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I start with two Beta distributions:

$$\mathrm{Beta_A}(p; \alpha_A, \beta_A) = \frac{p^{\alpha_A-1}\,(1-p)^{\beta_A-1}}{\mathrm{B}(\alpha_A, \beta_A)}$$

$$\mathrm{Beta_B}(p; \alpha_B, \beta_B) = \frac{p^{\alpha_B-1}\,(1-p)^{\beta_B-1}}{{\mathrm{B}(\alpha_B, \beta_B)}}$$

where in the context of Bernoulli trials, $\alpha$ can be interpreted as $1 + \mathrm{successes}$ and $\beta$ can be interpreted as $1 + \mathrm{fails}$. $\mathrm{B}$ is the Beta function.

I then define the 'difference' between $\mathrm{Beta_A}$ and $\mathrm{Beta_B}$ as:

$$F(x; \alpha_A, \beta_A, \alpha_B, \beta_B) = \mathrm{Beta_A}(p) - \mathrm{Beta_B}(p)$$

Questions

  • what is the PDF of $F(x)$?
  • what family of probability density distributions does $F(x)$ belong to?

Example and illustration

For example for $\alpha_A=3, \beta_A=9$ (2 successes from 8 Bernoulli trials) and $\alpha_A=1, \beta_A=5$ (0 successes from 4 Bernoulli trials) the distribution of values that $p$ can take is:

Beta_A and Beta_B

If I then take $n$ random values $X_A \sim \mathrm{Beta_A}$ and $X_B \sim \mathrm{Beta_B}$, and find the differences between each $i^\mathrm{th}$ element, $X_{A,i} - X_{B,i}$, and plot these $n$ differences in a histogram, I am essentially sampling $F(x)$ - the underlying distribution of $\mathrm{Beta_A} - \mathrm{Beta_B}$ which can only be defined for $x \in [-1,+1]$.

With $n = 5 \times 10^7$ random samples and bin widths of $\Delta x = 0.004$, $F(x)$ takes the following form:

Histogram of sampled differences

What is the PDF of $F(x)$?


Notes

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    $\begingroup$ Do you mean a convolution? math.la.asu.edu/~jtaylor/teaching/Fall2010/STP421/lectures/… $\endgroup$
    – Dave
    Nov 14 '19 at 12:32
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    $\begingroup$ I guess I meant to follow the logic of the derivation but applied to subtraction instead of addition. I don’t think there would be that bound on $x$, though working through the calculus should reveal if there is. $\endgroup$
    – Dave
    Nov 14 '19 at 13:15
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    $\begingroup$ The notations are awfully confusing, as they make $F(x)$ look like the difference of two Beta densities. Furthermore, $x$ and $p$ are not explicitly related. $\endgroup$
    – Xi'an
    Nov 14 '19 at 13:19
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    $\begingroup$ There is an answer on maths.stackexchange. $\endgroup$
    – Xi'an
    Nov 14 '19 at 13:20
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    $\begingroup$ Here are some papers also. $\endgroup$ Nov 14 '19 at 13:39
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I know this is a bit of an old question but for what it's worth there is an established closed-form solution to this problem, found by Pham-Gia, Turkkan, and Eng in 1993. It's a piecewise solution that relies on the Appell F1 hypergeometric function. Given $$ \begin{align} \theta_1 &= \text{beta}(\alpha_1, \beta_1) \\ \theta_2 &= \text{beta}(\alpha_2, \beta_2) \\ \theta_d &= \theta_0 - \theta_1 \end{align} $$

Then the probability of the difference of $\theta_d$ is piecewise over $\theta_d$. I've re-written it here, in case you can't access the paper. I'm using $\cdot$ to indicate multiplication when I need to break the equation over several lines.

Given $A = \text{Beta}(\alpha_1, \beta_1)\text{Beta}(\alpha_2, \beta_2)$.

For $-1 \leq \theta_d < 0$: $$ \begin{align} p(\theta_d) = &\text{Beta}(\alpha_2, \beta_1)\theta_d^{\beta_1 + \beta_2 - 1}(1 - \theta_d)^{\alpha_2 + \beta_1 - 1} \\ &\cdot F_1(\beta_1, \alpha_1 + \beta_1 + \alpha_2 + \beta_2 - 2, 1 - \alpha_1; \beta_1 + \alpha_2; 1 - \theta_d, 1 - \theta_d^2) \\ &/ A \end{align} $$

For $0 < \theta_d \leq 1$:

$$ \begin{align} p(\theta_d) = &\text{Beta}(\alpha_1, \beta_2)\theta_d^{\beta_1 + \beta_2 - 1}(1 - \theta_d)^{\alpha_1 + \beta_2 - 1} \\ &\cdot F_1(\beta_2, 1 - \alpha_2, \alpha_1 + \beta_1 + \alpha_2 + \beta_2 - 2; \alpha_1 + \beta_2; 1 - \theta_d^2, 1 + \theta_d) \\ &/ A \end{align} $$

And for $\theta_d = 0$, $\alpha_1 + \alpha_2 > 1$, $\beta_1 + \beta_2 > 1$: $$ p(\theta_d) = \text{Beta}(\alpha_1 + \alpha_2 - 1, \beta_1 + \beta_2 -1) / A $$

As for the family of this distribution, I'm honestly not sure. Somebody else may be able to jump in there.

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