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.
Best Answer
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.