Discussion between user159813 and BCLC

7:31 AM

In David Williams' Probability with Martingales, $\exists$ this exercise. What's fair about a fair game? Let $X_n$ be iid RVs s.t. $X_i = i^2 - 1$ with prob $1/i^2$ and $-1$ with prob $1-1/i^2$. I find it clear that $E(X_n)=0$. However, I do not understand why: if $S_n = \sum_{i=1}^{n} X_i$, ...

user159813

are you sure that you wrote it down right because isn't it true by strong law of large numbers that $P(\bar{X}=E(X))$ a.s. thus it should be $P(\frac{S_{n}}{n}=0)$ a.s.

or perhaps Im just misinterpreting the statement

BCLC

@user159813 The strong law is for N(0,1) I think. The exact notation in the book is "$\frac{S_n}{n} \to -1$ a.s."

oops. i realized i made another mistake though. thanks

user159813

Nope I think it applies to any rv, en.wikipedia.org/wiki/Law_of_large_numbers#Strong_law. plus remember $\bar{X}$ becomes normal as you increase n

BCLC

@user159813 Edited. ^_^

user159813

Oh and $E(X_{n})=-1$, not $0$, thats where my confusion was

BCLC

7:31 AM

@user159813 Good point. Weird. But I think that the point of the exercise is that sample average doesn't always converge to sample mean. Perhaps one of the assumptions of SLLN is violated?

@user159813 The expected value is 0. It is even stated in the book...

user159813

Oh shoot you are right my bad, stupid mistake

I apologize for my confusion, didn't know there could be exception to strong law (which there are some shown in wiki page)

Hey also your problem says the RVs are iid but the description of distribution shows they are not identical

BCLC

My first chat on stackexchange. Yay. ^_^ Anyway user159813, 1 are my calculations right? 2 Do you know what assumption of SLLN is violated? 3 Thanks for pointing out non-identicalness

user159813

I think that is why strong law of large numbers does not apply (it probably also doesn't pass the second criterion for series sum over variance to be finite as stated in wiki post

Yea the not identicalness is definitely playing a role here

BCLC

aaahhhhhhhh I see. right sorry about and thanks again

user159813

because the thing is even though you expect to get 0, for every $E(X_{i})$ the probability of getting -1 increase as i increase

BCLC

7:35 AM

anywaaayyyy how do you find my probability computations?

user159813

no Im the one that is sorry, I had it wrong form the beginning

BCLC

oooh nice intuitive explanation

no no i'm sorry too for saying they are iid when they are not :P

user159813

Oh no its fine its just a force of habit really for most of us

BCLC

soooo...how do you find my probability computations?

user159813

There was one question I had earlier today that may pertain to whether your proof is legitimate or not

Q: Moving Limit within Probability/Law of Large Numbers

Edits in Bold When I look up the strong law of large numbers it says that (Looking at Discrete Random Variables) $$P\left(\lim_{n\rightarrow\infty}\bar{X}_{n}=\mu\right)$$ That got me wondering are the following equivalent $$P\left(\lim_{n\rightarrow\infty}\bar{X}_{n}=\mu\right)=\lim_{n\rightar...

probability law-of-large-numbers

I feel that your computation relies on the fact the limit is outside the $P(.)$, but when a.s. defitnion has the limit inside $P(.)$,

BCLC

7:39 AM

checking out. thanks.

wait

I may have been confusing. My intention is to calculate the probability without the io

then use Borel-Cantelli 2

is that not right?

I mean, is the ff incorrect? "Proving $S_n/n \to -1$ a.s. is equivalent to proving $P(S_n/n = -1 \ i.o.) = 1$ or $P(S_n/n = -1 \ ev.) = 1$. I think?"

user159813

I don't think Im completely familiar with what i.o. and ev. means

unfortunately I have to get to bed, Im pretty sure someone who has more experience with measure theory and probability than me will answer your question sooner or later

sorry if i confused you at all

sometimes I just comment before really thinking lol

BCLC

8:00 AM

How presumptuous of me. My loss. Anyway, P(A_n i.o.) = P(limsup A_n). Thanks though for the insight on SLLN

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