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Using bayes theroem on two-way categorical data

Still revising, but I figured that in true Bayesian fashion, I’d update this dynamically as more information came in

It’s said that the best way to understand something is to teach it, and the huge number of explanations of Bayes’ Theorem suggest that many (like me!) have struggled to learn it. Here is my short description of the approach that ultimately led to some clarity for me:

Lets do the “draw a cookie from the jar” example from Allen Downey’s Think Bayes, with a bit more of a plausible backstory (one cannot be an actor, or act like a Bayesian, without understanding their character’s motivation). I made two 100-cookie batches, one with cilantro (Bowl 1) and one without (Bowl 2). Because my cilantro-hating friend prefers vanilla cookies, I made 75 of those (plus 25 chocolate to round out the batch) and put them in the cilantro-free bowl. I made 50 vanilla and 50 chocolate for the cilantro-added bowl.

Everyone comes over for my weird cookie party, but I forget to tell my cilantro-hating friend that they should only choose out of one of the bowls. <EDIT – have the bowls be mixed together, so that each cookie has an equal probability. Doesn’t change the problem, but removes need for an assumption of equal bowl probability> Being entirely too trusting, they just grab a cookie randomly from one of the bowls in the kitchen, and walk back to the living room. I stop them and say “Wait! Do you know which bowl that came out of?” and they say “Oh, no I wasn’t paying attention, but if you made different numbers of vanilla, which this cookie is, that should at least give us a probability of whether it came from the cilantro bowl. It wouldn’t be catastrophic to accidentally take a cilantro bite, so I’ll go for it if my chance of it being cilantro-free is greater than 55%”


Here’s what the situation would look like as a table

               | Vanilla | Chocolate | Total |
Cilantro-free  |   75    |    25     |  100  |
Cilantro-add   |   50    |    50     |  100  |
Total          |   125   |    75     |  200  |

Now, we already know that they had a 75% chance of getting a vanilla cookie if they chose the CF bowl. But that’s NOT the question at hand. The question at hand is related, but different: What is the chance that they chose the CF bowl if they got a vanilla cookie. Let’s watch a replay:

We know: Chance they have a vanilla cookie if cookie from CF bowl

We want:  Chance cookie from CF bowl if they have a vanilla cookie

The reason that I stress this is that the conventional method of “Null Hypothesis Significance Testing” (the whole concept of “the null hypothesis was rejected at p<.05” that you see in most papers) is analogous to the first statement, but we almost always want to make decisions based on the value of the second statement. To be even more direct: Most statistical analysis that we see leaves us with a number (p) that is one step short of what we can make decisions on.

Fortunately, there is an equation that can take us from what we’ve know to what we actually want. Unfortunately, it requires additional variables to solve. In this case, we have the additional information. In other cases, we would have to estimate those values, without any method of checking our estimate (until we get more data). Painfully, after all of this work to design a clean experiment, accurately measure results, and methodically run the numbers, out very last step requires us to irrevocably taint our objective results with an estimate that is picked out of the air. It’s so frustrating, and you can start to understand why people try to use the first number, whose definition is so close to sounding like what they need, but that’s just how the math works out. Let me know if you find an alternative.

Lets try it on this case. Here’s the simple derivation of Bayes Theroem:

p(A and B) = p(B and A)

p(A if B) x p(B) = p(B if A) x p(A)

Therefore: p(A if B) = p(B if A) x p(A) / p(B)

Remembering where we left off:

We know: Chance they have a vanilla cookie if cookie from CF bowl

We want:  Chance cookie from CF bowl if they have a vanilla cookie

If A = they have a vanilla cookie

and B =  cookie from CF bowl

Then p(cookie from CF bowl if they have a vanilla cookie) = p(they have a vanilla cookie if cookie from CF bowl) x p(cookie from CF bowl) / p(they have a vanilla cookie)

Note that the first term to the right of the equals sign is ‘we know’, and the final result is ‘we want’. Unfortunately, there are those two other unknowns to calculate, which is where a bit of subjectivity comes in:

p(cookie from CF bowl) means “The overall chance that any cookie (vanilla or chocolate) would be drawn from the CF bowl”. Since there are two bowls, and we don’t know of any reason one would be picked over another, we assume this is 50%. But this is an assumption, and many real-life problems would give alternatives that are clearly not 50/50, without giving clear guidance on whether they should be considered 45/55, 15/85 or .015/99.985. Note that, if  you assume each cookie was equally likely to be selected, this number could be calculated from the total number of cookies in each bowl on the far right column (ie 100 of the 200 total cookies are in the CF bowl)

p(they have a vanilla cookie) means “The overall chance that any cookie would be vanilla”. In this case, simply look at the total number of cookies of each type (the totals on the bottom row of the table) and see that vanilla makes up 125/200 of the total. (NOTE: does this change if the bowls are not equally likely to be selected?)

