Probably Overthinking It

If you like this article, you can read more about this kind of Bayesian analysis in Think Bayes.

Recently I found a copy of Probably Overthinking It at a local bookstore and posted a picture on Twitter. Aubrey Clayton replied with this question:

It’s a great question with what turns out to be an interesting answer. I’ll summarize the results here, but if you want to see the calculations, you can run the notebook on Colab.

Assumptions

Suppose you are the author of a book like Probably Overthinking It, and when you visit a local bookstore, like Newtonville Books in Newton, MA, you see that they have two copies of your book on display.

Is it good that they have only a few copies, because it suggests they started with more and sold some? Or is it bad because it suggests they only keep a small number in stock, and they have not sold. More generally, what number of books would you like to see?

To answer these questions, we have to make some modeling decisions. To keep it simple, I’ll assume:

• The bookstore orders books on some regular cycle of unknown duration.
• At the beginning of every cycle, they start with k books.
• People buy the book at a rate of λ books per cycle.
• When you visit the store, you arrive at a random time t during the cycle.

We’ll start by defining prior distributions for these parameters, and then we’ll update it with the observed data.

Priors

For some books, the store only keeps one copy in stock. For others it might keep as many as ten. If we would be equally unsurprised by any value in this range, the prior distribution of k is uniform between 1 and 10.

If we arrive at a random point in the cycle, the prior distribution of t is uniform between 0 and 1, measured in cycles.

Now let’s figure the book-buying rate is probably between 2 and 3 copies per cycle, but it could be substantially higher – with low probability. We can choose a lognormal distribution that has a mean and shape that seem reasonable. Here’s what it looks like.

From these marginal prior distributions, we can form the joint prior. Now let’s update it.

The update

Now for the update, we have to handle two cases:

• If we observe at least one book, n, the probability of the data is the probability of selling k-n books at rate λ over period t, which is given by the Poisson PMF.
• If there are no copies left, we have to add in the probability that the number of books sold in this period could have exceeded k, which is given by the Poisson survival function.

After computing these likelihoods for all possible sets of parameters, we do a Bayesian update in the usual way, multiplying the priors by the likelihoods and normalizing the result.

As an example, we’ll do an update with the hypothetically observed 2 books. Then, from the joint posterior, we can extract the marginal distributions of k and λ, and compute their means.

Seeing two books suggests that the store starts each cycle with 3-4 books and sells 2-3 per cycle. Here’s the posterior distribution of k compared to its prior.

And here’s the posterior distribution of λ.

Seeing two books doesn’t provide much information about the book-selling rate.

Optimization

Now let’s consider the more general question, “What number of books would you most like to see?” There are two ways we might answer:

• One answer might be the observation that leads to the highest estimate of λ. But if the book-selling rate is high, relative to k, the book will sometimes be out of stock, leading to lost sales.
• So an alternative is to choose the observation that implies the highest number of books sold per cycle.

Computing the second is a little tricky — you can see the details in the notebook. But with that problem solved, we can loop over possible values of n and compute for each one the posterior mean values of λ and the implied number of books sold per cycle.

Here’s the implied sales rate as a function of the observed number of books. By this metric, the best number of books to see is 0.

And here’s the implied number of books sold per cycle.

This result is a little more interesting. Seeing 0 books is still good, but the optimal value is around 5. The worst possibility is to see just one book.

Now, we should not take these values too literally, as they are based on a very small amount of data and a lot of assumptions – both in the model and in the priors. But it is interesting that the optimal point is neither 0 nor “as many as possible.

Thanks again to Aubrey Clayton for asking such an interesting question. If you are interesting in the history and future of statistical thinking, you might like his book, Bernoulli’s Fallacy.

previous

Whenever something unlikely happens, it is tempting to ask, “What are the odds?”

In some very limited cases, we can answer that question. For example, if someone deals you five cards from a well-shuffled deck, and you want to know the odds of getting a royal flush, we can answer that question precisely. At least, we can if you are clearly referring to just this one hand.

But if you’ve been playing poker regularly for a decade and then one night you are dealt a royal flush, it might not be clear, when you ask the question, whether you mean the odds of getting a royal flush on one deal, or one evening of play, or some time in your career, or once in all of the poker hands that have every been dealt. Those are different questions with very different answers — in fact, the first is close to 0 and the last is close to 1 (and known to be 1 in this universe).

