Timeline for Expectation of square root of binomial r.v.
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Nov 24, 2020 at 12:34 | comment | added | Ori Gurel-Gurevich | @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. | |
| Nov 22, 2020 at 4:14 | comment | added | Mark M. Wilde | 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... | |
| Nov 22, 2020 at 3:49 | comment | added | Mark M. Wilde | 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? | |
| Sep 8, 2013 at 0:46 | comment | added | Ori Gurel-Gurevich | 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 $a\ge 0$. | |
| Sep 6, 2013 at 21:00 | comment | added | Mohammad Alaggan | Can someone help be notice why $1 + \frac{x-1}{2} - \frac{(x-1)^2}{2} \le \sqrt{x} $? | |
| Feb 11, 2013 at 19:08 | vote | accept | Matilde Martins Santos | ||
| Feb 10, 2013 at 21:05 | history | answered | Ori Gurel-Gurevich | CC BY-SA 3.0 |