# Incomplete beta function increasing in $\alpha$

Let $$I_{x}(\alpha,\beta)$$ is the regularized incomplete $$\beta$$ function, aka the cdf for a random variable with distribution $$\text{Beta}(\alpha,\beta)$$.

In a recent paper (https://arxiv.org/abs/1911.06317) it is claimed that the function $$d \to I_{1-9/16d}\left(\frac{d+1}{2},\frac{1}{2}\right)$$ is increasing. This seems plausible, but they don't provide a reference or a proof. So, does anyone have a reference, or a proof, for this claim?

Let me add that one way to prove it may be to use the fact that by definition: $$I_{1-9/16d}\left(\frac{d+1}{2},\frac{1}{2}\right) = \frac{\int_{0}^{1-9/16d}t^{\frac{d-1}{2}}(1-t)^{-1/2}dt}{\int_{0}^{1}t^{\frac{d-1}{2}}(1-t)^{-1/2}dt}$$ and then differentiate, and hopefully show that this derivative is positive. This seems a bit labour-intensive (and I'm not certain it will work), so I wonder if there is a better way?

• Forgive me for not understanding, but if a function represents a CDF, it must perforce be monotonic increasing and cadlag. Unless you mean they claim strictly increasing? Also, which page in your linked paper? Commented Dec 16, 2021 at 21:55
• p. 15 at the bottom. It is not clear whether the function is increasing since the input parameter $d$ affects both the integration bound as well as the shape parameter of the beta distribution here. (Actually increasing $d$ here even decreases the integration bound) Commented Jan 14, 2023 at 16:58