# On the ratio of Beta-distributed random variables

Tagged:`CatBlogging`

/
`Math`

/
`Statistics`

/
`TheDivineMadness`

[Warning: Post contains full frontal nerdity. Bug reports appreciated!] I finally got a copy of Pham-Gia’s paper on the distribution of the ratio of 2 independent Beta-distributed random variables. While I still have some childhood trauma around hypergeometric functions like ${}_{2}F_{1}()$ and its scarier brother ${}_{3}F_{2}()$… time to face my fears.

## The three B’s: Bernoulli, Binomial, and Beta

Suppose you flip a loaded coin that has probability $p$ of coming up heads. That’s a Bernoulli distribution, with just 2 discrete values.

Now suppose you do that $N$ times, and observe that heads comes up $k$ times. That’s a binomial distribution:

\[\Pr(k | N, p) = \binom{N}{k} p^k (1-p)^{(N-k)}\]Next, suppose you don’t know how *much* the coin is loaded. Somebody lets you take $N$
flips, and you observe $k$ heads. What should you believe about $p$, the probability the
coin comes up heads? A naïve estimate would just give the single point value
$p = k/N$.

A better approach is to regard $p$ as a random variable, whose distribution you’re inferring from experiment. The Bayesian way to do this is to start with a prior distribution that summarizes your knowledge before the experiment. If you don’t know anything, then it’s hard to beat a uniform distribution on $[0, 1]$. This can conveniently be done with the Beta distribution of the first kind:

\[\Pr(p | \alpha, \beta) = \frac{p^{\alpha - 1} (1 - p)^{\beta - 1}}{B(\alpha, \beta)}\]where the normalization is $B(\alpha, \beta)$ is the complete Beta function.

It’s pretty clear that the uniform distribution is $\alpha = 1, \beta = 1$. At that point
it’s a pretty straightforward application of Bayes Rule to show that the posterior
distribution for $p$ will be Beta-distributed with parameters
$\alpha = k + 1, \beta = N - k + 1$. So all you have to do is *count* the number of
trials and successes to get a posterior probability distribution that reflects your
complete knowledge (and uncertainty!) of $p$.

## Why we’re interested: vaccine efficacy confidence intervals

The reason we’re interested in this is vaccine efficacy confidence intervals. Basically you have $N_v$ people enrolled in the vaccine arm of the trial, and see $I_v$ infections. At the same time, you have $N_c$ people enrolled in the control arm, and observe $I_c$ infections.

The coin we flipped above here is: heads you get the disease, tails you don’t. We’d like to know how much the vaccine lowers your risk of disease.

So point estimates of the probability of infection in each arm are:

\[\begin{align*} p_v &= I_v / N_v \\ p_c &= I_c / N_c \end{align*}\](We’d of course like to use a posterior Beta distribution instead of a point estimate here.)

Then the vaccine efficacy is how much the risk is lowered, compared to the unvaccinated control arm:

\[VE = \frac{p_c - p_v}{p_c} = 1 - \frac{p_v}{p_c}\]Now if we believe that $p_v$ and $p_c$ are Beta-distributed, given the clinical trial as a bunch of disease-catching coin flips, then vaccine efficacy is distributed as the ratio of a couple of independent Beta variables.

Ok, so what’s the distribution of a ratio of independent Beta variables? There are a variety of ways to approach this, and we’ll compare several of them in an upcoming post. For now, we’re going to fight our way through a paper which gives the exact Bayesian result.

## Exact solution: the probability distribution function (PDF) of a ratio of 2 Beta-distributed random variables

We’ll look specifically at the application of all this to the case of vaccine efficacies
in a later post. For now, let’s just examine the mathematical question of what the
distribution is for the ratio of two Beta-distributed variables. The exact solution was
published in 2000 by Pham-Gia. ^{[1]} It lives behind an
execrable paywall, and was thus devilishly difficult to acquire without paying ransom.
Fortunately, I know people who know people who know people; somebody or other in that
chain found it or had institutional access, and passed it back up the chain. Whoever you
are, my thanks.

### The problem

Consider 2 Beta-distributed variables $p_1$ and $p_2$:

\[\begin{align*} \Pr(p_1) &= \frac{p_{1}^{\alpha_{1}-1}(1-p_{1})^{\beta_{1}-1}}{B(\alpha_1, \beta_1)} \\ \Pr(p_2) &= \frac{p_{2}^{\alpha_{2}-1}(1-p_{2})^{\beta_{2}-1}}{B(\alpha_2, \beta_2)} \end{align*}\]We then ask: if we compute their ratio $R = p_1 / p_2$, then what is the PDF $\Pr(R)$? If we knew the answer, particularly the CDF (or even better the quantile function, which is the functional inverse of the CDF!), we could calculate 95% confidence intervals on the ratio.

