Here is the question:
Find the limiting distribution of the quartile ratio defined as $\frac{X_{(3n/4):n}}{X_{(n/4):n}}$ (the third quartile divided by the first quartile)for the exponential distribution.
So, using the multivariate delta theorem I found that
$\sqrt(n)(\frac{X_{(3n/4):n}}{X_{(n/4):n}}-\frac{X_{3/4}}{X_{1/4}})$ converges to $N(0, (1/y^2)Var(x)+\frac{x^2}{y^4}Var(y)-2\frac{x}{y^3}Cov(x,y))$.
Where $x=X_{(3n/4):n}$ and $y=X_{(n/4):n}$.
I'm pretty confident in my calculations for the var(x/y) and the multivariate theorem, but the part I am confused about how to solve for the exponential distribution. How can you find the expected value and variance of the first and third quartiles? Is this even what the question is asking? I know how to find the variance and expected value for maximum and minimum, but not quartiles.
Any help is appreciated. Thanks!
Best Answer
In general for a continuous distribution with probability density $f(x)$ and inverse cumulative distribution function $F^{-1}(p)$, you can say that $X_{(np)}$ (i.e. the $p$ quantile of a sample of size $n$) is asymptotically normal in that sense that the the distribution of $\sqrt{n} X_{(np)}$ converges to $N\left(F^{-1}(p), \frac{{p}(1-p)}{f(F^{-1}(p))}\right)$ as $n$ increases
So for an exponential distribution with rate $\lambda$, you might say for large $n$ that $X_{(n/4)}$ has a mean of about $\frac1{\lambda}\log_e(4/3)$ and variance about $\frac{1}{4\lambda^2 n}$, while $X_{(3n/4)}$ has a mean of about $\frac1{\lambda}\log_e(4)$ and variance about $\frac{3}{4\lambda^2 n}$