# Expectation of square root of binomial r.v.

Let $X\sim B(n,p)$ denote a binomial random variable. Is there any approximation available for the quantity $E(\sqrt{X})$? Clearly Jensen's inequality holds, but rudimentary tooling around with Maple hasn't turned up anything more substantial.

• If $np$ is really large, then $X$ is approximately normal. You can take a Taylor series expansion for $\sqrt x$ around the mean of $X$ and get a good approximation for $\mathbb E\sqrt X$. – Anthony Quas Feb 10 '13 at 18:26

We can use this to get an approximation of $$\E(\sqrt{X})\approx 1-\frac{\var(X)}{8} ,$$ which should be valid for any RV concentrated around an expectation of 1. Equivalently, $$\E(\sqrt{X})\approx \sqrt{\E(X)}\bigg(1-\frac{\var(X)}{8\E(X)^2}\bigg) ,$$ for any RV concentrated around its mean.
As you noted, we can use Jensen inequality to get $\E(\sqrt{X})\le \sqrt{\E(X)}$ for any nonnegative RV. We can tweak the Taylor expansion to get a lower bound, by noticing that $$1 + \frac{x-1}{2} - \frac{(x-1)^2}{2} \le \sqrt{x} \ .$$ Hence, we get $$\sqrt{\E(X)}\bigg(1-\frac{\var(X)}{2\E(X)^2}\bigg) \le \E(\sqrt{X}) ,$$ for any nonnegative RV.
In the case of $X\sim Bin(n,p)$ we get $$\sqrt{np}-\frac{1-p}{2\sqrt{np}} \le \E(\sqrt{X})\le \sqrt{np} .$$
• Can someone help be notice why $1 + \frac{x-1}{2} - \frac{(x-1)^2}{2} \le \sqrt{x}$? – Mohammad Alaggan Sep 6 '13 at 21:00
• Sure. Expanding and multiplying by 2 we get that the inequality is equivalent to $3x-x^2 \le 2 \sqrt{x}$. Writing $a=\sqrt{x}$ and dividing we get $3a -a^3 \le 2$ which is true for any $a\ge 0$. – Ori Gurel-Gurevich Sep 8 '13 at 0:46