So, even in a highly constrained environment like a poker game, answering questions like this can be tricky. It’s even worse in real life. Say you go to college in Massachusetts and then two years later you visit Paris, go for a walk in the Tuileries Garden, and run into a friend from college. What are the odds? Now we have to define both “in how many attempts?” and “odds of what?” Meeting this friend in this particular place? Or any old friend in any unexpected place?

Now let’s put all of this thinking to the test with an example, which is the most surprising thing that has happened to me since the time I ran into a college friend in Paris. Two days ago I was working on a heat vent in my house and wanted to attach this socket

to this screwdriver

But the socket takes a 1/4 inch square drive, and the screwdriver takes 1/4 inch hex bits. I figured there was probably an adapter that could connect them, but I didn’t have one. I thought about getting one, but then I found another way to do the job.

Two days later I went for a walk and about 30 yards from my house, in the middle of the street, I saw a small bit of metal that I picked up just to get it out of the way. And when I looked more closely at what is was — it was a 1/4 inch hex to 1/4 inch square drive adapter.

And here’s how it works.

So, what are the odds of that? I don’t know, but if you have a non-zero prior for the existence of a benevolent deity, you might want to update it.

In the preface of Probably Overthinking It, I wrote:

Sometimes interpreting data is easy. For example, one of the reasons we know that smoking causes lung cancer is that when only 20% of the population smoked, 80% of people with lung cancer were smokers. If you are a doctor who treats patients with lung cancer, it does not take long to notice numbers like that.

When I re-read that paragraph recently, it occurred to me that interpreting those number might not be as easy as I thought. To find out, I ran a Twitter poll. Here are the results:

Some of the people who chose “other” said that there is not enough information — we need to know the absolute risk for one or both of the groups.

I think that’s not right — with just these two numbers, we can compute the relative risk of the two groups. There are a few ways to do it, but a good way to get started is to check each of the multiple choice responses.

Off the bat, “60 percentage points” is just wrong. If the lifetime risk of cancer was 20% in one group and 80% in the other, we could describe that difference in terms of percentage points. But those are not the numbers we were given, and the actual risks are much lower.

But “a factor of 4” is at least plausible, so let’s check it. Suppose that the actual lifetime risk of lung cancer for non-smokers is 1% — in that case the risk for smokers would be 4%. In a group of 1000 people, we would expect 800 non-smokers and 8 cases among them, and we would expect 200 smokers and 8 cases among them. Under these assumptions 50% of people with lung cancer would be smokers, but the question says it should be 80%, so this check fails.

Let’s try again with “a factor of 16”. If the risk for non-smokers is 1%, the risk for smokers would be 16%. Among 800 non-smokers, we expect 8 cases again, but among 200 smokers, now we expect 32. Under these assumptions, 80% of people with lung cancer are smokers, so 16 is the correct answer.

Here are the same numbers in a table.

	Number	Risk	Cases	Percent
Non-smoker	800	1%	8	20%
Smoker	200	16%	32	80%

Now, you might object that I chose 1% and 16% arbitrarily, but as it turns out it doesn’t matter. To see why, let’s assume that the risk is x for non-smokers and 16x for smokers. Here’s the table with these unknown risks.

	Number	Risk	Cases	Percent
Non-smoker	800	x	800x	20%
Smoker	200	16x	3200x	80%

The percentage of cases among smokers is 80%, regardless of x.

Now suppose you are not satisfied with this guess-and-check method. We can solve the problem more generally using Bayes’s rule.

• We are given p(smoker) = 20%, which we can convert to odds(smoker) = 1/4.
• And we are given p(smoker | cancer) = 80%, which we can convert to odds(smoker | cancer) = 4.

Applying Bayes’s rule, we have

odds(smoker | cancer) = odds(smoker) * p(cancer | smoker) / p(cancer | non-smoker)

Rearranging terms, we can solve for the risk ratio:

p(cancer | smoker) / p(cancer | non-smoker) = odds(smoker | cancer) / odds(smoker) = 16

That’s the answer I had in mind, but let me address an objection raised by one poll respondent, who chose “Other” because, “You can’t draw casual inferences from observational data without certain assumptions which I’m unwilling to make.”

