Fixing Random, part 12

[Code is here.]

Last time on FAIC we implemented an efficient “conditioned” probability using the Where operator on distributions; that is, we have some “underlying” distribution, and we ask the question “if a particular condition has to be met, what is the derived distribution that meets that condition?” For discrete distributions we can compute that distribution directly and just return it.

There is another kind of conditional probability though, which is much more rich, complex and counter-intuitive, and that is exploring the relationship between “what is the probability of X?” and “what is the probability of Y given that we know X?

For example: pick a random person in the world who has a cold. What is the probability that they sneezed in the last 24 hours? Probably something like 85%.

Now pick a random person who does not have a cold. For them, the probability is maybe more like 3%. In months when I do not have a cold, I sneeze maybe one or two days.

So what we’ve got here is a rather more complex probability distribution; in fact we have two entirely different distributions, and which one we use depends on a condition.

Notice how this is clearly related to our recent discussion of conditioned probabilities, but different. With a “Where” clause we are saying make the support of this distribution smaller because some outcomes are impossible based on a condition. What we’re talking about here is choosing between two (or more) distributions depending on a condition.

The standard notation for this kind of probability in mathematics is a bit unfortunate. We would say something like P(sneezed|no cold ) = 0.03 to represent “3% chance that I sneezed if I didn’t have a cold” and P(sneezed|cold) = 0.85 to represent “85% chance that I sneezed if I had a cold”. That is, the syntax is P(A|B) means “what is the probability of A given that B happened?”

How might we represent this in our system? It seems like IDiscreteDistribution<T> is not rich enough. Let’s just start making some types and see what we can come up with.

“Has sneezed recently” and “has a cold” are Booleans, but I want the types of everything to be very clear in the analysis which follows, so I’m going to make my own custom types:

enum Cold { No, Yes }
enum Sneezed { No, Yes }

I want to be slightly abusive of notation here and say that P(Cold.Yes) and P(Cold.No) are the weights of a probability distribution that I’m going to call by the shorthand P(Cold). Similarly for P(Sneezed); that’s the probability distribution that gives weights to P(Sneezed.Yes) and P(Sneezed.No). Normally we think of P(something) as being a value between 0.0 and 1.0, but if you squint at it, really those values are just weights. It doesn’t matter what convention we use for weights; a bunch of integers that give ratios of probabilities and a bunch of doubles that give fractions have pretty much the same information content.

Plainly what I would very much like is to have IDiscreteDistribution<Cold> be the C# type that represents P(Cold).

But how can we represent our concept of “There’s a 3% chance I sneezed if I do not have a cold, but an 85% chance if I do have a cold?”

That sure sounds like precisely this:

IDiscreteDistribution<Sneezed> SneezedGivenCold(Cold c)
  var list = new List<Sneezed>() { Sneezed.No, Sneezed.Yes };
  return c == Cold.No ?
    list.ToWeighted(97, 3) :
    list.ToWeighted(15, 85);

That is, if we do not have a cold then the odds are 97 to 3 that we did not sneeze, and if we do have a cold, then the odds is 15 to 85 that we did not sneeze.

I’ve said that I want to represent P(Cold.Yes) and P(Cold.No) by the shorthand P(Cold), and that plainly this in our type system is IDiscreteDistribution<Cold>. Now I want to represent the notion of P(Sneezed) given a value of Cold as P(Sneezed|Cold), which is implemented by the function above. So, what type in our type system is that? Well, suppose we wanted to assign SneezedGivenCold to a variable; what would its type be? Clearly the type of P(Sneezed|Cold) is Func<Cold, IDiscreteDistribution<Sneezed>>!

How interesting! Conditional probabilities are actually functions.

This sort of function has a name; it is called a likelihood function. That is, given some condition, how likely is some outcome?

So that’s interesting, but how is this useful?

Let’s randomly choose a person in the world, where we do not know whether they have a cold or not. What is the probability that they sneezed recently? It depends entirely on the prevalence of colds! If 100% of the world has a cold, then there’s an 85% chance that a randomly chosen person sneezed recently, but if 0% of the world has a cold, then there’s only a 3% chance. And if it is somewhere in between, the probability will be different from either 85% or 3%.

To solve this problem we need to know the probability that the person we’ve chosen has a cold. The probability that some randomly chosen person has a cold is called the prior probability.

What if 10% of the world has a cold? Let’s work it out by multiplying the probabilities:

 Cold      Sneezed       Result
 (prior)   (likelihood)  (conditional)
 10% Yes   85% Yes       8.5% have a cold, and sneezed
           15% No        1.5% have a cold, did not sneeze
 90% No     3% Yes       2.7% do not have a cold, and sneezed
           97% No       87.3% do not have a cold, did not sneeze

Sure enough those probabilities in the right column add up to 100%. The probability that a randomly chosen person in the world sneezed recently (given that these numbers that I made up are accurate) is 8.5% + 2.7% = 11.2%.

