# Do two quantiles of a beta distribution determine its parameters?

If I give two quantiles $(q_1,q_2)$ and their corresponding locations $(l_1,l_2)$ (each) in the open interval $(0,1)$, can I always find parameters of a beta distribution that has those quantiles at the specified locations?

• No, basic counterexample (q1,q2) = (0,1) and (l1,l2) = (0,1) no matter of parameters. – Tim Sep 19 '16 at 10:25
• @Tim I think I see your point, but your counterexample does not satisfy the conditions I specified (for instance that the locations are in the open interval $(0,1)$). – Bota Sep 19 '16 at 10:37
• I think you can do it numerically (and that there will be a unique solution), but it would involve a little effort. – Glen_b Sep 20 '16 at 5:44
• I think too -- the numerical solving is not difficult, but it’s not easy to find an argument for the uniqueness. – Elvis Sep 20 '16 at 6:30
• @Elvis actually, I suspect that there might be a way to do it by looking at the logits of both variables (the OP's $l$ and $q$). – Glen_b Sep 21 '16 at 9:42

The answer is yes, provided the data satisfy obvious consistency requirements. The argument is straightforward, based on a simple construction, but it requires some setting up. It comes down to an intuitively appealing fact: increasing the parameter $$a$$ in a Beta$$(a,b)$$ distribution increases the value of its density (PDF) more for larger $$x$$ than smaller $$x$$; and increasing $$b$$ does the opposite: the smaller $$x$$ is, the more the value of the PDF increases.

The details follow.

Let the desired $$q_1$$ quantile be $$x_1$$ and the desired $$q_2$$ quantile be $$x_2$$ with $$1 \gt q_2 \gt q_1 \gt 0$$ and (therefore) $$1 \gt x_2 \gt x_1 \gt 0$$. Then there are unique $$a$$ and $$b$$ for which the Beta$$(a,b)$$ distribution has these quantiles.

The difficulty with demonstrating this is that the Beta distribution involves a recalcitrant normalizing constant. Recall the definition: for $$a\gt 0$$ and $$b \gt 0$$, the Beta$$(a,b)$$ distribution has a density function (PDF)

$$f(x;a,b) = \frac{1}{B(a,b)} x^{a-1}(1-x)^{b-1}.$$

The normalizing constant is the Beta function

$$B(a,b) = \int_0^1 x^{a-1}(1-x)^{b-1}\,\mathrm{d}x = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}.$$

Everything gets messy if we try to differentiate $$f(x;a,b)$$ directly with respect to $$a$$ and $$b$$, which would be the brute force way to attempt a demonstration.

One way to avoid having to analyze the Beta function is to note that quantiles are relative areas. That is,

$$q_i = F(x_i;a,b)=\frac{\int_0^{x_i} f(x;a,b)\,\mathrm{d}x}{\int_0^1 f(x;a,b)\,\mathrm{d}x}$$

for $$i=1,2$$. Here, for example, are the PDF and cumulative distribution function (CDF) $$F$$ of a Beta$$(1.15, 0.57)$$ distribution for which $$x_1=1/3$$ and $$q_1=1/6$$.

The density function $$x\to f(x;a,b)$$ is plotted at the left. $$q_1$$ is the area under the curve to the left of $$x_1$$, shown in red, relative to the total area under the curve. $$q_2$$ is the area to the left of $$x_2$$, equal to the sum of the red and blue regions, again relative to the total area. The CDF at the right shows how $$(x_1,q_1)$$ and $$(x_2,q_2)$$ mark two distinct points on it.

In this figure, $$(x_1,q_1)$$ was fixed at $$(1/3,1/6)$$, $$a$$ was selected to be $$1.15$$, and then a value of $$b$$ was found for which $$(x_1,q_1)$$ lies on the Beta$$(a,b)$$ CDF.

Lemma: Such a $$b$$ can always be found.

To be specific, let $$(x_1, q_1)$$ be fixed once and for all. (They stay the same in the illustrations that follow: in all three cases, the relative area to the left of $$x_1$$ equals $$q_1$$.) For any $$a\gt 0$$, the Lemma claims there is a unique value of $$b$$, written $$b(a),$$ for which $$x_1$$ is the $$q_1$$ quantile of the Beta$$(a,b(a))$$ distribution.

