Vtyjkyuk

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)

$$beginalign
&P(|frac S_n sqrt n|<epsilon, |Z|>2epsilon) \
=&P(|frac S_n sqrt n|<epsilon)P(|Z|>2epsilon) \
geq&(1 - frac sigma^2 nepsilon^2)P(|Z|>2epsilon) to P(|Z|>2epsilon) > 0
endalign$$

Let see if this works.

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

Define $W_n=fracS_2nsqrt2n$. These variables converge to Z in probability too.

Finally, take $T_n=fracS_2n-S_nsqrtn$.
T converges in distribution to a standard normal, but in probability to $(sqrt2-1) Z$, because $T_n=sqrt2W_n-V_n$, that has not standard normal distribution.

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

$$lim_nrightarrow infty Pleft(left|frac S_nsqrt n - frac W_nsqrt nright|> epsilonright) = lim_nrightarrow infty Pleft(left|frac 2X_1sqrt n right|> epsilonright) = 0$$

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.