That’s true. Even if the risk is 16x higher for smokers, that’s not enough to conclude that the entire difference, or any of it, is caused by smoking. It is still possible either:

(1) that the supposed effect is really the cause, or in this case that incipient cancer, or a pre-cancerous condition with chronic inflammation, is a factor in inducing the smoking of cigarettes, or (2) that cigarette smoking and lung cancer, though not mutually causative, are both influenced by a common cause, in this
case the individual genotype.

If you think that’s the stupidest thing you’ve ever heard, you can take it up with Sir Ronald Fisher, who actually made this argument with apparent sincerity in a 1957 letter to the British Medical Journal. I mention this in case you didn’t already know what an ass he was.

However, if we are willing to accept that smoking causes lung cancer, and is in fact responsible for all or nearly all of the increased risk, then we can use the data we have to answer a related question: if a smoker is diagnosed with lung cancer, what is the probability that it was caused by smoking?

To answer that, let’s assume that smokers are exposed at the same rate as non-smokers to causes of cancer other than smoking. In that case, their 16x risk would consist of 15x risk due to smoking and 1x risk due to other causes. So 15/16 cancers among smokers would be due to smoking, which is about 94%.

Some actual numbers

I rounded off the numbers in my example to make the math easy, so let’s see what the actual numbers are. “Smoking and Cancer“, one of the fact sheets published along with “The Health Consequences of Smoking—50 Years of Progress: A Report of the Surgeon General” includes this figure.

In the most recent interval, the relative risk was about 25x, and at that time about 20% of the U.S. population smoked. The lifetime risk of lung cancer is about 6%, including both smokers and non-smokers, so to find the lifetime risk for non-smokers, we can solve this equation for x:

p(smoker) x + p(non-smoker) (25 x) = 6%

0.8x + 0.2(25x) = 0.06

Which means the lifetime risk is about 1% for non-smokers and 25% for smokers. If we update the table with these numbers, we have

	Number	Risk	Cases	Percent
Non-smoker	800	1%	8	14%
Smoker	200	25%	50	86%

And with that, we can address another point raised by a Twitter friend:

By “smoking increases the risk of lung cancer” you could either mean relative to being a non-smoker or relative to the overall base rate of cancer (including a weighted average of smokers and non-smokers).

I meant the first (which is more common in epidemiology), but if we want the second, it’s about 25 / 6, which is a little more than 4.

Finally, looking at that figure you might wonder why the relative risk of smoking has increased so much. Based on my first pass through the literature, it seems like no one knows. There are at least three possibilities:

• Over this period, cigarettes have been reformulated in ways that might make them more dangerous.
• As the prevalence of smoking has decreased, it’s possible that the number of casual smokers has decreased more quickly, leaving a higher percentage of heavy smokers.
• Or maybe the denominator of the ratio — the risk for non-smokers — has decreased.

In what I’ve read so far, the first explanation seems to get the most attention, but there doesn’t seem to be a clear causal path for it.The second and third explanations seem plausible to me, but I haven’t found the data to support them.

Causes of lung cancer in non-smokers include radon, second-hand smoke, asbestos, heavy metals, diesel exhaust, and air pollution. I would guess that exposure to all of them has decreased substantially since the 1960s. But it seems like we don’t have good evidence that the risk for non-smokers has decreased. That’s surprising, and a possible topic for a future post.

Today is the official publication date of Probably Overthinking It! You can get a 30% discount if you order from the publisher and use the code UCPNEW. You can also order from Amazon or, if you want to support independent bookstores, from Bookshop.org.

I celebrated launch day by giving a talk at PyData Global 2023 called “Extremes, outliers, and GOATs: On life in a lognormal world“. In my opinion, it went well! Here’s the abstract:

“The fastest runners are much faster than we expect from a Gaussian distribution, and the best chess players are much better. In almost every field of human endeavor, there are outliers who stand out even among the most talented people in the world. Where do they come from?

“In this talk, I present as possible explanations two data-generating processes that yield lognormal distributions, and show that these models describe many real-world scenarios in natural and social sciences, engineering, and business. And I suggest methods — using SciPy tools — for identifying these distributions, estimating their parameters, and generating predictions.”

• My slides are available here
• The Jupyter notebook with the examples from Chapter 4 is here
• And the notebook with the additional analysis for the talk is here

When the video is available, I will post it here.