To see why, note first that as $$b$$ approaches zero, all the probability piles up near values of $$0$$, whence $$F(x_1;a,b)$$ approaches $$1$$. As $$b$$ approaches infinity, all the probability piles up near values of $$1$$, whence $$F(x_1;a,b)$$ approaches $$0$$. In between, the function $$b\to F(x_1;a,b)$$ is strictly increasing in $$b$$.

This claim is geometrically obvious: it amounts to saying that if we look at the area to the left under the curve $$x\to x^{a-1}(1-x)^{b-1}$$ relative to the total area under the curve and compare that to the relative area under the curve $$x\to x^{a-1}(1-x)^{b^\prime-1}$$ for $$b^\prime \gt b$$, then the latter area is relatively larger. The ratio of these two functions is $$(1-x)^{b^\prime-b}$$. This is a function equal to $$1$$ when $$x=0,$$ dropping steadily to $$0$$ when $$x=1.$$ Therefore the heights of the function $$x\to f(x;a,b^\prime)$$ are relatively larger than the heights of $$x\to f(x;a,b)$$ for $$x$$ to the left of $$x_1$$ than they are for $$x$$ to the right of $$x_1.$$ Consequently, the area to the left of $$x_1$$ in the former must be relatively larger than the area to the right of $$x_1.$$ (This is straightforward to translate into a rigorous argument using a Riemann sum, for instance.)

We have seen that the function $$b\to f(x_1;a,b)$$ is strictly monotonically increasing with limiting values at $$0$$ and $$1$$ as $$b\to 0$$ and $$b\to\infty,$$ respectively. It is also (clearly) continuous. Consequently there exists a number $$b(a)$$ where $$f(x_1;a,b(a))=q_1$$ and that number is unique, proving the lemma.

The same argument shows that as $$b$$ increases, the area to the left of $$x_2$$ increases. Consequently the values of $$f(x_2;a, b(a))$$ range over some interval of numbers as $$a$$ progresses from almost $$0$$ to almost $$\infty.$$ The limit of $$f(x_2;a,b(a))$$ as $$a\to 0$$ is $$q_1.$$

Here is an example where $$a$$ is close to $$0$$ (it equals $$0.1$$). With $$x_1=1/3$$ and $$q_1=1/6$$ (as in the previous figure), $$b(a) \approx 0.02.$$ There is almost no area between $$x_1$$ and $$x_2:$$

The CDF is practically flat between $$x_1$$ and $$x_2,$$ whence $$q_2$$ is practically on top of $$q_1.$$ In the limit as $$a\to 0$$, $$q_2 \to q_1.$$

At the other extreme, sufficiently large values of $$a$$ lead to $$F(x_2;a,b(a))$$ arbitrarily close to $$1.$$ Here is an example with $$(x_1,q_1)$$ as before.

Here $$a=8$$ and $$b(a)$$ is nearly $$10.$$ Now $$F(x_2;a,b(a))$$ is essentially $$1:$$ there is almost no area to the right of $$x_2.$$

Consequently, you may select any $$q_2$$ between $$q_1$$ and $$1$$ and adjust $$a$$ until $$F(x_2;a,a(b))=q_2.$$ Just as before, this $$a$$ must be unique, QED.

Working R code to find solutions is posted at Determining beta distribution parameters $\alpha$ and $\beta$ from two arbitrary points (quantiles) .

• This answer shows that if we have choose a fixed $a$ or $b$ we will find a unique corresponding value. It would be possible to construct functions that have a fixed area in $[0,x_1]$, $[x_1,x_2]$ and $[x_2,1]$. I do not immediately see why this would guarantee that the set of $\alpha$ and $\beta$ is unique. Would you be willing to elaborate and enlighten me? – Jan Jun 4 '19 at 14:17
• @Jan Could explain what you mean by the "set of $\alpha$ and $\beta$"? Those symbols do not appear anywhere in this thread. – whuber Jun 4 '19 at 14:35