Once you’ve gotten over the implications  The final calculation is easy:

.75 * (100/200) / (125/200) = .6

It’s also interesting to see that the .75 could be calculated in much the same was as the other two variables (percentage of the total in their row or column), along the top column. Specifically, “within the Cilantro-free column, what portion of cookies are vanilla?”, it’s simply the intersection of CF and Vanilla, divided by the total of the column.

               | Vanilla | Chocolate | Total |
<strong>Cilantro-free  |   75    |    25     |  100  |</strong>
Cilantro-add   |   50    |    50     |  100  |
Total          |   125   |    75     |  200  |

Let’s look at all the factors again in that light:

p(they have a vanilla cookie if cookie from CF bowl):

  • Numerator: # in the intersection of CF and Vanilla
  • Denominator: # of CF

p(cookie from CF bowl)

  • Numerator: # of CF
  • Denominator: # of total cookies

p(they have a vanilla cookie)

  • Numerator: # of Vanilla
  • Denominator: # of total cookies

This is all very symmetric with the definition of our result:

p(cookie from CF bowl if they have a vanilla cookie)

  • Numerator: # in the intersection of CF and Vanilla
  • Denominator: # of Vanilla

44848-keanu-reeves-whoa-gif-nOup

Inspiration: ist-socrates.berkeley.edu/~maccoun/PP279_Cohen1.pdf

Bayes in Excel

Why Excel?

Solving the Cookie Problem

As always, the row labels (Cilantro-Free and Cilantro-Add) are your hypotheses. You can’t know them directly, but the whole goal of this exercise is to use data (that’s the columns) to increase the difference in probability until you can feel comfortable that one hypothesis is likely enough to act on.

p(H|D) = p(D|H)p(H) / p(D)

p(D) = ∑p(D|Hi)p(Hi), or p(D & Hi) for all i, assuming hypotheses are MECE

∑: Sum of all (In this case, we’re saying that the probability of the data is p(D & H) for all possible hypotheses, assuming they are MECE

MECE: Mututally Exclusive and Collectively Exhaustive. This is a great term for a concept that we all intuit but don’t have a good word for. Generally, we should be formulating our set of hypotheses so that there’s no chance that the data could have come any other way. So I’m not saying that the cookie could have come from the Cilantro-Free, Cilantro-Add, or “an unknown variety of other sources”. If there are other sources, I need to make that a definite hypothesis, with a prior, and everything that my first two have.

In this simple case, ∑p(D|Hi) just means p(V & CF) =75 plus  p(V & CA) = 50 <EDIT it’s actually this divided by the total, 200>

Vanilla Chocolate Total
Cilantro-Free 75 25 100
Cilantro-Add 50 50 100
Total 125 75 200

So we have all of this information, and BAM – we draw a Vanilla cookie. That means we have a piece of data. Whereas our previous best guess at where a given cookie came from was 100/200 for each container (from the ‘total’ column, since we had no more specific information), we can leave those totals behind and focus on the ‘Vanilla’ column, where the probability is the percentage of vanilla cookies from each jar.

Vanilla
Cilantro-Free 0.6
Cilantro-Add 0.4

I did all of this with a count of cookies, to keep it intuitive. Let’s reply the exact same thing with probabilities, which will allow us to take the critical leap in our next section:

Vanilla Chocolate Total
Cilantro-Free 37.50% 12.50% 50.00%
Cilantro-Add 25.00% 25.00% 50.00%
Total 62.50% 37.50% 100.00%

This is exactly what we saw above, except that I’ve divided everything by 200 (the total number of cookies). Look at the lower right corner first. The probability that you draw a cookie of either type from either jar is 100%. Awesome. Look at the Total column – the probability that you draw it from the CF jar is 50%, and the same from the CA jar, because they had an equal number of cookies. That’s p(H), by the way. Similarly, p(D) is the probability that you drew that vanilla. It’s 62.5%, because there are more Vanilla cookies overall. So, returning to the central equation:

p(H|D) = p(D|H)p(H)/p(D)

p(H|D) = (37.5%/50%) x 50% / 62.5%

p(H|D) = 60%

What we really just did was say “Multiply the priors of each hypothesis by how likely they were to give the data, then divide everything by the sum to make them sum to one again”. In this case, you’re saying “Now that we know we have a vanilla cookie, the chances of the cookie coming from CF are 75 (number of vanilla cookies in CF) over 75 + 50 (number of vanilla cookies anywhere)”. That’s pretty intuitive when visualizing cookies, but it feels a little weirder when talking about probabilities. But those numbers are just the same, divided by the total number of cookies.

