2026-07-14 22:24:12
I’ve been thinking about ICD-10 codes; they come up a lot in my work.
The ICD-10-CM standard is divided into 21 chapters, which generally correspond to the first letter of a code. However, a chapter may contain blocks beginning with more than one letter, and codes starting with a single letter, namely D, can span two chapters.
Here’s a diagram I made to visualize the relationship between chapters and initial letters of codes.
Notice there’s no letter U on the diagram. That’s because U is reserved for special/provisional codes.
2026-07-13 03:26:29
A few days ago I wrote a post entitled Does additional data always reduce posterior variance?. In a nutshell, the answer is no, not always.
That led the previous post which looked at posterior means for three Bayesian models, showing how the posterior mean is a weighted average of the prior mean and the mean of the new data. The weights are precisions, which means something different for each model.
For the beta-binomial model, variance may increase when seeing unexpected data (details here), but precision always increases.
For the normal-normal model precision is the reciprocal of variance. Every new data point makes precision go up and posterior variance go down.
The Poisson-gamma model may be the most interesting. As stated in the previous post, if data has a Poisson distribution with parameter λ, and λ has a gamma(α0, β0) prior distribution, then the posterior distribution on λ after observing k events over time t has a gamma(α0 + k, β0 + t) posterior distribution. Therefore the posterior variance is
(α0 + k) / (β0 + t)².
Note the posterior variance is an increasing function of k and a decreasing function of t. This means that the posterior variance increases every time an event is observed, and it decreases quadratically between observations.
Here’s an illustration. I simulated data from a Poisson process with λ and used a gamma(1, 1) prior on λ. Here’s a plot of the posterior variance.
2026-07-13 01:48:14
Common sense says that what you believe after seeing new data should be some sort of compromise between what you believed before and what the new data says. You don’t want to ignore previous information or new information.
How much should new data change your prior beliefs? When prior judgment and new information are in conflict, which one should be given the benefit of the doubt?
Bayesian data models provide a framework for making such decisions quantitative and objective. The choice of a data model is somewhat subjective—whether it’s a Bayesian model or not—but given a Bayesian model, the rules for updating the representation of your beliefs are objective. As some put it, you “turn the Bayesian crank.” A likelihood model and a prior on parameters together specify how new data changes the prior distribution into a posterior distribution.
We will make this more concrete with three examples.
Suppose that data X has a normal distribution with unknown mean μ and known variance σ², and we assume that a priori μ has a normal distribution with mean μ0 and variance σ0².
After observing x, the posterior distribution on μ also has a normal distribution, but with a different mean and variance. Its mean is somewhere between the prior mean and x. We will ignore the change in the variance for this post.
The posterior mean of μ is
This equation becomes more understandable when we introduce precisons τ = 1/σ² and τ0 = 1/σ0².
Then we have
which you can read as saying the posterior mean is the weighted average of the prior mean and x, with the weights given by the precision. Intuitively, you take the weighted mean of your conclusions from previous data and new data, weighting the mean according to how much confidence you have in each.
Now let’s switch over to a different data model. Now assume X is a binary random variable, with probability of success p and probability of failure 1 − p, and we assume p has a beta(a, b) distribution.
After observing s successes and f failures, the posterior mean of the distribution on p becomes
We can rewrite this as
This says that the posterior mean is the weighted average of the prior mean a/(a + b) and the mean of the data s/n. The weights are the prior effective sample size a + b and the sample size of the new data n. In this example (effective) sample size is playing the role that precision played in the normal-normal model above.
Suppose data have a Poisson distribution with parameter λ, and λ has a gamma(α0, β0) prior distribution [1]. And suppose you observe k events over time t. Then the posterior distribution of λ given the data has a gamma(α0 + k, β0 + t) prior distribution and the mean of the posterior distribution is given by
As before, the posterior mean is a weighted average of the prior mean and new data, and the weights are interpretable as some sort of measure of confidence, namely time. The variable t is directly time and the parameter β0 is sort of an effective time, just as a + b is an effective sample size for the beta distribution.
In each example the posterior mean is the weighted average of the prior mean and the mean of the data, with the weights given by a precision. However, precision means something different in each example. In the normal-normal model, precision is the reciprocal of variance, but in the beta-binomial model precision is sample size and in the Poisson-gamma model precision is time.
What all three examples have in common is that they are conjugate models using distributions from the “exponential family” of probability distributions. In technical terms, precision is the multiplicative factor on the sufficient statistic in the exponent of the posterior kernel.
[1] There are multiple conventions for parameterizing the gamma distribution. Here we’re using the shape-rate parameterization, where the mean is α/β.
The post Posterior mean first appeared on John D. Cook.2026-07-12 05:30:11
Years ago I wrote about Gilbreath’s conjecture. It’s a simple conjecture; you could explain it to anyone who understands what prime numbers are. See the linked post for a description of the problem.
