the probability of choosing that side, which is (q.Weight(x) / 1000) / nq
  • The value of each side is g(x).
  • The expected value is the sum of g(x) * (q.Weight(x) / 1000) / nq but if you work that out, of course that is the sum of f(x) * p.Weight(x) / np / 1000

And once again, we’ve gotten back to the same sum by clever choice of function. If we can compute the expected value of gevaluated on samples from q, then we know the expected value of f evaluated on samples from p!

This means that we can choose our helper distribution q so that it is highly likely to pick values in regions we consider important. Let’s look at our graph of p.Weight and f*p.Weightagain:

Screen Shot 2019-05-21 at 9.17.06 AM

There are segments of the graph where the area under the blue line is very small but the area under the orange line is large, and that’s our black swan; what we want is a distribution that samples from regions where the orange area is large, and if possible skips regions where it is small. That is, we consider the large-area regions important contributors to the expected value, and the small-area regions unimportant contributors; we wish to target our samples so that no important regions are ignored. That’s why this technique for computing expected value is called “importance sampling”.


Exercise: The uniform distribution is pretty good on the key requirement that it never be small when the area under the orange line is large, because it is always the same size from 0.0 to 1.0; that’s why it is the uniform distribution, after all. It’s not great on our second measure; it spends around 30% of its time sampling from regions where the orange line has essentially no area under it.

Write some code that tries different distributions. For example, implement the distribution that has weight function x => (0 <= x && x <= 1) ? x : 0

(Remember that this is not a normalized distribution, so you’ll have to compute nq.)

Does that give us an even better estimate of the expected value of f?

Something to ponder while you’re working on that: what would be the ideal choice of distribution?


Summing up:

  • Suppose we have a weighted distribution of doubles p and a function from double to double f.
  • We wish to accurately estimate the average value of f when it is applied to a large set of samples from p; this is the expected value.
  • However, there may be “black swan” regions where the value of f is important to the average, but the probability of sampling from that region is low, so our average could require a huge number of samples to get an accurate average.
  • We can fix the problem by choosing any weighted distribution q that has the same support as pbut is more likely to sample from important regions.
  • The expected value of g (given above) over samples drawn from q is the same as the expected value of fover samples from p.

This is great, and I don’t know if you noticed, but I removed any restriction there that p or q be distributions only on 0.0 to 1.0; this technique works for weighted distributions of doubles over any support!


Aside: We can weaken our restriction that q have the same support as p; if we decide that q can have zero weight for particularly unimportant regions, where, say, we know that f(x)*p.Weight(x) is very small, then that’s still going to produce a good estimate.

Aside: Something I probably should have mentioned before is that all of the techniques I’m describing in this series for estimating expected values require that the expected value exists! Not all functions applied to probability distributions have an expected value because the average value of the function computed on a group of samples might not converge as the size of the group gets larger. An easy example is, suppose we have a standard normal distribution as our p​ and x => 1.0 / p.Weight(x) as our f. The more samples from p we take, the more likely it is that the average value of f gets larger!


However, it’s not all sunshine and roses. We still have two problems, and they’re pretty big ones:

  • How do we find a good-quality q distribution?
  • We need to know the normalization constants for both distributions. If we do not know them ahead of time (because, say, we have special knowledge that the continuous uniform distribution is normalized) then how are we going to compute them?  Area(p.Weight) or Area(q.Weight) might be expensive or difficult to compute.It seems like in the general case we still have to solve the calculus problem. 😦

Aside: The boldface sentence in my last bullet point contains a small but important error. What is it? Leave your guesses in the comments; the answer will be in an upcoming episode.


I’m not going to implement a general-purpose importance sampling algorithm until we’ve made at least some headway on these remaining problems.

Next time on FAIC:  It’s Groundhog Day! I’m going to do this entire episode over again; we’re going to make a similar argument — things equal to the same are equal to each other — but starting from a different place. We’ll end up with the same result, and deduce that importance sampling works.