Where it gets really interesting is if you did this cookie draw again. Imagine that the vanilla cookie was put back in its original jar, and then another cookie was randomly drawn (I can’t think of a story that would justify this, but work with me here). Here’s the deal. You can actually think of those two draws as one piece of data: Vanilla-Vanilla, Vanilla-Chocolate, Chocolate-Vanilla, or Chocolate-Chocolate. You simply multiply the original probabilities together (so p(Vanilla Chocolate | CF) = 37.5%x 12.5%). Here’s how that plays out:

First Draw Vanilla Chocolate
Second Draw Vanilla Chocolate Vanilla Chocolate Total
Cilantro-Free 14.06% 4.69% 4.69% 1.56% 25.00%
Cilantro-Add 6.25% 6.25% 6.25% 6.25% 25.00%
Total 20.31% 10.94% 10.94% 7.81% 50.00%

<NOTE: Explain why this doesn’t sum to 100% – I think it’s because this table makes the assumption that you’re drawing from the same jar twice, which is a 50% chance)

Or, dividing everything by the total-total again:

First Draw Vanilla Chocolate
Second Draw Vanilla Chocolate Vanilla Chocolate Total
Cilantro-Free 28.13% 9.38% 9.38% 3.13% 50.00%
Cilantro-Add 12.50% 12.50% 12.50% 12.50% 50.00%
Total 40.63% 21.88% 21.88% 15.63% 100.00%

So if you drew Vanilla-Vanilla from the same jar, your probability of it coming from CF is 28.13% / 40.63% = 69.23%. Note that this is higher than the 60% certainty you had with one draw, and any further straight vanilla draws would increase the chance even further.

But we’ve already tortured this analogy as far as we can. Cookies in a jar are easy to visualize for one draw, but lets move on to something that lends itself more easily to repeated experimentation

Solving the Coin Problem (Simplified)

One confusing part about going from the cookie example to this coin problem is that the hypotheses are now numeric, which is convenient for calculation, but  a little confusing. Let’s say that we encounter a weird-shaped coin, and we have literally no idea what the odds are of ‘heads’ versus ‘tails’ (however that’s defined on this coin). Instead of the hypotheses being that you’re drawing from one jar instead of another (both of which exist and have a definite number of cookies of each type), it’s that you’re flipping one of several possible coins (only one of which exists, and has a single attribute “probability of heads”). To make this somewhat more concrete, here’s a table like the one we saw for cookies, but for a set of “probability of heads” possibilities:

0 Heads, 0 Tails
Probability of Heads Heads Tails Total
0% 0 0.2 0.2
25% 0.05 0.15 0.2
50% 0.1 0.1 0.2
75% 0.15 0.05 0.2
100% 0.2 0 0.2
Total 0.5 0.5 1

It’s critical that we keep in mind that the left most column is for hypotheses, not results. So “25%” is just like “Cilantro Free” or “Cilantro Added”, in that it’s the thing we want to figure out, and which we’re going to build confidence in based on data.

Speaking of data, we still have two possible outcomes, now Heads and Tails instead of Vanilla and Chocolate.

We start by looking at our chances if we’ve have 0 flips of the coin. We’re assuming that all hypotheses are equally likely, including the possibility that the coin could never turn up Heads, as well as the possibility that it will always turn up heads.

Let’s look down the ‘Heads’ column (remember, each of these numbers represents the probability that both the hypothesis and the data are true). We shouldn’t be surprised to see that the there’s no chance that both our hypothesis that Heads is impossible, and that our flip ends up Heads. To look at it the other way, if our flip ends up Heads, we have eliminated this hypothesis forever, even if we get straight Tails after that.