Gilbreath’s conjecture is simple, but it’s also kinda weird. As I wrote before,
Paul Erdős speculated that Gilbreath’s conjecture is true but it would be 200 years before anyone could prove it. I find Erdős’s conjecture more interesting than Gilbreath’s conjecture.
The conjecture is hard in a way that, say, solving a nasty-looking differential equation is not. Over the last three centuries, mathematics has developed quite a toolbox for solving differential equations. But Gilbreath’s conjecture is just odd enough that it’s not at all clear what kind of tool might be useful in approaching it.
Terence Tao has a new blog post announcing a paper that he and two coauthors wrote on a random model intended to mimic Gilbreath’s calculation on primes. This random model is more sophisticated than the little game Gilbreath was playing, but it’s also much more amenable to analysis by established techniques. Tao’s post gives a heuristic explanation for why Gilbreath’s conjecture is plausible, but then adds
However, it seems well beyond current technology to try to make these heuristics rigorous; even the first step … is far out of reach.
2026-07-06 22:22:35
I ran across a geometry theorem with the following diagram.
The theorem corresponding to the diagram is interesting, but I found reproducing the diagram more interesting.
The segment AB is a diameter and the line CD is perpendicular to the diameter.
Assume the outer circle is a unit circle. I guessed C = (cos(1), sin(1)) and made the following diagram.
I guessed the value of C by eyeballing it, but in retrospect this would have been a convenient value for the creator of the original diagram to have chosen.
Drawing the blue circle inscribed in the triangle was easy using the equations for the center and radius from this post. Drawing the other two circles, the green and orange circles, was harder. They are also inscribed circles, but not inscribed in a triangle. They’re inscribed in a three-sided figure with two perpendicular sides and a circular arc.
The radius r of the green circle is the distance from the center of the circle to each of its tangent lines. Also, the distance from the origin to the center of the circle must be 1 − r. This is enough information to set up a quadratic equation for r. The same reasoning applies to the orange circle.
The original diagram comes from [1] and the theorem it illustrates says the diameter of the blue circle equals the sum of the radii of the green and orange circles.
In case you’re interested, here’s the code that created the diagram.
#!/usr/bin/env -S uv run --script
# /// script
# dependencies = ["numpy", "matplotlib"]
# ///
import numpy as np
import matplotlib.pyplot as plt
def connect(A, B, color='gray'):
plt.plot([A[0], B[0]], [A[1], B[1]], color=color, linewidth=2)
def circle(c, r, color='gray'):
t = np.linspace(0, 2*np.pi)
plt.plot(c[0] + r*np.cos(t), c[1] + r*np.sin(t), color=color, linewidth=2)
def quadratic(a, b, c):
det = b**2 - 4*a*c
return ((-b - det**0.5)/(2*a), (-b + det**0.5)/(2*a))
A = np.array([-1, 0])
B = np.array([ 1, 0])
C = np.array([np.cos(1), np.sin(1)])
a = np.linalg.norm(B - C)
b = np.linalg.norm(A - C)
c = np.linalg.norm(B - A)
s = (a + b + c)/2
circle([0,0], 1)
connect(A, B,)
connect(A, C)
connect(C, B)
connect(C, C*np.array([1, -1]))
center = (a*A + b*B + c*C)/(2*s)
radius = 0.5*a*b/s
circle(center, radius, 'C0')
Ex = C[0]
roots = quadratic(1, 2 + 2*Ex, Ex**2 - 1)
r = roots[1] # Smaller root is negaive
print(roots)
center = (r + Ex, -r)
circle(center, r, 'C1')
roots = quadratic(1, 2 - 2*Ex, Ex**2 - 1)
r = roots[1] # Smaller root is negaive
center = (Ex - r, -r)
circle(center, r, 'C2')
plt.gca().set_aspect("equal")
plt.axis("off")
plt.show()
[1] Leon Bankoff. A Geometrical Coincidence. Mathematics Magazine, Vol. 37, No. 5 (Nov., 1964), p. 324.
The post Reproducing a geometry theorem diagram first appeared on John D. Cook.2026-07-06 20:22:29
I ran across the approximation
e ≈ 2721/1001
recently. What makes this remarkable is its accuracy relative to the size of the denominator.
You can create a trivial approximation just by truncating a decimal expansion
e ≈ 2718/1000
but this is only good to four significant figures, but 2721/1001 is good to seven, almost eight, significant figures.
e = 2.71828182… 2721/1001 = 2.71828171…
The comparison is more impressive in binary.
$ bc -l >>> obase=2 >>> 2721/1001 10.10110111111000010100… >>> e(1) 10.10110111111000010101…
The denominator is a 10-bit number but the approximation is accurate to 21 bits.
The post e approximation first appeared on John D. Cook.