The rightmost column of the table that I’ve sketched out here is called the joint probability, which we’ll notate as P(Cold&Sneezed).

We can write this table more compactly like this:

             Cold Yes    Cold No   Total
Sneezed Yes     8.5%        2.7%   11.2%
Sneezed No      1.5%       87.3%   88.8%
Total            10%         90%    100%

The rightmost column of this table is called the marginal probability, so-called because of the way the sums end up at the margins of the table.

What if we expressed the marginal probability as integers? The odds that a random person has sneezed is 11.2% to 88.8%, which if you work out the math, is exactly odds of 14 to 111.

Aside: when I was debugging the code to compute the weights that we will see in a future episode, I got “111” printed out when I was primed to see “112”, having just computed “11.2%” by hand. I almost went on a lengthy bug hunt looking for the non-existing off-by-one error. Fortunately I stopped and double-checked my work, and realized that the 111 represents the 88.8%, not the 11.2%.

How can we do this math given the set of types we’ve created so far? Let’s start with the prior:

var colds = new List<Cold>() { Cold.No, Cold.Yes };
IDiscreteDistribution<Cold> cold = colds.ToWeighted(90, 10);

We’ve got the prior, and we’ve got the likelihood function SneezedGivenCold. We would like to get the marginal probability IDiscreteDistribution<Sneezed>​.

We could implement such a distribution by first sampling from the prior, then calling SneezedFromCold, and then sampling from the returned distribution. Let’s implement it.

Aside: We are of course assuming that the likelihood function is pure.

public sealed class Combined<A, R> : IDiscreteDistribution<R>
  private readonly List<R> support;
  private readonly IDiscreteDistribution<A> prior;
  private readonly Func<A, IDiscreteDistribution<R>> likelihood;
  public static IDiscreteDistribution<R> Distribution(
      IDiscreteDistribution<A> prior,
      Func<A, IDiscreteDistribution<R>> likelihood) =>
    new Combined<A, R>(prior, likelihood);
  private Combined(
    IDiscreteDistribution<A> prior,
    Func<A, IDiscreteDistribution<R>> likelihood)
    this.prior = prior;
    this.likelihood = likelihood;
    var q = from a in prior.Support()
            from b in this.likelihood(a).Support()
            select b; = q.Distinct().ToList();
  public IEnumerable<R> Support() => => x);
  public R Sample() =>
  public int Weight(R r) => WE’LL COME BACK TO THIS ONE

We haven’t implemented Weight, but we don’t need it to run a histogram. Let’s try it out:

Combined<Cold, Sneezed>.Distribution(cold, SneezedGivenCold)


Sure enough, it looks like there is about an 11% chance that a randomly chosen person sneezed, given these distributions.

Now, of course as I have done throughout this series, let’s make a little helper function to make the call sites look a little nicer:

public static IDiscreteDistribution<R> MakeCombined<A, R>(
    this IDiscreteDistribution<A> prior,
    Func<A, IDiscreteDistribution<R>> likelihood) => 
  Combined<A, R>.Distribution(prior, likelihood);

Once again, that should look very familiar! I should change the name of this helper.

If you are still surprised at this point, you have not been paying attention. I’ve already made Select and Where, so the obvious next step is…

public static IDiscreteDistribution<R> SelectMany<A, R>(
    this IDiscreteDistribution<A> prior,
    Func<A, IDiscreteDistribution<R>> likelihood) => 
  Combined<A, R>.Distribution(prior, likelihood);

the bind operation on the probability monad.

And the inelegant call site above is now the much more clear:


Coming up on FAIC: We’ll verify that the distribution type really is a monad, and make a few tweaks to get it working with query comprehension syntax. Then we’ll figure out how to implement the Weight​ function.

6 thoughts on “Fixing Random, part 12

  1. I’ve always found the query syntax with SelectMany to be a little headache inducing. But strangely enough, it feels quite intuitive when it comes to distributions:

    from a in SDU.Distribution(1, 6)
    from b in SDU.Distribution(1, 6)
    select a + b

    What really blows my mind is that this works too:

    from a in SDU.Distribution(1, 6)
    from b in SDU.Distribution(1, 6)
    from c in SDU.Distribution(1, 6)
    select a + b + c

    How the compiler is able to make sense of that and translate it into two SelectMany calls is pretty amazing. Might be worth a blog post on its own.

  2. Pingback: Dew Drop – March 12, 2019 (#2917) | Morning Dew

  3. Pingback: Fixing Random, part 13 | Fabulous adventures in coding

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