Suppose that a random vector $X=(X_1,X_2,X_3)$ follows a Dirichlet distribution with a shape parameter $(a_1,a_2,a_3).$

What I want to calculate is the probability of $X_1>X_2$ and I want to check whether I'm on the right track.

What I have done so far is:

Step 1) Find out the joint distribution of $(X_1,X_2)$. Because $x_i$'s should sum up to 1, This is simply given by $$(X_1,X_2)\sim Dir(a_1,a_2,a_3).$$ Say that the distribution is $$Kx_1^{a_1-1}x_2^{a_2-1}(1-x_1-x_2)^{a_3-1},$$ where $K$ is the constant part.

Step 2) Find the probability: $$P[X_1>X_2]=\int^\frac{1}{2}_0\int^{x_1}_0f_{X_1X_2}(x_1,x_2)dx_2dx_1+\int^1_\frac{1}{2}\int^{1-x_1}_0f_{X_1X_2}(x_1,x_2)dx_2dx_1$$ Here, using the integration by parts, the first term of the LHS is $$Kx_1\int^{x_1}_0x_1^{a_1-1}x_2^{a_2-1}(1-x_1-x_2)^{a_3-1}dx_2\bigg|^{x_1=\frac{1}{2}}_0-K\int^\frac{1}{2}_0x_1x_1^{a_1+a_2-2}(1-2x_1)^{a_3-1}dx_1\\=K\bigg(\frac{1}{2}\bigg)^{a_1}\int^\frac{1}{2}_0x_2^{a_2-1}(\frac{1}{2}-x_2)^{a_3-1}dx_2-K2^{a_3-1}\int^\frac{1}{2}_0x_1^{a_1+a_2-1}(\frac{1}{2}-x_1)^{a_3-1}dx_1.$$ Letting $\frac{1}{2}y=x_2$ and $\frac{1}{2}z=x_1$, the above expression is the same as $$K\bigg(\frac{1}{2}\bigg)^{a_1+a_2+a_3-1}\int^1_0y^{a_2-1}(1-y)^{a_3-1}dy-K\bigg(\frac{1}{2}\bigg)^{a_1+a_2}\int^1_0z^{a_1+a_2-1}(1-z)^{a_3-1}dz.$$ So, the value is $$K\bigg(\frac{1}{2}\bigg)^{a_1+a_2+a_3-1}B(a_2,a_3)-K\bigg(\frac{1}{2}\bigg)^{a_1+a_2}B(a_1+a_2,a_3).$$ Similarly, using the integration by parts, the second term of the LHS is $$Kx_1\int^{1-x_1}_0x_1^{a_1-1}x_2^{a_2-1}(1-x_1-x_2)^{a_3-1}dx_2\bigg|^{x_1=1}_\frac{1}{2}-0\\=K\bigg(\frac{1}{2}\bigg)^{a_1+a_2+a_3-1}B(a_2,a_3).$$ so that we can conclude $$P[X_1>X_2]=K\bigg(\frac{1}{2}\bigg)^{a_1+a_2+a_3}B(a_2,a_3)-K\bigg(\frac{1}{2}\bigg)^{a_1+a_2}B(a_1+a_2,a_3).$$ I'm not really good at finding probability. Is this approach correct? or is there any other approach which is correct and easier than this to find out $P[X_1>X_2]?$

  • 1
    $\begingroup$ The form of the answer suggests it cannot be right. Because swapping $a_1$ and $a_2$ should change the answer (call it $p$) to $1-p,$ you will have obtained a general relation for the Beta function whose coefficients are powers of $1/2:$ that's not plausible. Glancing up through the work, it looks like your integrations by parts are incorrect (even accounting for the typographical errors). To get a feel for correct calculation, consider the simpler problem with just two variables: it amounts to evaluating the CDF of a Beta distribution at $1/2.$ $\endgroup$
    – whuber
    Aug 30, 2019 at 12:38

2 Answers 2


The answer:

\begin{equation} P_{X_1 > X_2} = \frac{B(\frac{1}{2};a_2,a_1)}{B(a_1,a_2)}, \end{equation}

where $B(\alpha,\beta) = \int_0^1 t^{\alpha-1}(1-t)^{\beta-1}dt$ is the Beta function and $B(x; \alpha,\beta) = \int_0^x t^{\alpha-1}(1-t)^{\beta-1}dt$ is the incomplete Beta function.