 

Fixing Random, part 33

Last time on FAIC I showed why our naïve implementation of computing the expected value can be fatally flawed: there could be a “black swan” region where the “profit” function f is different enough to make a big difference in the average, but the region is just small enough to be missed sometimes when we’re sampling from our distribution p

It looks so harmless [Photo credits]

The obvious solution is to work harder, not smarter: just do more random samples when we’re taking the average! But doesn’t it seem to be a little wasteful to be running ten thousand samples in order to get 9990 results that are mostly redundant and 10 results that are extremely relevant outliers?

Perhaps we can be smarter.

We know how to compute the expected value in a discrete non-uniform distribution of doubles: multiply each value by its weight, sum those, and divide by the total weight. But we should think for a moment about why that works.

If we have an unfair two-sided die — a coin — stamped with value 1.23 on one side, and -5.87 on the other, and 1.23 is twice as likely as -5.87, then that is the same as a fair three sided die — whatever that looks like — with 1.23 on two sides and -5.87 on the other. One third of the time we get the first 1.23, one third of the time we get the second 1.23, and one third of the time we get -5.87, so the expected value is 1.23/3 + 1.23/3 – 5.87/3, and that’s equivalent to (2 * 1.23 – 5.87) / 3. This justifies our algorithm.

Can we use this insight to get a better estimate of the expected value in the continuous case? What if we thought about our continuous distribution as just a special kind of discrete distribution?


Aside: In the presentation of discrete distributions I made in this series, of course we had integer weights. But that doesn’t make any difference in the computation of expected value; we can still multiply values by weights, take the sum, and divide by the total weight.


Our original PDF — that shifted, truncated bell curve — has support from 0.0 to 1.0. Let’s suppose that instead we have an unfair 1000-sided die, by dividing up the range into 1000 slices of size 0.001 each.

  • The weight of each side of our unfair die is the probability of rolling that side.
    • Since we have a normalized PDF, the probability is the area of that slice. 
    • Since the slices are very thin, we can ignore the fact that the top of the shape is not “level”; let’s just treat it as a rectangle.
    • The width is 0.001; the height is the value of the PDF at that point.
    • That gives us enough information to compute the area.
  • Since we have a normalized PDF, the total weight that we have to divide through is 1.0, so we can ignore it. Dividing by 1.0 is an identity.
  • The value on each side of the die is the value of our profit function at that point.

Now we have enough information to make an estimate of the expected value using our technique for discrete distributions.


Aside: Had I made our discrete distributions take double weights instead of integer weights, at this point I could simply implement a “discretize this distribution into 1000 buckets” operation that turns weighted continuous distributions into weighted discrete distributions.

However, I don’t really regret making the simplifying choice to go with integer weights early in this series; we’re immediately going to refactor this code away anyways, so turning it into a discrete distribution would have been a distraction.


Let’s write the code: (Code for this episode is here.)

 public static double ExpectedValue(
    this IWeightedDistribution<double> p,
    Func<double, double> f) =>
  // Let’s make a 1000 sided die:
  Enumerable.Range(0, 1000)
  // … from 0.0 to 1.0:
  .Select(i => ((double)i) / 1000)
  // The value on the “face of the die” is f(x)
  // The weight of that face is the probability
// of choosing this slot

  .Select(x => f(x) * p.Weight(x) / 1000)
  .Sum();
  // No need to divide by the total weight since it is 1.0.

And if we run that:

Console.WriteLine($”{p.ExpectedValue(f):0.###});

we get

0.113

which is a close approximation of the true expected value of this profit function over this distribution. Total success, finally!

Or, maybe not.

I mean, that answer is correct, so that’s good, but we haven’t solved the problem in general.

The obvious problem with this implementation is: it only works on normalized distributions whose support is between 0.0 and 1.0. Also, it assumes that 1000 is a magic number that always works. It would be nice if this worked on non-normalized distributions over any range with any number of buckets.

