# The limit of random variable is not defined

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1. Nov 16, 2015

### Mike.B

Let $X_i$ are i.i.d. and take -1 and +1 with probability 1/2 each. How to prove $\lim_{n\rightarrow\infty}{\sum_{i=1}^{n}{X_i} }$does not exsits (even infinite limit) almost surely.
My work:
I use cauchy sequence to prove it does not converge to a real number.
But I do not how to prove it does not converge to infinity. Can some one give hints?

Last edited: Nov 16, 2015
2. Nov 16, 2015

### andrewkirk

Denote $\sum_{k=1}^j X_k$ by $S_j$

Then If the sum $S_j$ converges in probability to infinity then
$$\forall M>0\ \forall \epsilon>0:\ \exists N>0 \textrm{ such that } j\geq N\Rightarrow Pr(S_j>M)>1-\epsilon\ \ \ \ \ (1)$$

For this not to be the case we negate the previous line to obtain:

$$\exists M>0\ \exists \epsilon>0:\ \forall N>0: \exists j\geq N\Rightarrow Pr(S_j>M)<1-\epsilon\ \ \ \ \ (2)$$

That's what has to be proved. One way to do that (not necessarily the quickest) is to use the central limit theorem. Note that $S_j=2B_j-j$ where $B_j$ is binomial with parameters $j,0.5$, which has mean $\frac{j}{2}$ and variance $\frac{j}{4}$. The Central Limit Theorem tells us that $B_j$ converges in probability to a random variable with distribution $N(\frac{j}{2},\frac{j}{4})$, so that $S_j$ converges in probability to a RV $Z_j$ with distn $N(0,j)$.

It's easy enough to prove that the sequence of RVs $(Z_j)_{j=1}^\infty$ satisfies (2). You then just need to put that together with the convergence in probability of $S_j$ to $Z_j$ to get the required result.

Last edited: Nov 16, 2015
3. Nov 16, 2015

### Mike.B

$\sum_{j=1}^n X_j$ by $S_j$?

4. Nov 16, 2015

### andrewkirk

Make that $S_n$. I'll correct it above.

5. Nov 16, 2015

### Mike.B

What is the relation between almost surely and in probability? How to use here?

Last edited: Nov 16, 2015
6. Nov 17, 2015

### andrewkirk

That's correct. Almost surely is a stronger type of convergence than in probability.

7. Nov 17, 2015

### Mike.B

I have difficulty about combining the final results, is there any inequality involved?

Last edited: Nov 17, 2015
8. Nov 17, 2015

### Mike.B

I got you about (2) [<1/2]

9. Nov 17, 2015

### andrewkirk

Hmm. On reflection, making the connection from the convergence in distribution to a normal, to a conclusion that the set of points $\omega\in\Omega$ for which $\lim_{j\to\infty}S_j(\omega)=\infty$ has zero measure is not as straightforward as I initially thought. I will think more about it when I get a decent slab of free time.

10. Nov 18, 2015

### WWGD

Isn't this a random walk based, say at 0 , in the Real line? I don't know how to describe convergence in a random walk.

11. Nov 18, 2015

### andrewkirk

Yes. That's how I've been thinking about it.

The formal statement of the problem, as I understand it, is a request to prove that $$Pr(\lim_{j\to\infty} S_j=\infty)=0$$

where we interpret $\lim_{j\to\infty}S_j=\infty$ to mean that $\forall M\in\mathbb{N}\ \exists n\in\mathbb{N}\$ such that $j>n\Rightarrow S_j>M$. All the instances of $S_j$ in that statement should really be written $S_j(\omega)$ where $\omega$ is an element of the sample space $\Omega$ of infinite sequences, in order to clarify that the statements are about specific sequences $(S_j(\omega))_{j\in\mathbb{N}}$rather than about the random variables $S_j$.

If we label by $A$ the set of all sequences/walks $S_j$ that have that property, which we call 'diverging to infinity' then we are trying to prove that $Pr(A)=0$.

I'm starting to think that my Normal Approximation suggestion won't help though.

Another possible avenue of attack would be if we could show that $A$is countable. Then we would know it must have measure zero, since the set $\Omega$ of all sequences is uncountable. We can also think about the sequences as binary expansions of real numbers in [0,1] and I'm wondering if any known results about the measure of subsets of that interval might help,

12. Nov 19, 2015

### Samy_A

13. Nov 19, 2015

### Mike.B

14. Nov 19, 2015

### Samy_A

Yes, that's what I meant. Could that work, as $(V=\infty) \cap A =\emptyset$ (I think)?

15. Nov 19, 2015

### Mike.B

Define $B=\{\omega\in \Omega:V(\omega)=\infty\}$? And show $A\cap B$=0\$?

16. Nov 19, 2015

### Samy_A

If $\omega\in A$ then $\forall M\in\mathbb{N}\ \exists n\in\mathbb{N}\$ such that $\forall j>n\Rightarrow S_j(\omega)>M$.
Take $M>0$. Doesn't that imply that $\omega$ can return to 0 at most $n$ times, so $\omega \not\in B$?

Last edited: Nov 19, 2015
17. Nov 19, 2015

### Mike.B

I think You are right!

18. Nov 19, 2015

### WWGD

Isn't it also possible to use a recursion in this case , whose solution gives a closed form for the general probability?

19. Nov 19, 2015

### Mike.B

What do you mean by "recursion"?

20. Nov 19, 2015