# Is the PDF of the compound distribution of an exponential distribution with a beta-distributed parameter analytic?

Say we have a random variable which acts like an exponential random variable but its rate parameter is drawn from a Beta distribution. Now, according to Wikipedia (yeah, maybe I should find a book on this stuff....), the pdf of this compound distribution can be found as follows

$p_H(x|\alpha) = {\displaystyle \int\limits_\theta p_F(x|\theta)\,p_G(\theta|\alpha) \operatorname{d}\!\theta}$

(where $p_F$ is the pdf of an exponential distribution, $p_G$ is the same for a Beta distribution, $\alpha$ is the parameters of the Beta distribution (in reality this will be $\alpha, \beta$) and $\theta$ is the variable parameterising the exponential distribution and ranges over the positive reals).

I want to check whether this is analytic, because if it is then I can use certain fast numerical integration methods to approximate probabilities with it.

Turning again to Prof. 'Pedia, I learn that

The sums, products, and compositions of analytic functions are analytic. The reciprocal of an analytic function that is nowhere zero is analytic.

Now, given that the exponential and Gamma functions are analytic and nowhere zero, with the above I infer that the exponential and Beta pdfs are both analytic, and so should be their product, the integrand above.

I am inferring from some lecture notes I found online (<- link to .pdf file) that the integral is therefore also analytic:

From this I am tentatively concluding that the above pdf is actually analytic. Could anyone please let me know if this conclusion is right, and help me formalise an argument for it?

• analytic in $x$ ? if $p_F(z |\theta)$ is analytic in $z$ and your integral $F(z)$ converges uniformly for $|z-z_0| < r$ then $F(z)$ is analytic on $|z-z_0| < r$ (Morera's theorem : that the integral for $F(z)$ converges uniformly on some region means we have $\int_C F(z) dz = 0$ for $C$ any closed contour in this region and hence $F(z)$ is analytic there) – reuns Sep 7 '16 at 10:30
• Yes, analytic in $x$ on the real interval $(0, \infty)$ – jfhc Sep 7 '16 at 10:42
• no, complex analytic in $x$ on $Re(x) >0, |Im(x) | < T$ – reuns Sep 7 '16 at 10:45
• I mean that for the purposes of my question the function is defined in the real interval I stated. I am dealing with probability distributions defined only on the positive reals. – jfhc Sep 7 '16 at 10:54
• No, that's what I meant, for using the complex analysis theorem, you need to deal with the (natural) analytic continuation of your probability distributions. You'll agree that the exponential distribution has a very natural continuation to $x$ complex (and the beta distribution too) – reuns Sep 7 '16 at 10:57