# Math Help - limiting distribution

1. ## limiting 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.. !

2. Originally Posted by gconfused
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.. !
Is Xn(1) defined to be $\min \{X_1, \, X_2, \, .... \, X_n \}$ where $X_1, \, X_2, \, .... \, X_n$ are i.i.d. random variables with pdf given by f(x)?

3. 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..

4. Originally Posted by gconfused
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..
Limiting distribution of what?

Neither of your two posts in this thread make any sense to me. Please post the complete and entire question, exactly as it's written in your textbook or wherever it is that you got it from.

5. 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)."

6. Originally Posted by gconfused
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)."
I'm guessing that X(1) is $\min \{X_1, \, X_2, \, .... \, X_n \}$. To get the pdf of nX(1):

$\Pr(n X(1) \leq x) = \Pr\left( X(1) \leq \frac{x}{n} \right) = 1 - \Pr\left( X(1) > \frac{x}{n} \right)$

$= 1 - \Pr\left( X_1 > \frac{x}{n} \right) \cdot \Pr\left( X_1 > \frac{x}{n} \right) \cdot .... \cdot \Pr\left( X_n > \frac{x}{n} \right)$

$= 1 - \left[1 - F\left( \frac{x}{n}\right) \right]^n$.

But $F\left( \frac{x}{n}\right) = \int_0^{x/n} \frac{1}{(1 + u)^2} \, du = 1 - \frac{n}{x + n}$.

Therefore $\Pr(n X(1) \leq x) = 1 - \left[\frac{n}{x + n} \right]^n$.

Differentiation of this gives the pdf of n X(1): $\left[\frac{n}{x + n}\right]^{n+1}$.

7. Originally Posted by mr fantastic
I'm guessing that X(1) is $\min \{X_1, \, X_2, \, .... \, X_n \}$. To get the pdf of nX(1):

$\Pr(n X(1) \leq x) = \Pr\left( X(1) \leq \frac{x}{n} \right) = 1 - \Pr\left( X(1) > \frac{x}{n} \right)$

$= 1 - \Pr\left( X_1 > \frac{x}{n} \right) \cdot \Pr\left( X_1 > \frac{x}{n} \right) \cdot .... \cdot \Pr\left( X_n > \frac{x}{n} \right)$

$= 1 - \left[1 - F\left( \frac{x}{n}\right) \right]^n$.

But $F\left( \frac{x}{n}\right) = \int_0^{x/n} \frac{1}{1 + u}^2 \, du = 1 - \frac{n}{x + n}$.

Therefore $\Pr(n X(1) \leq x) = 1 - \left[\frac{n}{x + n} \right]^n$.
the limit is... $\Pr(n X(1) \leq x) = 1 - \left[\frac{n}{x + n} \right]^n\to 1-e^{-x}$.

And whats with that floating 2 in that integrand. It looks suspicious.

8. Originally Posted by matheagle
the limit is... $\Pr(n X(1) \leq x) = 1 - \left[\frac{n}{x + n} \right]^n\to 1-e^{-x}$.

And whats with that floating 2 in that integrand. It looks suspicious.
I guess it was filled with helium gas .....

9. Originally Posted by mr fantastic
I'm guessing that X(1) is $\min \{X_1, \, X_2, \, .... \, X_n \}$. To get the pdf of nX(1):

$\Pr(n X(1) \leq x) = \Pr\left( X(1) \leq \frac{x}{n} \right) = 1 - \Pr\left( X(1) > \frac{x}{n} \right)$

$= 1 - \Pr\left( X_1 > \frac{x}{n} \right) \cdot \Pr\left( X_1 > \frac{x}{n} \right) \cdot .... \cdot \Pr\left( X_n > \frac{x}{n} \right)$

$= 1 - \left[1 - F\left( \frac{x}{n}\right) \right]^n$.

But $F\left( \frac{x}{n}\right) = \int_0^{x/n} \frac{1}{1 + u}^2 \, du = 1 - \frac{n}{x + n}$.

Therefore $\Pr(n X(1) \leq x) = 1 - \left[\frac{n}{x + n} \right]^n$.

Differentiation of this gives the pdf of n X(1): $\left[\frac{n}{x + n}\right]^{n+1}$.

10. I posted it, look up^^^^^
This converges in distribution to an exponential with mean 1.

hot air 4 sure

11. Originally Posted by matheagle
I posted it, look up^^^^^
This converges in distribution to an exponential with mean 1.

hot air 4 sure
We make a good team - I do the heavy lifting and baldeagle writes down the answer.

Thread closed before it drifts (like a helium filled 2) off-topic (sorry, I like to get the last word ).