1
$\begingroup$

I've just got a quick question.

Suppose $X_1, X_2 ,\dots,X_n$ are iid with $\mathrm{poi} (\mu)$, and $\bar{X_n} = \frac{1}{n}\sum_{i=1}^{n}{X_i}$ then by central limit theorem we can have $\frac{\sqrt{n}(\bar{Xn}-\mu)}{\sqrt{\mu}} → Z$.

But by law of large numbers $\bar{X_n}\to \mu $, and by convergence theory $\frac{\sqrt{n}(\bar{X_n}-\mu)}{\sqrt{n}} \to \frac{\sqrt{n}(\mu-\mu)}{\sqrt{n}}$, which is $0$ (if I understand correctly), which is kinda inconsistent with the CLT. How did I get this wrong?

After thinking for I while I just figured that I misunderstood the concepts of convergence in probability and convergence in distribution.

$\endgroup$
1
  • 1
    $\begingroup$ Why do you think there is an inconsistency? Also,can you fix your formulas? $\endgroup$ Nov 16 '16 at 21:36
4
$\begingroup$

Breaking down the types of convergence helps here. Let's denote by $A_n$ the random variable

$$ A_n = \frac{X_1 + X_2 + \cdots + X_n}{n} $$

In the law of large numbers, the convergence is in probability. This means that, as $n \rightarrow \infty$, it becomes increasingly unlikely that the value of $A_n$ is far from $\mu$:

$$ P( \mu - \epsilon < A_n \leq \mu + \epsilon) \rightarrow 1 $$

The above holds for every positive number $\epsilon$.

In the central limit theorem the convergence is in distribution. This essentially means that probability statements about $A_n$ can be well approximated in the limit by assuming that $A_n$ follows a normal distribution:

$$ A_n \approx N \left( \mu, \frac{\sigma}{\sqrt{n}} \right) $$

Note though, that the variance of the approximating normal distribution shrinks as $n \rightarrow \infty$. This means that, using the central limit theorem:

$$ P( \mu - \epsilon < A_n \leq \mu + \epsilon) \approx P \left( \mu - \epsilon < N \left( \mu, \frac{\sigma}{\sqrt{n}} \right) \leq \mu + \epsilon \right) \rightarrow 1 $$

So the two results are completely consistent.

$\endgroup$
2
  • 1
    $\begingroup$ "$\bar X \rightarrow_{D} N \left( \mu, \frac{\sigma}{\sqrt{n}} \right)$" makes no sense because the right side depends on $n$. Fix this by studying $\sqrt{n}(\bar X - \mu)$ instead of $\bar X$ itself. $\endgroup$
    – whuber
    Nov 19 '16 at 20:29
  • $\begingroup$ @whuber I edited to make the informal nature of the argument more clear (hopefully). $\endgroup$ Nov 19 '16 at 20:53

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.