Median of ratio of independent variates with Beta distributions

Let $X, Y$ be independent random variables where $X \sim Beta(\alpha_1,\beta_1)$, $Y \sim Beta(\alpha_2,\beta_2)$, and $Z = X/Y$.

Recall $X, Y$ are supported on $(0,1)$, so $z > 0$.

I've computed several $pdf$'s of the distribution of $Z$ (for varying Beta's) using a formula from a paper I found and paid for online, and have discovered to my dismay that the median of $Z$ is $1$.

I would love to submit my code but don't want to give away this formula because it costs \$48, but let's just say I am very sure there are no errors in it. For one thing, the area under the curve on$(0,\infty)$is approximately$1$(within 1e-8 according to Python's scipy.integrate.quad function) when I test several values for$\alpha_i,\beta_i$. My question is: Is this correct? Expected? Obvious even? Possibly obvious from the [hidden] formula which I'd assume most academics have access to? *Note: The reason this is surprising is because while the mean of the distribution behaves as one might expect:$\mathbb{E}X_1/\mathbb{E}Y_1 \leq \mathbb{E}X_2/\mathbb{E}Y_2 \implies \mathbb{E}Z_1 \leq \mathbb{E}Z_2$, the median does not seem to budge. Happy to provide screenshots if anyone is interested. • Formulas are not secrets. If you want this question answered, then please be clear and state the formula. – whuber Jul 17 '15 at 10:18 • I don't believe the claim in your question; I expect there's some omitted detail. – Glen_b Jul 17 '15 at 10:39 • It's not hard to show that the median is a differentiable function of the four parameters with everywhere nonzero gradient. Consequently, the Implicit Function Theorem implies the set of parameters for which$Z$has median$1$is a codimension-one submanifold of$(0,\infty)^4$. Consequently, it is almost never the case that the median of$Z$is$1$. Where you write about the "area under the curve is approximately$1$," it appears you have rediscovered the fact that the total probability of any random variable must be unity. – whuber Jul 17 '15 at 12:18 • @whuber I will try to respond in more detail a bit later today, but for now will clarify that when I stated above the "the area under the curve is 1" was nothing to do with rediscovering the first axioms of what a probability measure was, but rather I was using this to spot check the formula for the pdf. I appreciate the vote of confidence though. – user3659451 Jul 17 '15 at 14:56 • Thank you for that clarification--I certainly was wondering what that remark in your question had to do with the median! That kind of calculation is indeed a smart way of checking a numerical integration routine. It still would be nice to see a specific example of a (nontrivial) quartet of parameters for which your calculation indicates the median is$1$. – whuber Jul 17 '15 at 15:57 2 Answers Without providing details and the precise meaning, the random variable$Y$goes to$0$when$\beta_2 \to \infty$and$X/Y$goes to$\infty$. Hence the median of$X/Y$should go to$\infty$too, and it can't be always$1$. There is a case in which your statement is easily seen to be true. The median is$1$means$\Pr(X/Y>1)=\frac12$or in other words$\Pr(X>Y)=\frac12$. Then the statement is true when$X$and$Y$have the same distributions ($\alpha_1=\alpha_2$and$\beta_1=\beta_2$), because of symmetry (thus a possible error in your 48\$ code is that it always sets $\alpha_1=\alpha_2$ and $\beta_1=\beta_2$).

• I needed to sleep on this and went and saw my code was listing the same parameters accidentally. – user3659451 Jul 17 '15 at 15:09

As I mentioned in comments, I don't believe the claim in your question (that the median of the ratio of two beta variates is 1); I expect there's some omitted detail.

Here's two such ratios, a Beta(10,1)/Beta(1,10) and Beta(1,10)/Beta(10,1) -- they have very different medians, neither anywhere near 1.

(R code)

x=rbeta(100000,10,1)
y=rbeta(100000,1,10)
plot(ecdf(x/y),xlim=c(0,20),ylim=c(0,1))
lines(ecdf(y/x),col=2)
abline(h=.5,col=8,lty=2)
abline(v=c(median(y/x),median(x/y)),col=c(2,1),lty=3)

• Hi Glen_b, what are you using to generate this. Where do the functions rbeta, plot, ecdf, lines, median, and abline come from? – user3659451 Jul 27 '15 at 18:18
• Also,, how do we know that the ratio you computed assumes independence of the random variables? – user3659451 Jul 27 '15 at 18:20
• Sorry, I should have said that's R. Those are all R functions. Variates are generated independently* by default ... you have to make them dependent. *(more strictly, pseudorandom numbers are not truly perfectly independent, but they're constructed in such a way that make them behave for essentially any purpose like this as if they were independent; they'll be as independent as scipy's random generation will be) – Glen_b Jul 27 '15 at 18:55
• Thanks -- I'm trying to do the same thing in Python but having some trouble.... will figure it out! – user3659451 Jul 27 '15 at 20:41
• the essential steps of my algorithm were 1. generate $n$ x-values from a beta (skewed left) 2. generate $n$ y-values from a beta (skewed right). 3. Compute the ratios ($U_i=X_i/Y_i$) and find the median, (and plot). I also computed $V_i=Y_i/X_i$ to get the distribution of another ratio. $X_i$ and $Y_i$ will be effectively independent, as will $X_i$ and $X_j$ (and similarly for $Y$). So now $U_i$ and $U_{j\,\{j\neq i\}}$ should be independent as well. (While $U_i$ and $V_i$ will be dependent, that doesn't really impact what we're checking for here.) – Glen_b Jul 28 '15 at 0:31