Fortunately, we can solve these problems by making our implementation only slightly more complicated:

public static double ExpectedValue(
  this IWeightedDistribution<double> p,
  Func<double, double> f,
  double start = 0.0,
  double end = 1.0,
  int buckets = 1000)
{
  double sum = 0.0;
  double total = 0.0;
  for (int i = 0; i < buckets; i += 1)
  {
    double x = start + (end  start) * i / buckets;
    double w = p.Weight(x) / buckets;
    sum += f(x) * w;
    total += w;
  }
  return sum / total;
}

That totally works, but take a closer look at what this is really doing. We’re computing two sums, sum and total, in exactly the same manner. Let’s make this a bit more elegant by extracting out the summation into its own method:

public static double Area(
    Func<double, double> f,
    double start = 0.0,
    double end = 1.0,
    int buckets = 1000) =>
  Enumerable.Range(0, buckets)
    .Select(i => start + (end  start) * i / buckets)
    .Select(x => f(x) / buckets)
    .Sum();
public static double ExpectedValue(
    this IWeightedDistribution<double> p,
    Func<doubledouble> f,
    double start = 0.0,
    double end = 1.0,
    int buckets = 1000) =>
  Area(x => f(x) * p.Weight(x), start, end, buckets) /
    Area(p.Weight, start, end, buckets);

As often happens, by making code more elegant, we gain insights into the meaning of the code. That first function should look awfully familiar, and I’ve renamed it for a reason. The first helper function computes an approximation of the area under a curve; the second one computes the expected value as the quotient of two areas. 

It might be easier to understand it with a graph; here I’ve graphed the distribution p.Weight(x) as the blue line and the profit times the distribution f(x) * p.Weight(x)as the orange line:

Screen Shot 2019-05-21 at 9.17.06 AM.png

The total area under the blue curve is 1.0; this is a normalized distribution.

The orange curve is the blue curve multiplied by the profit function at that point. The total area under the orange curve — remembering that area below the zero line is negative area — divided by the area of the blue curve (1.0) is the expected value.


Aside: You can see from the graph how carefully I had to contrive this “black swan” scenario. I needed a region of the graph where the area under the blue line is close to 0.001 and the profit function is so negative that it makes a large negative area there when multiplied, but without making a large negative area anywhere else.

Of course this example is contrived, but it is not unrealistic; unlikely things happen all the time, and sometimes those unlikely things have important consequences.

An interesting feature of this scenario is: look at how wide the negative region is! It looks like it is around 10% of the total support of the distribution; the problem is that we sample from this range only 0.1% of the time because the blue line is so low here. We’ll return to this point in the next episode.


Aside: The day I wrote this article I learned that this concept of computing expected value of a function applied to a distribution by computing the area of the product has a delightful name: LOTUS, which stands for the Law Of The Unconscious Statistician.

The tongue-in-cheek name is apparently because statistics students frequently believe that “the expected value is the area under this curve” is the definition of expected value. I hope I avoided any possible accusation of falling prey to the LOTUS fallacy. We started with a correct definition of expected value: the average value of a function applied to a bunch of samples, as the size of the bunch gets large. I then gave an admittedly unrigorous, informal and hand-wavy justification for computing it by approximating area, but it was an argument.


We’ve now got two ways of computing an approximation of the expected value when given a distribution and a function:

  • Compute a bunch of samples and take their average.
  • Compute approximate values of two areas and divide them.

As we know, the first has problems: we might need a very large set of samples to find all the relevant “black swan” events, and therefore we spend most of our time sampling the same boring high-probability region over and over.

