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I'm asking this question based off of Example 10.2.3 in Casella & Berger. We have a sample $X_1, ..., X_n$ from a population with pdf $f$ and (differentiable) cdf $F$. Let $\mu$ be the population median and $M_n$ be the sample median. We wish to compute

$$ \lim_{n \to \infty} P(\sqrt{n}(M_n - \mu) \leq a) $$

for some $a$. We then define $Y_i$ by

$$ Y_i = \begin{cases} 1 & \mbox{if } X_i \leq \mu + a/\sqrt{n} \\ 0 & o.w. \end{cases}$$

Clearly the $Y_i$ are Bernoulli with success probability $p_n = F(\mu + a/\sqrt{n})$. If we assume that $n$ is odd, then $\{M_n \leq \mu + a/\sqrt{n}\}$ is equivalent to $\{\sum Y_i \geq (n+1)/2\}$. Now, the book states "A straightforward limit calculation will also show that"

$$ \frac{(n+1)/2 - np_n}{\sqrt{np_n(1-p_n)}} \to -2aF'(\mu) = -2af(\mu) $$.

I follow completely up until this limit calculation at which point I'm completely lost. I tried substituting in the definition of $p_n$, separating, etc. I'm probably just overlooking something simple but this "straightforward" calculation has been eluding me for a couple hours now.

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Welp, I've had an epiphany soon after posting. Here's what I have,

\begin{align*} \lim_{n \to \infty} \frac{(n+1)/2 - np_n}{\sqrt{np_n(1-p_n)}} &= \lim_{n \to \infty} \frac{(n+1)F(\mu) - nF(\mu + a/\sqrt{n})}{\sqrt{np_n(1-p_n)}} \\ &= \lim_{n \to \infty} \frac{n(F(\mu) - F(\mu + a/\sqrt{n}))}{\sqrt{n}\sqrt{p_n(1-p_n)}} + \lim_{n \to \infty}\frac{F(\mu)}{\sqrt{np_n(1-p_n)}} \\ &= \lim_{n \to \infty} \frac{F(\mu) - F(\mu + a/\sqrt{n})}{1/\sqrt{n}}\frac{1}{\sqrt{p_n(1-p_n)}} \\ &= \lim_{n \to \infty} a\frac{F(\mu) - F(\mu + a/\sqrt{n})}{a/\sqrt{n}}\frac{1}{\sqrt{p_n(1-p_n)}} \\ &= -aF'(\mu)\frac{1}{\sqrt{.25}} \\ &= -2af(\mu) \end{align*}

That took an embarrassingly long time to figure out but it's nice to understand it.

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  • $\begingroup$ Nice answer. Thanks for taking the time to answer your own question. $\endgroup$ – Glen_b Apr 18 '17 at 0:48
  • $\begingroup$ Posting does that to a person. (+1). $\endgroup$ – Alecos Papadopoulos May 28 '17 at 13:12

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