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 $1$?

  • 1
    $\begingroup$ If you write out the integral and solve it yourself, you will see it does integrate to 1! $\endgroup$
    – jbowman
    Commented Jul 31, 2020 at 20:56
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    $\begingroup$ 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. $\endgroup$
    – whuber
    Commented Aug 1, 2020 at 12:24

1 Answer 1


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).


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