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.. !

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- Apr 29th 2009, 11:56 PMgconfusedlimiting distribution
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.. ! - Apr 30th 2009, 02:09 AMmr fantastic
- Apr 30th 2009, 02:15 AMgconfused
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..

- Apr 30th 2009, 04:17 AMmr fantastic
- Apr 30th 2009, 04:22 AMgconfused
umm well the whole question is:

"Let X1, X2, .. Xn be a random sample from a Pareto distribution which has probability density function

f(x) = 1/(1+x)^2, x>0

Find the limiting distribution of Yn = nX(1)." - Apr 30th 2009, 04:52 AMmr fantastic
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): $\displaystyle \left[\frac{n}{x + n}\right]^{n+1}$. - May 4th 2009, 08:36 PMmatheagle
- May 4th 2009, 08:57 PMmr fantastic
- May 4th 2009, 09:11 PMaus_devoe
- May 4th 2009, 09:35 PMmatheagle
I posted it, look up^^^^^

This converges in distribution to an exponential with mean 1.

hot air 4 sure - May 4th 2009, 10:47 PMmr fantastic