### Ranges

It’s important to understand the ranges of each of the variables $p_1$, $p_2$, and $R$ so that as we transform variables we don’t accidentaly step outside reality. This will help us keep the integration limits straight. Because $p_1$ and $p_2$ are from the standard Beta distribution, we have:

\[0 \le p_1, p_2 \le 1\](Usually these are proportions or probabilities, so we certainly want to stay in $[0, 1]$!)

$R$, on the other hand, is a bit more gnarly. Since both $p_1$ and $p_2$ are non-negative, then surely 0 is a lower bound for $R$. But the denominator $p_2$ can of course be arbitrarily near 0, so if at the same time the numerator $p_1$ is finite (stepping carefully around the land mine at 0/0), then the upper bound must be infinite:

\[0 \le R \le +\infty\](Values of $R \gt 1$ will, when we apply this to vaccine efficacies, result in *negative*
efficacies. Is that meaningful? Sure: your vaccine could make things *worse*, making the
vaccinees *more* susceptible to disease… which is surely negative efficacy, no?)

### Changes of variables

Our strategy is to start with the joint density $\Pr(p_1, p_2)$ and then do various tortured changes of variables to eliminate one of $p_1$ or $p_2$ in favor of $R$, and express the residual integral in terms of a hypergeometric function ${}_{2}F_{1}()$.

How complicated can it be? Well, there’s lots of detail coming up, but really it comes down to the fact that with 3 variables $p_1, p_2, R$ there are only 2 ways to eliminate one of them in favor of $R$. Either we substitute out $p_1$ in favor of $p_2, R$ via:

\[\begin{align*} p_1 &= p_2\, R \\ dp_1 &= p_2\, dR \end{align*}\]This is appropriate for $0 \le R \le 1$, since if $R \gg 1$ it could force a value of $p_1 > 1$, which takes us out of its allowed range.

Or we substitute out $p_2$ in favor of $p_1, R$ via:

\[\begin{align*} p_2 & = \frac{p_1}{R} \\ dp_2 &= -\frac{p_1}{R^2} dR \end{align*}\]This is correspondingly appropriate for $1 \lt R$, as it guarantees $p_2 \le 1$, as the range requires. (We’ll eventually lose the minus sign, taking absolute value of Jacobians.)

So we’ll need to do *both* transformations, piecewise over the range of $R$.

Double the workload. Le sigh… who coulda seen *that* coming?

### Reading off the distribution of $R$

I like to do these changes of variable by looking at the normalization integral for the joint distribution. That way, as you change variables, the integration measure will force you to pick up the Jacobian properly. The joint distribution of $p_1$ and $p_2$ is, since they’re assumed independent, just the product of their individual distributions. So the normalization integral is pretty straightforward to write down:

\[1 = \frac{1}{B(\alpha_1, \beta_1) B(\alpha_2, \beta_2)} \int_0^1\!\!\!\!dp_1 \int_0^1\!\!\!\!dp_2\, p_1^{\alpha_1 - 1} (1-p_2)^{\beta_1 - 1} p_2^{\alpha_2 - 1} (1-p_2)^{\beta_2 - 1}\]Next, we’ll use *both* the transformations above to get the integral in 2 pieces, one
using $(p_2, R)$ and another using $(p_1, R)$. The first will involve an integral over
$R$ from 0 to 1, while the second will integrate from 1 to $+\infty$. Then we’ll do a
little razzle-dazzle high school algebra to pull the $R$ integrations to the left, and
thus be able to read off the distribution of $R$. It’ll be a piecewise function, with one
piece for $0 \le R \le 1$ and another for $R \gt 1$:

