2
$\begingroup$

So, just to begin with I feel like this is a problem I am massively overthinking, and the solution is very simple. That said, it has been a while since I've taken a math class, and so some of my fundamentals are a little fuzzy. In addition, it seems that my probability theory class that covered this topic was taught in a somewhat idiosyncratic fashion ... Anyway, background aside, let's get to my question.

Say we have a sequence of independent, identically distributed random variables, $X_1,...,X_n$ with $E(X_i)=0$ and $Var(X_i)=1$. It is trivial to show that by the central limit theorem,

$\sqrt{n}\bar{X}_n\xrightarrow[]{d}N(0,1)$; more rigorously of course this is $\sqrt{n}\frac{\bar{X}_n-0}{\sqrt{1}}\xrightarrow[]{d}N(0,1)$.

But by the weak law of large numbers, we also have $\bar{X}_n\xrightarrow{p}\mu$, where in this case $\mu=0$. Convergence in probability implies convergence in distribution, so then we have $\bar{X}_n\xrightarrow{d}0$. By the continuous mapping theorem, if $g$ is a continuous function, then $X_n\xrightarrow[]{d}X$ implies $g(X_n)\xrightarrow[]{d}g(X)$. So applying that we would say $g(y)=\sqrt{n}y$, and thus $g(\bar{X}_n)\xrightarrow{d}g(0)$; so $\sqrt{n}\bar{X}_n\xrightarrow{d}0$. Which is NOT equivalent to saying that $\sqrt{n}\bar{X}_n\xrightarrow[]{d}N(0,1)$, the result we get from applying the central limit theorem.

So for this question, am I applying one (or both) of these theorems incorrectly? Which is the correct interpretation in this situation, or is there another interpretation entirely I am missing?

EDIT: It occurs to me that my problem may be that $\sqrt{n}y$ may not be a continuous function, and so it is inappropriate to invoke the continuous mapping theorem in that context. All the same, I am having some trouble reconciling the predictions made by the CLT and WLLN, respectively.

$\endgroup$
3
$\begingroup$

The choice of continuous function $g$ should not depend on $n$ for continuous mapping theorem. Because then the continuous function changes with every $n$. What do you think it converges to?

Edit: $\sqrt{n}y$ is a continuous function. It is just that it varies with n.

$\endgroup$
  • $\begingroup$ Literally just edited my questions because I thought of that seconds after posting! Thanks for noticing that faster than I did, though. $\endgroup$ – Ryan Simmons Aug 7 '14 at 17:07
  • $\begingroup$ You say you have trouble understanding WLLN and CLT? $\endgroup$ – Gautam Shenoy Aug 7 '14 at 17:13
  • $\begingroup$ I was at the time. Now that I look back it is fairly obvious. I think it was a combination of being out-of-practice and generally fatigued. Thanks for your help! $\endgroup$ – Ryan Simmons Aug 11 '14 at 15:49
1
$\begingroup$

Looking here (http://en.wikipedia.org/wiki/Proofs_of_convergence_of_random_variables) we see that convergence in distribution implies for a continous bounded function $g(x)$ that $$E(g(X_{n}))\rightarrow E(X)=0$$

I think this may be the fault in your logic

$\endgroup$
  • 1
    $\begingroup$ I don't think that is the issue here. He made an error in applying continuous mapping theorem. Observe his function isn't bounded. And he isn't taking expectations. $\endgroup$ – Gautam Shenoy Aug 7 '14 at 17:17
1
$\begingroup$

You have indeed overlooked that the continuous mapping theorem requires a transformation function that does not depend on $n$. Why? Let's consider a function that depends on $n$. So we are contemplating $g_n(\bar X_n)$. This is a composition of functions, and the limit of the composition does not necessarily equal the composition of the limits (even when the respective limits do exist and are finite). And in your particular case you had to choose $g_n(t) = \sqrt n \cdot t$ -does this function has a finite limit?

Apart from that, try to think about the LLN in the following informal, if not overly utilitarian and playful way: The LLN provides probabilistic "certainty" (convergence in probability to a constant), but we suddenly realize that this "certainty" comes at a price -because we need uncertainty to make finite-sample inference using asymptotic results -we need a distribution... and the CLT comes to the rescue. The CLT "keeps alive" a distance (between the variable and its probability limit) that the LLN tends to nullify. The CLT does not let the LLN complete its workings. So it is not so much a matter of reconciling the results, but more of realizing how the CLT "postpones for ever" the result of the LLN.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.