What’s interesting is that the opposite is not true. Toward the bottom of the column, we see that the assumption that the coin will always flip Heads does not necessarily go to 100% if we get a heads. That would be true even if we got 1,000 Heads in a row (or a million, or an billion). As long as there are other hypotheses that could also explain that string of Heads (even if the hypothesis that Heads  is 50% likely, and you’ve just gotten exceedingly lucky), you can’t say for sure that the 100% Heads hypothesis is true. This is the mathematical reasoning behind the statement “No amount of data can prove a theory, but one data point can refute it”, or the story that people assumed that all swans were white, until someone saw a single black swan and *poof*, obliterated the theory that had seemed clear for hundreds of years.

Let’s say we got Heads on that first flip. That’s like filtering out the Tails column, leaving just the Heads (note that there’s now a probability of 1 at the bottom of the Heads column, since that’s what actually happened):

1 Heads, 0 Tails (end of first flip)
Probability of Heads Heads Tails Total
0% 0 0
25% 0.1 0.1
50% 0.2 0.2
75% 0.3 0.3
100% 0.4 0.4
Total 1 1

So what if we got a Heads on our first flip, and we wanted to flip again? This is analogous to the end of the Cookie section, where we drew a second cookie from the same jar. Here, we’re getting a second result from the same coin – another output of the same unknown system.

We constructed a somewhat awkward table for the Cookie, with a two layered column header, which showed the probability of every pair of outcomes. For simplicity’s sake, I’m going to ignore the columns where Tails came up first, since we’ve already gotten one Heads:

1 Heads, 0 Tails
Heads Tails
0% 0 0 0
25% 0.0125 0.0375 0.05
50% 0.05 0.05 0.1
75% 0.1125 0.0375 0.15
100% 0.2 0 0.2
0.375 0.125 0.5

What’s interesting here is that the equations are exactly the same, but the prior was copy-pasted from the Heads column of the last table. That’s because, when we flipped that first Heads, we learned more about the coin. For example, we learned that it was impossible that it could have a 0% chance of heads. Now I take that prior, multiply it by the probability of getting another Heads for each hypothesis, and get the probability of Heads-Heads for each of them. Note that the Total-Total is just .5, instead of 1. That’s because we’re looking at a sub-universe, in which we got heads for that first flip (which had a probability of .5, given our priors). I didn’t do the re-balancing that we saw in the one-column table before, so each box here is the probability of Heads-Heads, as judged before we’ve flipped anything.

Here’s where we start to see the future – we can just as easily do this a third time:

2 Heads, 0 Tails
Heads Tails
0% 0 0 0
25% 0.003125 0.009375 0.0125
50% 0.025 0.025 0.05
75% 0.084375 0.028125 0.1125
100% 0.2 0 0.2
0.3125 0.0625 0.375

Again, the Heads column here represents Heads-Heads-Heads, as judged from the standpoint of not having flipped the coin at all. We can basically say, before flipping anything, “There is a .025 chance that the coin is 50% likely to be heads, and we get Heads-Heads-Heads”.

Let’s say that we did get  Heads-Heads-Heads. then the Total at the bottom will be 1, and everything else in the column scales up by the same factor. So it would look like this:

2 Heads, 0 Tails
Heads
0% 0
25% 0.01
50% 0.08
75% 0.27
100% 0.64
1

So, if we got Heads-Heads-Heads, we’d say there was a .64 chance that the coin was 100% likely to come up heads.

Solving the Coin Problem

This ‘copy-paste the old posterior as the new prior’ thing makes some intuitive sense (each flip is like a new test, starting with the results from the last), but it gets awkward. Particularly if you end up with a mixture of heads and tails. But if you remember the fundamental equation that we’ve got going here:

p(H|D) = p(D|H)p(H)/p(D)

We can connect it with the extremely large set of tools for calculating p(D|H). In this case, it would be the binomal distribution.

Probability of first Heads, as calculated using =BINOMDIST(1,1,HYPOTHESIS,False)*PRIOR

Heads Tails
0 0 0.2 0.2
0.25 0.05 0.15 0.2
0.5 0.1 0.1 0.2
0.75 0.15 0.05 0.2
1 0.2 0 0.2

This matches the other method

HHH
0 0
0.25 0.003125
0.5 0.025
0.75 0.084375
1 0.2
0.3125

Then, normalizing

H-H-H
0% 0
25% 0.01
50% 0.08
75% 0.27
100% 0.64
1

BAM – exactly what we got the other way

Let’s extend it to 150 heads and 10 Tails. Because, we can!

Heads
0% 0
25% 0.0000002081245439
50% 0.004546550037
75% 0.9954532418
100% 0

100%, which had been in the lead, is thrown out by that single Tails. 75% is now the clear favorite.