However, the second has some problems too:

  • We need to know the support of the distribution; in my contrived example I chose a distribution over 0.0 to 1.0, but of course many distributions have much larger ranges.
  • We need to make a guess about the appropriate number of buckets to get an accurate answer.
  • We are doing a lot of seemingly unnecessary math; between 0.0 and, say 0.3, the contribution of both the blue and orange curves to the total area is basically zero. Seems like we could have skipped that, but then again, skipping a region with total probability of one-in-a-thousand led to a bad result before, so it’s not entirely clear when it is possible to save on work.
  • Our first algorithm was fundamentally about sampling, which seems appropriate, since “average of a set of samples” is the definition of expected value. This algorithm is just doing an approximation of integral calculus; something seems “off” about that.

It seems like we ought to be able to find an algorithm for computing expected value that is more accurate than our naive algorithm, but does not rely so much on calculus.


Next time on FAIC: We’ll keep working on this problem!

Porting old posts, part 4

I’m continuing with my project to port over, reformat and update a decade of old blog posts. Today, a few days in mid-October 2003; this is still my second month of blogging and I am writing at what I would consider today to be a ridiculous rate.


Why is there no #include?

Once again I fail to understand the customer mindset.


How do I script a non-default dispatch?

First, COM is too complicated. Second, sometimes defense in depth can go too far.


What everyone should know about character encoding

I had been meaning to write this post, and then Joel ended up doing a far better job than I would have. That’s a win in my opinion.


It never leaks but it pours

Some thoughts on memory leaks and language design to avoid them.


I take exception to that

Long jumps considered way more harmful than exceptions

Whether you should use exception handling or not is a decision to be made on its merits; but trying to mix code that uses exception handling with code that uses return codes is going to be a mess. Pick one.


Designing JScript .NET

The first of what will be a lot of articles describing the JScript .NET language design process.

Microsoft’s under-funding of this language and the long-term strategic ramifications of that choice continues to be the corporate strategy decision affecting me directly that I most strongly disagreed with. It’s super irritating to have a vision of the future, try to build towards it, and have all those efforts rebuked, only to see many aspects of that vision realized a decade later at many times the cost. I’m glad Microsoft got there eventually, but it could have been a lot sooner, a lot cheaper.

Expect to see a bunch more of this rant as I port over those articles; you’ve been warned.


Wrox is dead, long live Wrox
Digging a security hole all the way to China
Dead trees vs bits

Some thoughts on the economics of publishing computer books in 2003, before the e-book revolution.


VBScript is to VB as ping pong is to volleyball

Is it more important to learn the language, or the object model? Plus, some musings on the 2003 mobile phone operating system business.


How bad is good enough?

The first of many rants about how to think about improving performance of software.


More soon!

 

 

 

Fixing Random, part 32

Last time on FAIC we reviewed the meaning of “expected value”: when you get a whole bunch of samples from a distribution, and a function on those samples, what is the average value of the function’s value as the number of samples gets large?

I gave a naive implementation:

public static double ExpectedValue<T>(
    this IDistribution<T> d,
    Func<T, double> f) =>
  d.Samples().Take(1000).Select(f).Average();

public static double ExpectedValue(
    this IDistribution<double> d) =>
  d.ExpectedValue(x=>x);

Though short and sweet, this implementation has some problems; the most obvious one is that hard-coded 1000 in there; where did it come from? Nowhere in particular, that’s where.

It seems highly likely that this is either too big, and we can compute a good estimate in fewer samples, or that it is too small, and we’re missing some important values.

Let’s explore a scenario where the number of samples is too small. The scenario will be contrived, but not entirely unrealistic.

Let’s suppose we have an investment strategy; we invest a certain amount of money with this strategy, and when the strategy is complete, we have either more or less money than we started with. To simplify things, let’s say that the “outcome” of this strategy is just a number between 0.0 and 1.0; 0.0 indicates that the strategy has completely failed, resulting in a loss, and 1.0 indicates that the strategy has completely succeeded, resulting in the maximum possible return.