As $x_1+x_2+x_3 = 1$, conditional on a realisation $X_3=x_3$, condition $X_1 > X_2$ becomes $X_2 < \frac{1 - x_3}{2}$. The probability then is computed according to the equation

\begin{align} P_{X_1 > X_2} = &\int_0^1 \int_0^{\frac{1-x_3}{2}} K x_3^{a_3-1}x_2^{a_2-1} (1-x_2-x_3)^{a_1-1} dx_2 dx_3=\\ = &K \int_0^1 x_3^{a_3-1}\int_0^{\frac{1-x_3}{2}} x_2^{a_2-1} (1-x_2-x_3)^{a_1-1} dx_2 dx_3, \end{align}

where $K:= \frac{1}{B(a_1,a_2,a_3)}$, and $B(a_1,a_2,a_3)$ is the multivariate Beta function.

The inner integral is computed by the change of variables:

\begin{align} & \int_0^{\frac{1-x_3}{2}} x_2^{a_2-1}(1-x_3-x_2)^{a_1-1}dx_2 = \\ =& \int_0^{\frac{1-x_3}{2}} (1-x_3)^{a_1+a_2-1} (\frac{x_2}{1-x_3})^{a_2-1}(1 - \frac{x_2}{1-x_3})^{a_1-1}d\frac{x_2}{1-x_3} =\\ =& (1-x_3)^{a_1+a_2-1} \int_0^{\frac{1}{2}} t^{a_2-1}(1-t)^{a_1-1} dt =\\ =& (1-x_3)^{a_1+a_2-1} \cdot B(\frac{1}{2}; a_2, a_1). \end{align}

The outer integral computation is straightforward:

\begin{align} P_{X_1 > X_2} = &K \int_0^1 x_3^{a_3-1} (1-x_3)^{a_1+a_2-1} B(\frac{1}{2}; a_2, a_1) dx_3 = \\ = &\frac{B(\frac{1}{2}; a_2, a_1) B(a_3, a_1+a_2)}{B(a_1,a_2,a_3)} \end{align}

Using the property relating Beta and Gamma functions,

\begin{equation} B(a_1,...,a_n) = \frac{\prod_{i=1}^n \Gamma (a_i)}{\Gamma(\sum_{i=1}^n a_i)}, \end{equation}

we can further simplify the final answer as

\begin{align} P_{X_1>X_2} = &B(\frac{1}{2}; a_2, a_1) \frac{\Gamma(a_3)\Gamma(a_1+a_2) \Gamma(a_1+a_2+a_3)}{\Gamma(a_1+a_2+a_3)\Gamma(a_1)\Gamma(a_2)\Gamma(a_3)} = \\ =& \frac{B(\frac{1}{2};a_2,a_1)}{B(a_1,a_2)}. \end{align}

  • 1
    $\begingroup$ It would be a good idea to verify your result somehow. I am suspicious because when I contemplate letting $a_1$ grow small, it looks like your formula grows without bound and therefore cannot be a valid probability. $\endgroup$
    – whuber
    Oct 31, 2019 at 15:17
  • $\begingroup$ I found an error in my reasoning, indeed $a_1$ must not appear in the denominator. Hope you like the updated version ! $\endgroup$
    – Konstantin
    Oct 31, 2019 at 17:03
  • 1
    $\begingroup$ The reasoning in Hunaphu's reply demonstrates the answer cannot depend on $a_3.$ $\endgroup$
    – whuber
    Oct 31, 2019 at 19:04
  • $\begingroup$ It does not, indeed. $\endgroup$
    – Konstantin
    Oct 31, 2019 at 19:56
  • 2
    $\begingroup$ +1 Now it looks right! Thanks for your persistence. $\endgroup$
    – whuber
    Oct 31, 2019 at 20:16

For a two dimensional dirichlet we have $P(X_1 \leq X_2) = P(X_1 \leq 1 - X_1) = P(X_1 \leq \tfrac{1}{2})$. Thus, $$ P(X_1 \leq \tfrac{1}{2}) =\tfrac{1}{B(a_1, a_2)} \int_0^{1/2} x^{a_1 - 1}(1-x)^{a_2-1}\,dx = \tfrac{B_{1/2}(a_1, a_2)}{B(a_1, a_2)}. $$ So $P(X_1 > X_2) = 1-\tfrac{B_{1/2}(a_1, a_2)}{B(a_1, a_2)} = \tfrac{B_{1/2}(a_2, a_1)}{B(a_1, a_2)}$.

If $(Y_i)_i$ are independent Gamma and their sum is $V$ then $(Y_i)_i/V$ is Dirichlet. Therefore $P(X_1 \leq X_2) = P(Y_1 \leq Y_2)$ and, since this does not depend on $V$, any variable beyond the second is irrelevant.


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