limiting distribution

**gconfused** · April 29th 2009, 11:56 PM

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.. !

**mr fantastic** · April 30th 2009, 02:09 AM

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)?

**gconfused** · April 30th 2009, 02:15 AM

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..

**mr fantastic** · April 30th 2009, 04:17 AM

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.

**gconfused** · April 30th 2009, 04:22 AM

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)."

**mr fantastic** · April 30th 2009, 04:52 AM

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): $\left[\frac{n}{x + n}\right]^{n+1}$ .

**matheagle** · May 4th 2009, 08:36 PM

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.

**mr fantastic** · May 4th 2009, 08:57 PM

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 .....

**aus_devoe** · May 4th 2009, 09:11 PM

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): $\left[\frac{n}{x + n}\right]^{n+1}$ .

So is that the answer?

**matheagle** · May 4th 2009, 09:35 PM

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

hot air 4 sure

**mr fantastic** · May 4th 2009, 10:47 PM

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

).

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