From this we can directly read off the PDF for $R$, piecewise for $0 \le R \le 1$ and similarly for $R \gt 1$, respectively from each of the 2 terms:

\[\left\{ \begin{alignat*}{6} \Pr(R | 0 \le R \le 1) &= \frac{1}{B(\alpha_1, \beta_1) B(\alpha_2, \beta_2)} & \cdot & R^{\alpha_1 - 1} & \cdot & \int_0^1\!\!\!\!dp_2\, p_2^{\alpha_1 + \alpha2 - 1} (1-p_2)^{\beta_2 - 1} (1-Rp_2)^{\beta_1 - 1} \\ \Pr(R | R \gt 1) &= \frac{1}{B(\alpha_1, \beta_1) B(\alpha_2, \beta_2)} & \cdot & \frac{1}{R^{\alpha_2 + 1}} & \cdot & \int_0^1\!\!\!\!dp_1\, p_1^{\alpha_1 + \alpha_2 -1} (1-p_1)^{\beta_1 - 1} \left(1 - \frac{p_1}{R}\right)^{\beta_2 - 1} \end{alignat*} \right.\]These still contain a residual $p$-integral each, but we’ll see next how to interpret those in terms of the hypergeometric function ${}_{2}F_{1}()$ with various tortured arguments.

### Hypergeometric functions

Ok, so what’s this hypergeometric thingummy I’ve been mumbling about? I approach this
subject with some caution, due to some childhood trauma around hypergeometric
functions. ^{[2]}

Hypergeometric functions got their start in the 1600s, but hit it big in the 1800s with major players like Euler, Gauss, Riemann, and Kummer coming up for bat. They’re at once mind-numbingly complex (at least to me, since they’re kind of my kryptonite) with a bajillion special cases, and amazingly versatile. You can express almost all the special functions of theortical physics (Bessel functions and the like) in terms of hypergeometric functions.

So there’s a temptation: if you can just learn everything about hypergeometric functions,
you can master nearly all of 19th century physics. The bug, of course, is that *nobody*
can master all of the lore of hypergeometric functions. Least of all me!

Like most magical artifacts, they are best approached by wizards and avoided by mundanes. I am not a wizard in these matters, and thus approach with some fear and considerable respect for the virtue of keeping one’s fingers out of the gears.

As a matter of definition, in the unit disk $|z| \lt 1$ in the complex plane, the hypergeometric function of interest at the moment is defined by a power series (and by analytic continuation outside the disk, stepping carefully around the branch points at 1 and infinity):

\[{}_2F_1(a, b; c; z) = \sum_{n=0}^{+\infty} \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!} = 1 + \frac{ab}{c} \frac{z}{1!} + \frac{a(a+1)b(b+1)}{c(c+1)} \frac{z^2}{2!} + \cdots\]If $c$ is a non-positive integer, it’s undefined or infinite. The funny parenthetical dingus is the rising Pochammer symbol:

\[(q)_n = \left\{ \begin{array}{ll} 1 & n = 0 \\ q(q+1)\cdots(q+n-1) & n > 0 \end{array} \right. = \frac{\Gamma(q+n)}{\Gamma(q)}\]An interesting special case for us will be when $a$ or $b$ are nonpositive integers (as with counts in a clinical trial), in which case the Pochammer symbols eventually include a 0 and the series thus terminates, resulting a polynomial. True, it will be a polynomial of very high order (say 15,000 participants in a trial arm), but a polynomial nonetheless!

That’s all… fine, if you like that sort of thing. But what does it have to do with the integrals we have to do above? Well, it turns out that ${}_{2}F_{1}()$ has an integral representation, as well, apparently due to Euler:

\[{}_2F_1(a,b;c;x) = \frac{1}{B(a, c-a)} \int_0^1\!\!\!\!du\, u^{a-1} (1-u)^{c-a-1} (1-xu)^{-b}\]This is the trick that Pham-Gia used to get the density distribution in closed form (at least, if you regard ${}_{2}F_{1}()$ as “closed”…), by recognizing the integrals above as special cases of this.

### Expressing the residual $p$-integrals in hypergeometric terms

Basically we take the above equations for $\Pr(R)$ with residual integrals , and recognize that the annoying integral in them can, with a suitable change of variables, be made identical to the integral representation of ${}_{2}F_{1}()$.

#### Case $0 \le R \le 1$:

The dictionary of variables to recognize the hypergeometric integral is:

\[\begin{align*} u &= p_2 \\ a &= \alpha_1 + \alpha_2 \\ b &= 1 - \beta_1 \\ c &= \alpha_1 + \alpha_2 + \beta_2 \end{align*}\]That gives the final result for small $R$ of:

\[\Pr(R | 0 \le R \le 1) = \frac{B(\alpha_1 + \alpha_2, \beta_2)}{B(\alpha_1, \beta_1) B(\alpha_2, \beta_2)} R^{\alpha_1 - 1} {}_2F_1(\alpha_1 + \alpha_2, 1 - \beta_1; \alpha_1 + \alpha_2 + \beta_2; R)\]And that’s Pham-Gia’s first result on p. 2698.

