Hi!! Um I'm really lost with limiting distributions, so can someone please explain to me slowly how to solve this problem, or just help me get a start on it?
Find the limiting distribution of Yn = nXn(1)
f(x) = 1/ (1+x)^2
THank you.. !
Hi!! Um I'm really lost with limiting distributions, so can someone please explain to me slowly how to solve this problem, or just help me get a start on it?
Find the limiting distribution of Yn = nXn(1)
f(x) = 1/ (1+x)^2
THank you.. !
Thank you for replying. but ummm well the question only says "Let X1, X2, .., Xn be a random sample from a Pareto distribution. Find the Limiting distribution". So I'm really not sure what I'm supposed to do, coz my textbook doesnt say anything about limiting distributions..
I'm guessing that X(1) is $\displaystyle \min \{X_1, \, X_2, \, .... \, X_n \}$. To get the pdf of nX(1):
$\displaystyle \Pr(n X(1) \leq x) = \Pr\left( X(1) \leq \frac{x}{n} \right) = 1 - \Pr\left( X(1) > \frac{x}{n} \right)$
$\displaystyle = 1 - \Pr\left( X_1 > \frac{x}{n} \right) \cdot \Pr\left( X_1 > \frac{x}{n} \right) \cdot .... \cdot \Pr\left( X_n > \frac{x}{n} \right)$
$\displaystyle = 1 - \left[1 - F\left( \frac{x}{n}\right) \right]^n$.
But $\displaystyle F\left( \frac{x}{n}\right) = \int_0^{x/n} \frac{1}{(1 + u)^2} \, du = 1 - \frac{n}{x + n}$.
Therefore $\displaystyle \Pr(n X(1) \leq x) = 1 - \left[\frac{n}{x + n} \right]^n$.
Differentiation of this gives the pdf of n X(1): $\displaystyle \left[\frac{n}{x + n}\right]^{n+1}$.