Before we go on, I want to talk a bit about that “resulting in a loss” part. If you’re a normal, sensible investor like me and you have $100 to invest, you buy a stock or a mutual fund for $100 because you believe it will increase in value. If it goes up to $110 you sell it and pocket the $10 profit. If it goes down to $90, you sell it and take the $10 loss. But in no case do you ever lose more than the $100 you invested. (Though of course you do pay fees on each trade whether it goes up or down; let’s suppose those are negligible.) Our goal is to get that 10% return on investment.

Now consider the following much riskier strategy for spending $100 to speculate in the market: suppose there is a stock currently at $100 which we believe will go down, not up. We borrow a hundred shares from someone willing to lend them to us, and sell them for $10000. We pay the lender $100 interest for their trouble. Now if the stock goes down to $90, we buy the hundred shares back for $9000, return them to the owner, and we’ve spent $100 but received $1000; we’ve made a 900% return on the $100 we “invested”. This is a “short sale”, and as you can see, you get a 900% return instead of a 10% return.

(I say “invested” in scare quotes because this isn’t really investing; it’s speculation. Which is a genteel word for “gambling”.)

But perhaps you also see the danger here. Suppose the stock goes down but only to $99.50. We buy back the shares for $9950, and we’ve only gained $50 on the trade, so we’ve lost half of our $100; in the “long” strategy, we would only have lost fifty cents; your losses can easily be bigger with the short strategy than with the long strategy.

But worse, what if the stock goes up to $110. We have to buy back those shares for $11000, so we’ve “invested” $100 and gotten a “return” of -$1000, for a total loss of $1100 on a $100 investment. In a short sale if things go catastrophically wrong you can end up losing more than your original investment. A lot more!

As I often say, foreshadowing is the sign of a quality blog; let’s continue with our example.

We have a process that produces values from 0.0 to 1.0 that indicates the “success level” of the strategy. What is the distribution of success level? Let’s suppose it’s a straightforward bell-shaped curve with the mean on the “success” side:

Screen Shot 2019-05-20 at 1.14.22 PM.png

This is a PDF, and extra bonus, it is normalized: the total area under the curve is 1.0. The vast majority of the time — 99.9% — our investment strategy produces an outcome between 0.48 and 1.0, so that’s the area of the graph I’m going to focus on.


Aside: I know it looks a little weird seeing a bell-shaped curve that just cuts off abruptly at 1.0, but let’s not stress about it. Again, this is a contrived example, and we’ll see why it is irrelevant later.


Now, I don’t intend to imply that the “success level” is the return: I’m not saying that most of the time we get a 75% return. Let’s suppose that we’ve worked out a function that translates “strategy outcome level” to the percentage gain on our investment. That is, if the function is 0.0, we’ve not gained anything but we’ve not lost anything either. If it is 0.5 then our profits are 50% of our investment; if it’s -0.25 then we’ve taken a net loss of 25% of our investment, and so on.

Let’s propose such a function, and zoom in on the right side of the diagram where the 99.9% of cases are found: from 0.46 to 1.0. Here’s our function:

Screen Shot 2019-05-20 at 1.34.22 PM.png

If our process produces an outcome of 0.54 or larger, we make between a 0% and an 18% return, which seems pretty good; after all the vast majority of the bulk of the distribution is to the right of 0.54. Now, in the unlikely case where we get an outcome from 0.48 to 0.54, we lose money; on the left hand end, we’re losing almost 200%; that is, if we put in $100 we not only fail to make it back, we end up owing $100.

The question at hand, of course, is what is the average value of the “profit function” given the distribution of outcomes? That is, if we ran this strategy a thousand times, on average what return would we get?

Well, we already wrote the code, so let’s run it: (Code can be found here.)

var p = Normal.Distribution(0.75, 0.09);
double f(double x) =>
  Atan(1000 * (x  .45)) * 20  31.2;
Console.WriteLine($”{p.ExpectedValue(f):0.##});


Aside: I’m ignoring the fact that if we get a sample out of the normal distribution that is greater than 1.0 or less than 0.0 we do not discard it. The region to the left of 0.0 is absurdly unlikely, and the region to the right of 1.0 has low probability and the value of the profit function is very flat. Writing a lot of fiddly code to correct for this problem would not change the estimated expected value much and would be a distraction from the far deeper problem we’re about to discover.