#### Case $1 \le R$:

Here the dictionary is slightly different:

\[\begin{align*} u &= p_1 \\ a &= \alpha_1 + \alpha_2 \\ b &= 1 - \beta_2 \\ c &= \alpha_1 + \alpha_2 + \beta_1 \end{align*}\]That gives the final result for large $R$ of:

\[\Pr(R | 1 \lt R) = \frac{B(\alpha_1 + \alpha_2, \beta_1)}{B(\alpha_1, \beta_1) B(\alpha_2, \beta_2)} \frac{1}{R^{\alpha_2 + 1}} {}_2F_1(\alpha_1 + \alpha_2, 1 - \beta_2; \alpha_1 + \alpha_2 + \beta_1; 1/R)\]And that’s Pham-Gia’s second result on p. 2699.

To summarize the PDF result:

\[\left\{ \begin{alignat*}{6} \Pr(R | 0 \le R \le 1) &= \frac{B(\alpha_1 + \alpha_2, \beta_2)}{B(\alpha_1, \beta_1) B(\alpha_2, \beta_2)} &\cdot& R^{\alpha_1 - 1} &\cdot& {}_2F_1(\alpha_1 + \alpha_2, 1 - \beta_1; \alpha_1 + \alpha_2 + \beta_2; R) \\ \Pr(R | 1 \lt R) &= \frac{B(\alpha_1 + \alpha_2, \beta_1)}{B(\alpha_1, \beta_1) B(\alpha_2, \beta_2)} &\cdot& \frac{1}{R^{\alpha_2 + 1}} &\cdot& {}_2F_1(\alpha_1 + \alpha_2, 1 - \beta_2; \alpha_1 + \alpha_2 + \beta_1; 1/R) \end{alignat*} \right.\]To map between the 2 solutions, swap $\alpha_1 \leftrightarrow \alpha_2$, swap $\beta_1 \leftrightarrow \beta_2$, and swap $R \leftrightarrow 1/R$.

### Continuity at $R = 1$

Pham-Gia did not address in his paper whether the 2 different expressions for $\Pr(R)$ matched up at $R = 1$, i.e., that the probability distribution is continuous. We can show that the above expressions for $\Pr(R | 0 \le R \le 1)$ and $\Pr(R | 1 \lt R)$ are equal in the left and right limits approaching $R = 1$, establishing continuity at that point.

We need 2 identities:

- From Wikipedia’s entry on ${}{2}F_{1}()$ and values at special points comes an identity for evaluating ${}{2}F_{1}()$ at 1, provided $\Re(c) \gt \Re(a+b)$. For the positive values of $\alpha_i, \beta_i$ we’re considering, this is the case so:

- Also, we need to decompose complete Beta functions into Gamma functions, via the standard identity we all learned at our mother’s knees:

#### Case $0 \le R \le 1$:

\[\begin{align*} \Pr(R | 0 \le R \le 1) &\xrightarrow[R \to 1^{-}]{} \frac{B(\alpha_1 + \alpha_2, \beta_2)}{B(\alpha_1, \beta_1) B(\alpha_2, \beta_2)} \frac{\Gamma(\alpha_1 + \alpha_2 + \beta_2) \Gamma(\beta_1 + \beta_2 - 1)}{\Gamma(\beta_2) \Gamma(\alpha_1 + \alpha_2 + \beta_1 + \beta_2 - 1)} \\ &= \frac{1}{B(\alpha_1, \beta_1) B(\alpha_2, \beta_2)} \frac{\Gamma(\alpha_1 + \alpha_2) \Gamma(\beta_2)}{\Gamma(\alpha_1 + \alpha_2 + \beta_2)} \frac{\Gamma(\alpha_1 + \alpha_2 + \beta_2) \Gamma(\beta_1 + \beta_2 - 1)}{\Gamma(\beta_2) \Gamma(\alpha_1 + \alpha_2 + \beta_1 + \beta_2 - 1)} \\ &= \frac{1}{B(\alpha_1, \beta_1) B(\alpha_2, \beta_2)} \frac{\Gamma(\alpha_1 + \alpha_2)\Gamma(\beta_1 + \beta_2 - 1)}{\Gamma(\alpha_1 + \alpha_2 + \beta_1 + \beta_2 - 1)} \\ &= \frac{B(\alpha_1 + \alpha_2, \beta_1 + \beta_2 - 1)}{B(\alpha_1, \beta_1) B(\alpha_2, \beta_2)} \end{align*}\]#### Case $1 \le R$:

