Does there exist a nicer form for $\beta(x + a, y + b) / \beta(a, b)$?

Question

I have the expression

$$\displaystyle\frac{\beta(x + a, y + b)}{\beta(a, b)}$$

where $\beta(a_1,a_2) = \displaystyle\frac{\Gamma(a_1)\Gamma(a_2)}{\Gamma(a_1+a_2)}$.

I have a feeling this should have a closed-form which is intuitive and makes less heavy use of the Beta function. Can someone describe to me whether this is true?

Here, $x$ and $y$ are integers larger than $0.$

Answer 1

$$ \beta(1+a,b) = \frac{\Gamma(1+a)\Gamma(b)}{\Gamma(1+a+b)} = \frac{a\Gamma(a)\Gamma(b)}{(a+b)\Gamma(a+b)} = \frac{a}{a+b} \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} = \frac{a}{a+b} \beta(a,b). $$ If you have, for example $\beta(5+a,8+b)$, just repeat this five times for the first argument and eight for the second: $$ \frac{(4+a)(3+a)(2+a)(1+a)\cdot(7+b)(6+b)\cdots (1+b)b}{(12+a+b)(11+a+b)\cdots (1+a+b)(a+b)}\beta(a,b). $$

Answer 2

I think Michael Hardy gives your answer. I only want to notify that if $y=0$ then

$x$th raw moment of Beta distribution $\mu_x=E(T^x)=\frac{\int_0^1 t^{x+a-1}(1-t)^{b-1}~dt}{\beta(a,b)}=\frac{\beta(a+x,b)}{\beta(a,b)}$ if $x>-a$

Answer 3

You can also write this as a ratio of rising factorials. The rising factorial is defined as $(a)_k = a(a+1)\cdots(a+k-1)$, with $(a)_0 = 1$. Then use the recurrence relationship for the Gamma function to reduce the ratio of the beta functions to $$\frac{\beta(a+x,b+y)}{\beta(a,b)} = \frac{(a)_x(b)_y}{(a+b)_{(x+y)}}.$$ I think that's as concise as it gets. Of course, some folks aren't too fond of rising factorial notation...