# How does the uniform Dirichlet PDF integrate to 1?

For the uninformative 3-dimensional Dirichlet prior $${\rm Dir}(1, 1, 1)$$, I understand that the probability density function (PDF) evaluates uniformly to 2, and the support are all three-dimensional vectors which sum to $$1$$. In this case, the support is the standard 2-simplex, which has an area of $$\sqrt{3}/2$$.

I also understand that the integral of a PDF over its support should be $$1$$. My question is, if the aforementioned Dirichlet PDF is uniformly $$2$$, how does this integrate to $$1$$?

• If you write out the integral and solve it yourself, you will see it does integrate to 1! Jul 31, 2020 at 20:56
• Isn't the support the set of $(x_1,x_2)\in\mathbb{R}^2$ for which both coordinates are non-negative with a sum less than or equal to $1$? If not, please supply an explicit expression for the PDF.
– whuber
Aug 1, 2020 at 12:24

This is somewhat related to this question, but you are asking about the 2-simplex instead of the 1-simplex. The answer is similar, however, in that the confusion is caused by one of the variables being redundant because of the constraint $$\sum_{i=1}^3 x_i = 1$$. Instead of integrating over the 2-simplex sitting in $$\mathbb{R}^3$$ defined over all three variables $$x_1$$, $$x_2$$, and $$x_3$$ given by $$T = \{ (x_1, x_2, x_3) \in \mathbb{R}^3 : x_1 \ge 0, x_2 \ge 0, x_3 \ge 0, x_1 + x_2 + x_3 = 1 \},$$ you should instead be integrating over the projection down to two variables, i.e. the set $$S = \{ (x_1, x_2) \in \mathbb{R}^2 : x_1 \ge 0, x_2 \ge 0, x_1 + x_2 \le 1 \},$$ which has area $$1/2$$. Then $$(2)(1/2) = 1$$ gives you the correct value for the integral of a pdf.
Addendum: The real confusion here is caused by the fact that the standard $$2$$-simplex $$T$$ (defined above) is a $$2$$-dimensional manifold embedded in $$\mathbb{R}^3$$. Now suppose that we want to define a probability density function for the Dirichlet distribution on $$\mathbb{R}^3$$, say $$f \colon \mathbb{R}^3 \to [0, \infty)$$. This function would be supported only on $$T$$, i.e. $$f(x) = 0$$ for all $$x \notin T$$. Because the triangle $$T$$ has a $$3$$-dimensional Lebesgue measure (volume) of zero, this function $$f$$ will not be a density, because if you integrated over all of $$\mathbb{R}^3$$ (or any subset of it) you would just get zero, even if $$f(x) = 2$$ for all $$x \in T$$. If you want to consider $$T \subset \mathbb{R}^3$$ as the support of the distribution, this would be a singular distribution. What we do instead is use a $$2$$-dimensional parametrization, namely the parametrization $$g : S \to T$$ given by $$g(x_1, x_2) = (x_1, x_2, 1 - x_1 - x_2)$$, to define a proper (non-singular) density because $$S \subset \mathbb{R}^2$$ has positive $$2$$-dimensional Lebesgue measure (area).