\[\begin{align*} \Pr(R | 0 \le R \le 1) &\xrightarrow[R \to 1^{+}]{} \frac{B(\alpha_1 + \alpha_2, \beta_1)}{B(\alpha_1, \beta_1) B(\alpha_2, \beta_2)} \frac{\Gamma(\alpha_1 + \alpha_2 + \beta_1)\Gamma(\beta_1 + \beta_2 - 1)}{\Gamma(\beta_1)\Gamma(\alpha_1 + \alpha_2 + \beta_1 + \beta_2 - 1)} \\ &= \frac{1}{B(\alpha_1, \beta_1) B(\alpha_2, \beta_2)} \frac{\Gamma(\alpha_1 + \alpha_2)\Gamma(\beta_1)}{\Gamma(\alpha_1 + \alpha_2 + \beta_1)} \frac{\Gamma(\alpha_1 + \alpha_2 + \beta_1)\Gamma(\beta_1 + \beta_2 - 1)}{\Gamma(\beta_1)\Gamma(\alpha_1 + \alpha_2 + \beta_1 + \beta_2 - 1)} \\ &= \frac{1}{B(\alpha_1, \beta_1) B(\alpha_2, \beta_2)} \frac{\Gamma(\alpha_1 + \alpha_2)\Gamma(\beta_1 + \beta_2 - 1)}{\Gamma(\alpha_1 + \alpha_2 + \beta_1 + \beta_2 - 1)} \\ &= \frac{B(\alpha_1 + \alpha_2, \beta_1 + \beta_2 -1)}{B(\alpha_1, \beta_1) B(\alpha_2, \beta_2)} \end{align*}\]These 2 expressions being identical, we have established continuity at $R = 1$.

### Smoothness at $R = 1$

We’d like to believe that in addition to being continuous at $R = 1$, the PDF is also smooth, i.e., some rather large number of derivatives are also continuous. There is no particular reason to expect a kink in the PDF here, so it would be nice to know that our piecewise representation of the PDF has not introduced a kink.

This requires differentiating ${}_{2}F_{1}(a, b; c; z)$ with respect to $z$. According to Wikipedia’s hypergeometric function entry, this is pretty straight forward with some nice-looking derivative relations:

\[\begin{alignat*}{4} \frac{d}{dz} &{}_{2}F_{1}(a,b;c;z) &&= \frac{ab}{c} &\times&{}_{2}F_{1}(a+1, b+1; c+1; z) \\ \frac{d^2}{dz^2} &{}_{2}F_{1}(a,b;c;z) &&= \frac{a(a+1)b(b+1)}{c(c+1)} &\times&{}_{2}F_{1}(a+2, b+2; c+2; z) \\ & && \vdots & \\ \frac{d^n}{dz^n} &{}_{2}F_{1}(a,b;c;z) &&= \frac{(a)_n (b)_n}{(c)_n} &\times&{}_{2}F_{1}(a+n, b+n; c+n; z) \end{alignat*}\]… where the last expression for the derivative in the general case again uses the rising Pochammer symbols, just as above in the series definition of ${}_{2}F_{1}(a, b; c; z)$.

We have not performed this analysis yet.

## Ok, what about the cumulative distribution function (CDF)?

That gives us the PDF (probability distribution function); if we want the CDF (cumulative
distribution function) to calculate quantiles, we’ll have to go beyond Pham-Gia’s paper.
Now, it turns out that Julian Saffer has already worked this out, and what’s more put a
library in Python on Github. ^{[3]} Now, I’m not so much with
the Python; I’m more of an R guy. But let’s have a look.

Saffer relies on 2 integrals of hypergeometric functions times algebraic functions, with
results using *generalized* hypergeometric functions ${}_{3}F_{2}()$… even more
fearsome than ${}_{2}F_{1}()$! I don’t know quite how to derive these, though I did manage to
confirm the first at Wolfram:

Now, I have to admit I don’t know where these integrals come from and don’t feel up to trying to derive them myself, at least not today. Maybe another day that hasn’t been spent facing my fears…

But I *can* see that, given these identities, the cumulative distribution functions (the
running integral from 0 to some value of the PDF) will give us the given CDFs. (In
Saffer’s notation, what we’ve called $R$ he calls $w$.)

