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.

  • 1
    $\begingroup$ Why do you think there is an inconsistency? Also,can you fix your formulas? $\endgroup$ Commented Nov 16, 2016 at 21:36

1 Answer 1


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.

  • 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
    Commented Nov 19, 2016 at 20:29
  • $\begingroup$ @whuber I edited to make the informal nature of the argument more clear (hopefully). $\endgroup$ Commented Nov 19, 2016 at 20:53

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