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Let $Y_2$ denote the second smallest item of a random sample of size $n$ from a distribution of the continuous type that has $\text{cdf }F(x)$ and $\text{pdf }f(x) = F'(x)$. Find the limiting distribution of $Wn = nF(Y_2)$.

I am not sure where to start.

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    $Pr[W_n\le w]=Pr[F(Y_2)\le\frac{w}{n}]=Pr[Y_2\le F^{-1}(\frac{w}{n})]$. Now $Y_2$ is sometimes called an ordinal statistic and IIRC you can use a binomial argument to calculate this probability: how many ways of choosing one of $n$ of the $X_i$ (if $Y_2$ is the second smallest of $\{X_1,\dots,X_n\}$) to be less than $Y_2$... [Ross](http://books.google.ch/books/about/A_first_course_in_probability.html?id=Bc1FAQAAIAAJ&redir_esc=y) has a good treatment of it.2012-04-18
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    @bgins Isn't $W_n=nF(Y_2)$? In which case, your first equality is a bit off?2012-04-18
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    @KannappanSampath: thank you, just in time!2012-04-18

1 Answers 1

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For a distribution of continuous type, we can neglect the possibility that two of the samples are equal, since this has probability $0$. Thus the probability density for the second smallest item being $y_2$ is the probability of getting one sample below $y_2$ times the probability density at $y_2$ times the probability of getting $n-2$ samples above $y_2$, times a factor accounting for the permutations. All $(n-2)!$ permutations of the $n-2$ samples are already being counted, so we just have to account for the $n(n-1)$ ordered choices of the smallest and second smallest item. Thus the probability density is

$$n(n-1)F(y_2)f(y_2)(1-F(y_2))^{n-2}\;.$$

We can integrate this to get rid of the derivative $f=F'$:

$$ \begin{eqnarray} P(Y_2\gt y_2) &=& \int_{y_2}^\infty n(n-1)F(y)f(y)(1-F(y))^{n-2}\mathrm dy \\ &=& \int_{y_2}^\infty n(n-1)F(y)(1-F(y))^{n-2}\frac{\mathrm dF(y)}{\mathrm dy}\mathrm dy \\ &=& \int_{F(y_2)}^1n(n-1)F(1-F)^{n-2}\mathrm dF \\ &=& \int_0^{1-F(y_2)}n(n-1)(1-u)u^{n-2}\mathrm du \\ &=& n(1-F(y_2))^{n-1}-(n-1)(1-F(y_2))^n\;. \end{eqnarray} $$

With $W_n=nF(Y_2)$, this becomes

$$ \begin{eqnarray} P(W_n\gt w) &=& n\left(1-\frac wn\right)^{n-1}-(n-1)\left(1-\frac wn\right)^n \\ &=& \left(1-\frac wn\right)^{n-1}\left(n-(n-1)\left(1-\frac wn\right)\right) \\ &=& \left(1-\frac wn\right)^{n-1}\left(1+w-\frac wn\right)\;. \end{eqnarray} $$

Thus the limit distribution is

$$ \lim_{n\to\infty}P(W_n\gt w)=(1+w)\mathrm e^{-w}\;, $$

or

$$ \lim_{n\to\infty}P(W_n\lt w)=1-(1+w)\mathrm e^{-w}=\frac12w^2+O\left(w^3\right)\;. $$

Here's a plot.

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    Why do you multiply by the probability density of y2?2012-04-19
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    @lord12: You can use $\TeX$ on this site. Inline formulas are enclosed in single dollar signs, displayed equations in double dollar signs. Patrick formatted your question for you. If you don't know the $\TeX$ commands for something you see on the site, you can right-click on it and select "Show Math As ... TeX commands". For instance, the code for $y_2$ is `y_2`.2012-04-19