Independant RV's and the Borel-Cantelli Lemmas

[Firepanda]

The first part I can do no problem!

The second part... well I've really no idea what to do with that S term, I assume it has something to do with the strong law of large numbers? (w is an element in the set of outcomes)

Borel-Cantelli Lemmas

Law of Large Numbers

Does anyone know where to start?

Thanks

edit:

I have an exmple of strong convergence showing that

if P(w : lim X_n(w) = X(w) )

then this is identically equal to

P(X_n -> X) = 1 , almost surely

So I guess I just have to show that P(X_n -> X) = 1,

So show: P(S_n(w)/n -> -1) = 1?

Probably going the wrong way around it.. not sure how Borel Cantelli links to it.

 

[Ray Vickson]

The first part I can do no problem!

The second part... well I've really no idea what to do with that S term, I assume it has something to do with the strong law of large numbers? (w is an element in the set of outcomes)

Borel-Cantelli Lemmas

Law of Large Numbers

Does anyone know where to start?

Thanks

edit:

I have an exmple of strong convergence showing that

if P(w : lim X_n(w) = X(w) )

then this is identically equal to

P(X_n -> X) = 1 , almost surely

So I guess I just have to show that P(X_n -> X) = 1,

So show: P(S_n(w)/n -> -1) = 1?

Probably going the wrong way around it.. not sure how Borel Cantelli links to it.

It is probably simpler to re-write the problem a bit: let $X_n = -1 + Y_n,$ where $$\Pr\{Y_n = 0 \} = 1 - \frac{1}{n^2}, \mbox{ and } \Pr \{Y_n = n^2 \} = \frac{1}{n^2}.$$ Essentially, you want to show that $\sum_{k=1}^n Y_k / n \rightarrow 0$ w.p. 1. If $A = \{ Y_n > 0 \mbox{ infinitely often } \},$ can you show that Pr{A} = 0?

RGV

 

[Firepanda]

Hi RGV, you always answer my questions it seems! :)

So I need to show given the event E_n = Y_n>0

that the sum from n=1 to inf of P(E_n) is finite (Borel Cantelli)

So this is the sum of n=1 to inf of 1/n²?

Which is pi^2/6, and so the probability is 0!

Correct?

 

[Firepanda]

Also any idea on the second part of this question?

It looks like the same structure..

Thanks!

 

[Firepanda]

bump..!