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Let $X_1,X_2,\ldots,X_n$ be i.i.d. random variables with cdf $F$ on $\mathbb{R}$ and $M_n$ be the sample median. The proof starts off with defining for any $a\in\mathbb{R}$, $S_n = \#\text{ of }X_i\text{s}>a/\sqrt{n}$. Then $S_n\sim\text{Bin}(n,p)$. We're interested in finding $P(M_n\le a/\sqrt{n})$. Why is the set $\{M_n\le a/\sqrt{n}\}$ equivalent to $\{S_n\le (n-1)/2\}$?

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With a slight augmentation of your notation, define $$S_{n,a/\sqrt{n}}:=\sum\limits_{i=1}^{n}{\bf 1}_{X_i>a/\sqrt{n}}=\sum\limits_{i=1}^{n}(1-{\bf 1}_{X_i\leqslant a/\sqrt{n}})=n-\sum\limits_{i=1}^{n}{\bf 1}_{X_i\leqslant a/\sqrt{n}}\,,$$ where we have used the indicator functions $\ {\bf 1}_{X_i\leqslant a/\sqrt{n}}\ $.

Now, we simply observe the following:

$$ \begin{align} \omega\in\{M_n\leqslant a/\sqrt{n}\} &\iff \text{ at least }\frac{n+1}{2}\ \text{ of the }X_i\text{'}s\ \text{ are such that}\ X_i(\omega)\leqslant \frac{a}{\sqrt{n}}\\&\iff\ S_{n,a/\sqrt{n}}(\omega)=n-\sum\limits_{i=1}^{n}{\bf 1}_{X_i\leqslant a/\sqrt{n}}(\omega)\leqslant n-\frac{n+1}{2}=\frac{n-1}{2}\\ &\iff \omega \in \{S_{n,a/\sqrt{n}}\leqslant (n-1)/2\}\,. \end{align} $$

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  • $\begingroup$ @stats134711, You are welcome. $\endgroup$ – ki3i Apr 5 '15 at 3:54

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