For $0 \le w \le 1$:

And for $w > 1$:

## Computational realization for practical use

At some point soon, I’d like to implement this in R, perhaps mirroring Saffer’s Python implementation. There are some gnarly issues with numerical roundoff. Even though the hypergeometric series terminates as a polynomial for integer $\beta$’s, one simply cannot naïvely compute a polynomial of order 15,000 for a large clinical trial and expect to get anything other than nonsense!

I note that Saffer sensibly computes everything in log space first, and then exponentiates. This does, though, run the risk of exponentiating the roundoff in a catastrophic fashion. Some rigorous test cases with known answers are required before I’ll trust it.

That’s work for another day.

## The Weekend Conclusion

In a spirit of proper collegiality, I wish to acknowledge warmly the assistance of the Weekend Publisher at several points in this analysis. He provided encouragement when I wanted to give up. He is shown here in mid-critique, providing valuable purr review.

That acknowledgement having been made, we’ve worked our way through the relevant parts of Pham-Gia’s paper (though there’s a lot more there!), and confirmed the primary result that the PDF of the ratio of 2 independent Beta-distributed random variables is a variety of hypergeometric function ${}_{2}F_{1}()$.

Somewhat beyond Pham-Gia’s paper, we’ve also proven the continuity of the PDF at $R = 1$, i.e., that the expression for $0 \le R \le 1$ and the one for $R \gt 1$ match up at $R = 1$.

However, there are several things we *haven’t* done:

- While we have
*continuity*at $R = 1$, we haven’t proven*smoothness*. We don’t strictly need that, but as there’s no particular reason for a kink at $R = 1$, it would be nice to show the derivative from the left equals the derivative from the right. We suspect it’s more than first derivative smooth at $R = 1$, and there may be a way to parlay the derivative relation on ${}_{2}F_{1}()$ into a derivative recurrence relation on $\Pr(R)$ to show that to all orders, if we’re ambitious. - We haven’t confirmed all the details of how to get the CDF; there’s probably some bit of hypergeometric lore that will tell us the integral identities above from which the CDFs follow straightforwardly (or as straightforwardly as things go here).
- We also haven’t wrestled with all the numeric issues of implementing this, say in R, though Saffer’s Python code is probably a good guide.
- Furthermore, we haven’t investigated the quantile function (functional inverse of the CDF) which would let us read off quantiles directly. Looking at Saffer’s code, he hasn’t either: he’s using Newton’s method to solve the relevant equation directly from the CDF, which is totally fair.

So we’ve got a bit more work to do to make this useable in a practical sense.

## Notes & References

1: T Pham-Gia, “Distributions of the ratios of independent beta variables and applications”, *Comm Stat: Theory & Methods*, 29:12, 2693-2715. DOI: 10.1080/03610920008832632. Since it was so hard to get, it’s archived here.

**NB:** We believe there are several errata in this paper which make it much harder to read
than need be. We’ve worked through the details, and with these corrections, obtain the
same eventual result in terms of ${}_{2}F_{1}()$. Specifically:

- p. 2696, in the definition of the Pochammer coefficients: $K$ should be $\ldots$
- p. 2698, in the equation for the marginal density $f(w)$:
- The upper limit of integration should be $+1$, not $+\infty$
- The exponent of $(1-x_2)$ should be $\beta_{2}-1$, not $\beta_{2}$

- p. 2698, in the integral for $f(w)$ for $0 \le w \le 1$:
- the integration measure is missing, and should be $dx_2$
- in the rightmost expression, the exponent of $(1-x_2)$ should again be $\beta_{2}-1$, not $\beta_{2}$

- p. 2703, in the variables for a Dirichlet distribution, $K$ should again be $\ldots$

While there may or may not be similar typos in the rest of the paper, we haven’t checked carefully since it does not bear directly on our interests. But with the corrigenda above, we were able to reproduce Pham-Gia’s main result, the piecewise PDF on pp. 2698-2699. ↩

2: OK, the truth is that I was actually 23 years old and in my first year of physics grad school at MIT. I got wrapped around the axle pretty tight because the notation between a couple texts was confusingly and subtly different. I thought I’d suddenly become stupid! It took years to get past that, and now even 40+ years later it’s a sensitive spot. Still… time to face my fears. ↩

3: J Saffer, “Beta Quotient Distribution”, *GitHub Repository*, last committed 2020-Jun-06, retrieved 2021-Sep-13. ↩

*Written*Mon 2021-Sep-13

## Gestae Commentaria