What is the Mean of a Cumulative Beta Distribution Function?

Question

What I mean is that if we normalize the beta cdf to unity and treat it as a pdf, could we have an analytical expression for the mean of this distribution in terms of alpha and beta parameters of the Beta distribution function?

Answer 1

It's an unusual question (with no evident applications), but it has been clearly stated, so I will answer it as written.

The mean is the expectation. Expectations for a distribution $F$ can be found by integrating against the survival function $1-F$.

Consider any random variable $X$ whose distribution is supported on the interval $[0,1]$ with no jumps at the endpoints (like all Beta distributions). Then according to the foregoing result,

$$\mu_{k+1}=\mathbb{E}(X^{k+1}) = \int_0^1 (1-F(x))\left((k+1) x^{k}\right)dx = 1 - (k+1)\int_0^1 x^kF(x)dx.$$

Easy algebra gives raw moments of the unnormalized CDF in terms of moments of $X$:

$$\int_0^1 x^k F(x)dx = \frac{1}{k+1}\left(1 - \mu_{k+1})\right).\tag{1}$$

The question asks to find the first normalized moment $m$, which is found by plugging $k=0$ and $k=1$ into $(1)$:

$$m = \frac{\int_0^1 xF(x)dx}{\int_0^1 F(x)dx} = \frac{(1-\mu_2)/2}{1-\mu_1}.\tag{2}$$

Usually we can look up the central moments. In terms of them, $\mu_2$ is the variance plus the square of the mean $\mu_1$. For the Beta Distribution,

$$\mu_1 = \frac{\alpha}{\alpha+\beta}$$

and

$$\mu_2 = \frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)} + \mu_1^2 = \frac{\alpha(1+\alpha)}{(\alpha+\beta)(1+\alpha+\beta)}.$$

Plug those expressions into $(2)$ (and simplify if you wish):

$$m = \frac{1 + 2\alpha+\beta}{2(1+\alpha+\beta)}= 1 - \frac{1+\beta}{2(1+\alpha+\beta)} = \frac{1}{2} + \frac{\alpha}{2(1+\alpha+\beta)}.$$