0
$\begingroup$

Suppose random variable $X$ follows Binomial distribution $\mathsf{B}(n,p)$. Prove the following statement:

\begin{align} \sup_{p\in [0,1]} \left |\mathbb{E}\left( \left( p - \frac{X}{n} \right) \left( \sqrt{\frac{(X+1)}{n}} - \sqrt{\frac{X}{n}} \right)\right) \right| = O(n^{-3/2}) \end{align}

  • 0
    Missed a bracket? $$E\left(p - \frac{X}{n}\right) = p - \frac{E[X]}{n} = p - \frac{np}{n} = 0$$ hence $$\left |\mathbb{E} \left( p - \frac{X}{n} \right) \left( \sqrt{(X+1)/n} - \sqrt{X/n} \right) \right| = 0$$ and so the claim is trivial… – 2017-01-11
  • 0
    @Gono Thanks for the editing that makes it better! Note that even in the previous form one cannot take expectations like that since the random variable X appears both in (p-X/n) and the other part. These two parts are not independent. – 2017-01-11
  • 0
    Not without the brackets! Before the edit it was a product of an expectation and the r.v. $\left( \sqrt{(X+1)/n} - \sqrt{X/n} \right)$ so brackets are important! It's a mess to write $EX$ for $E[X]$ because for expressions like EXY it's absolutely not clear if $E[XY]$ or $E[X]Y$ is meant… this has nothing to do with independeny! – 2017-01-11

0 Answers 0