Ytukyg

I'm given a sequence $(X_n)$ of i.i.d. random variables with mean $0$ and finite variance $sigma^2$. Let $S_n=X_1 + ... + X_n$. I have to show that $S_n/sqrt n$ does not converge in probability. Here's what I did.

Since $S_n/sqrt n$ converges in distribution to a normal random variable $Z$ with mean zero, if $S_n/sqrt n$ converges in probability at all it must be to $Z$. But

$$P(|frac {S_n} {sqrt n} - Z| > epsilon) geq P(|frac {S_n} {sqrt n}|<epsilon, |Z|>2epsilon)$$

Now I get to the main point I'm not sure of. Can I say that the random variables $S_nsqrt n$ (for any $n$) and $Z$ are independant? It seems like they might be, since in some sense we can't tell what the limit of a sequence will be from any initial segment of it.

If so, then I can continue (this all seems correct to me)

$$begin{align}
&P(|frac {S_n} {sqrt n}|<epsilon, |Z|>2epsilon) \
=&P(|frac {S_n} {sqrt n}|<epsilon)P(|Z|>2epsilon) \
geq&(1 - frac {sigma^2} {nepsilon^2})P(|Z|>2epsilon) to P(|Z|>2epsilon) > 0
end{align}$$

Let see if this works.

Set $sigma^2=1$ (if sigma is 0 the statment is true) and assume $V_n=S_n/sqrt{n}$ converges in probability to a r.v. Z.
By the CLT, Z has standard normal distribution.

Define $W_n=frac{S_{2n}}{sqrt{2n}}$. These variables converge to Z in probability too.

Finally, take $T_n=frac{S_{2n}-S_n}{sqrt{n}}$.
T converges in distribution to a standard normal, but in probability to $(sqrt{2}-1) Z$, because $T_n=sqrt{2}W_n-V_n$, that has not standard normal distribution.

I have a partial answer. I think what you need is the Cauchy criterion for convergence in probability, which says:

The sequence $left(frac{S_n}{sqrt{n}}right)_{ngeq 1}$ converges in probability if and only if
begin{align}
Pleft(biggvertfrac{S_n}{sqrt{n}}-frac{S_m}{sqrt{m}}biggvert>epsilonright)stackrel{n,mtoinfty}{longrightarrow} 0quad text{for every }epsilon>0.
end{align}

However, notice that
begin{align}
&Pleft(biggvertfrac{S_n}{sqrt{n}}-frac{S_m}{sqrt{m}}biggvert>epsilonright)=Pleft(frac{S_n}{sqrt{n}}-frac{S_m}{sqrt{m}}>epsilonright)+Pleft(frac{S_n}{sqrt{n}}-frac{S_m}{sqrt{m}}<-epsilonright)\
&geq Pleft(frac{S_n}{sqrt{n}}>2epsilon,frac{S_m}{sqrt{m}}leepsilonright)+Pleft(frac{S_n}{sqrt{n}}le-2epsilon,frac{S_m}{sqrt{m}}>-epsilonright)\
&geq Pleft(frac{S_n}{sqrt{n}}>2epsilonright)+Pleft(frac{S_n}{sqrt{n}}le-2epsilonright)+Pleft(frac{S_m}{sqrt{m}}leepsilonright)+Pleft(frac{S_m}{sqrt{m}}>-epsilonright)-2\
&stackrel{n,mtoinfty}{longrightarrow}2left(Q(2epsilon)+Q(-epsilon)-1right),
end{align}
where $Q(x)=intlimits_{x}^{infty} mathcal{N}(0,1),dx$.

I do not know how to find an $epsilon>0$ such that the term inside the brackets is strictly positive. I am not quite sure if we can find one at all. As I said, this is just a partial solution that I have worked out.

Please feel free to suggest any more additions, or strengthening of arguments.

One can easily construct an entity to which we can say that "$S_n/sqrt n$ converges in probability".

Take for example $W_n = -X_1+X_2+X_3+...+X_n$

Then

$$lim_{nrightarrow infty} Pleft(left|frac {S_n}{sqrt n} - frac {W_n}{sqrt n}right|> epsilonright) = lim_{nrightarrow infty} Pleft(left|frac {2X_1}{sqrt n} right|> epsilonright) = 0$$

and the criterion for convergence in probability is satisfied.

So I suspect that "$S_n/sqrt n$ does not converge in probability" must have a more specific and narrow sense in the OP's case.

On another front, the established phrase "$S_n/sqrt n$ converges in distribution to a random variable Z" sometimes makes us forget that the phenomenon described by "convergence in distribution" is that the sequence of distribution functions $F_n$ of $S_n/sqrt n$ converges to a certain distribution function $F$. There is really no $Z$ "at the end of the journey" waiting to "become one" with $S_n/sqrt n$ . $Z$ is a random variable, a separate entity from the distribution that characterizes it (which characterizes also an infinite number of other such $Z$'s). If there is no random variable, the question "is $S_n/sqrt n$ independent of $Z$?" cannot even be posed.