Also, I’m limiting the distribution to a support of 0.0 to 1.0 for pedagogical reasons; as we’ll see in later episodes, we will eventually remove this restriction, so let’s just ignore it now.


We get the answer:

0.14

Super, we will get on average a 14% return on our investment with this strategy. Seems like a good investment; though we have a small chance of losing big, the much larger chance of getting a solid 0% to 18% return outweighs it. On average.


Aside: It is important to realize that this 14% figure does not necessarily imply that typically we get a 14% return when we run this strategy, any more than the expected value of 3.5 implies that we’ll typically get 3.5 when we roll a d6. The expected value only tells you what the average behaviour is; it doesn’t tell you, for example, what the probability is that you’ll lose everything. It tells you that if you ran this strategy a million times, on average you’d get a 14% return, and that’s all it tells you. Be careful when making decisions relying upon expected values!


This 14% expected result totally makes sense, right? We know most of the time the result will be between 0% and 18%, so 14% seems like a totally reasonable expected value.

As one of my managers at Microsoft used to say: would you bet your car on it?

Let’s just run the code again to be sure.

0.14

Whew. That’s a relief.

You know what they say: doing the same thing twice and expecting different results is the definition of practicing the piano, but I’m still not satisfied. Let’s run it another hundred times:

0.14, 0.08, 0.09, 0.14, 0.08, 0.14, 0.14, 0.07, 0.07, 0.14, ...

Oh dear me.

The expected return of our strategy is usually 14%; why it is sometimes half that? Yes, there is randomness in our algorithm, but surely it should not cause the value computed to vary so widely!

It seems like we cannot rely on this implementation, but why not? The code looks right.

As it turns out, I showed you exactly the wrong part of the graphs above. (Because I am devious.) I showed you the parts that make up 99.9% of the likely cases. Let’s look at the graph of the profit-or-loss function over the whole range of outcomes from 0.0 to 1.0:

Screen Shot 2019-05-20 at 5.09.29 PM.png

OUCH. It’s an approximation of a step function; in the original graph that started at 0.46, we should have noticed that the extremely steep part of the curve was trending downwards to the left at an alarming rate. If you have an outcome of 0.48 from our process, you’re going to lose half your money; 0.46, you lose all your money and a little bit more. But by the time we get to 0.44, we’re down to losing over 60 times your original investment; this is a catastrophic outcome.

How likely is that catastrophic outcome? Let’s zoom in on just that part of the probability distribution. I’ll re-scale the profit function so that they fit on the same graph, so you can see where exactly the step happens:

2019-05-20 (3)

This is a normalized PDF, so the probability of getting a particular outcome is equal to the area of the region under the curve. The region under the curve from 0.38 to 0.45 is a little less than a triangle with base 0.07 and height 0.02, so its area is in the ballpark of 0.0007. We will generate a sample in that region roughly once every 1400 samples.

But… we’re computing the expected value by taking only a thousand samples, so it is quite likely that we don’t hit this region at all. The expected losses in that tiny area are so enormous that it changes the average every time we do sample from it. Whether you include a -60 value or not in the average is a huge influence as to whether the average is 0.14 or 0.07. 


Aside: if you run the expected value estimate a few thousand times or more it just gets crazier; sometimes the expected value is actually negative if we just by luck get more than one sample in this critical region.


I contrived this example to produce occasional “black swan events“; obviously, our naïve algorithm for computing expected value does not take into account the difficulty of sampling a black swan event, and therefore produces inaccurate and inconsistent results.


Next time on FAIC: We know how to solve this problem exactly for discrete distributions; can we use some of those ideas to improve our estimate of the expected value in this scenario?