# A question for the statisticians amongs us

1. Feb 9, 2010

### Apteronotus

Suppose Zj are N(0,1).

$$0\cdot \left(lim_{n \rightarrow \infty} \sum_{j=0}^n Z_j \right)$$

Is this zero? or even defined at all?

2. Feb 9, 2010

### torquil

In order to multiply two quantities, both must be individually well-defined. In your case, the quantity in the parenthesis is not well-defined, since the sum does not converge (unless you are working in some strange space where the sum actually converges...?)

Torquil

3. Feb 9, 2010

### CRGreathouse

$$0\cdot \left(lim_{n \rightarrow \infty} \sum_{j=0}^n Z_j \right)$$
is not well-defined, but of course
$$lim_{n\to\infty}0\cdot \sum_{j=0}^n Z_j=0$$

4. Feb 12, 2010

### EnumaElish

And of course,
$$\lim_{n\to\infty}{\left(\frac 1{f(n)} \cdot \sum_{j=0}^n Z_j\right)}$$
is well-defined for $f(n) = \sqrt n$ and for $f(n) = n$.

Last edited: Feb 12, 2010
5. Feb 13, 2010

### Apteronotus

But if $$f(n)=\sqrt{n}$$ then

$$lim_{n \rightarrow \infty}\frac{1}{\sqrt{n}}\sum_{j=0}^n Z_j =lim_{n \rightarrow \infty}\frac{\sqrt{n}}{n}\sum_{j=0}^n Z_j =\left(lim_{n \rightarrow \infty}\sqrt{n}\right) \cdot \left(lim_{n \rightarrow \infty}\frac{1}{n}\sum_{j=0}^n Z_j\right) =\left(lim_{n \rightarrow \infty}\sqrt{n}\right) \cdot \mu$$

by law of large numbers, where $$\mu$$ is the mean of the Z's. And consequently the limit diverges!

6. Feb 13, 2010

### torquil

Yes, that is as expected. E.g. the mean for finite $$n$$ is $$\mu\sqrt{n}$$, so this cannot converge.

This is unrelated to the expression of the Wiener process as a limit of finite-time difference increments, since in that case $$\mu=0$$ and therefore you cannot split the limit expression in the same way since you would then get "$$\infty\times 0$$".

Torquil

7. Feb 16, 2010

### Apteronotus

So
1. why/how is the expression $$\lim_{n\to\infty}{\left(\frac 1{f(n)} \cdot \sum_{j=0}^n Z_j\right)}$$
well defined for $$f(n)=sqrt n$$ as EnumaElish stated above?
and
2. what is its value?

8. Feb 16, 2010

### torquil

Assume that Exp[Z] = 0 and Var[Z] = 1, for all the Z's.

If you use f(n) = n then you get zero in the limit. You can prove this using the law of large numbers, as you did for one of the parenthesis you had above (just put your \mu=0).

If f(n) = sqrt(n), the limit is not a single number, but a gaussian random variable W ~ N(0,1). Notice that for each finite n, the expectations and variance of the truncated sum W_n is:

Exp[W_n] = 0, Var[W_n] = 1

This is true for finite n, and independent of n, so it is also true in the limit.