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.
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asked Feb 10, 2013 at 18:12
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3$\begingroup$ 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 $\mathbb E\sqrt X$. $\endgroup$– Anthony QuasCommented Feb 10, 2013 at 18:26
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$\newcommand{\E}{\mathbf{E}}$ $\renewcommand{\P}{\mathbf{P}}$ $\DeclareMathOperator{\var}{Var}$ If we use Taylor expansion (as Anthony suggested) for $\sqrt{x}$ around 1, we get: $$\sqrt{x}\approx 1 + \frac{x-1}{2} - \frac{(x-1)^2}{8} .$$
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 $$ \sqrt{np}-\frac{1-p}{2\sqrt{np}} \le \E(\sqrt{X})\le \sqrt{np} .$$
answered Feb 10, 2013 at 21:05
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1$\begingroup$ Can someone help be notice why $1 + \frac{x-1}{2} - \frac{(x-1)^2}{2} \le \sqrt{x} $? $\endgroup$ Commented Sep 6, 2013 at 21:00
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4$\begingroup$ 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 $a\ge 0$. $\endgroup$ Commented Sep 8, 2013 at 0:46
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$\begingroup$ I don't see how the general inequality $\sqrt{E(X)}\bigg(1-\frac{Var(X)}{2 E(X)^2}\bigg) \le E(\sqrt{X})$ for an arbitrary non-negative RV $X$ follows from $1 + \frac{x-1}{2} - \frac{(x-1)^2}{2} \le \sqrt{x} $. Could you please clarify this point? Is it obvious? $\endgroup$ Commented Nov 22, 2020 at 3:49
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$\begingroup$ I can see how it follows from $\sqrt{x} \geq \sqrt{a} +\frac{x-a}{2\sqrt{a}} - \frac{(x-a)^2}{2a^{3/2}}$, which Mathematica tells me is true for all $x \geq 0$ and $a>0$. I suppose there is some simple translation from the inequality you wrote to this more general one? If you consider this too simple or obvious, please don't bother... $\endgroup$ Commented Nov 22, 2020 at 4:14
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$\begingroup$ @MarkM.Wilde, You apply the inequality to the RV Y=X/E(X) and take expectation. This is basically equivalent to the inequality you wrote. $\endgroup$ Commented Nov 24, 2020 at 12:34