I'm having some trouble with convergence in distribution and convergence in probability, mainly because I'm getting different results that seem to contradict each other using the Central limit theorem and the Law of large numbers.

Let $(X_1,...,X_n)$ be a $n$ independent random variables such that $\forall i\in\{{1,...,n}\}: X_i\sim Ber(\frac{1}{5})$, and let $\bar{X_n}=\frac{1}{n}\sum_{i=1}^n X_i$. Calculate $\lim_{n \to \infty} P(\sum_{i=1}^n X_i > \frac{1}{5}n+\frac{\sqrt n}{5})$

Now here is my problem, The result of LLN is that the sample mean converges in probability to $E(X)=\frac{1}{5}$, So that also means that $(*)\bar{X_n}\xrightarrow{d} \frac{1}{5}$

Ultimately I can use that to get

$P(\sum_{i=1}^n X_i > \frac{1}{5}n+\frac{\sqrt n}{5}) = P(\frac{1}{n}\sum_{i=1}^n X_i > \frac{1}{5}+\frac{1}{5\sqrt n}) = 1- P(\bar{X_n} \leq \frac{1}{5}+\frac{1}{5\sqrt n})$ and since $\frac{1}{5}+\frac{1}{5\sqrt n} > \frac{1}{5}$ and $(*)$ I get that the limit is equal to $1-1=0$.

However, using CLT I could've written that probability as follows:

$P(\sum_{i=1}^n X_i > \frac{1}{5}n+\frac{\sqrt n}{5}) = P(\frac{\bar{X_n} -\frac{1}{5}}{\frac{2}{5 \sqrt n}} \leq \frac{1}{2})$

and I know that the limit of that expression should be $\phi(\frac{1}{2})$ which is definitely not 0.

One thing I notice is that doing it the first way, the RHS of is bigger than $\frac{1}{5}$ but as $n$ tends to $\infty$ that is not the case anymore so maybe that could be the reason of why it would be wrong doing it that way. But generally, should these two methods be equivalent? Or is one different than the other, and if so which one should I use? And how does that not contradict the fact that $\xrightarrow{p}$ implies $\xrightarrow{d}$?



1 Answer 1


The LLN is not strong enough to tell you the value of $\lim_n P(\sum X_i > n/5+\sqrt{n}/5);$ you can't use limits that way.

Knowing that $\bar X_n\stackrel{p}{\to}1/5$ tells you that $\lim_n P(\bar X_i > 1/5+\epsilon)\to 0$ for every fixed $\epsilon>0$, but $\sqrt{n}/5/n=1/(5\sqrt{n})$ is not a fixed $\epsilon>0$; it's a decreasing sequence.

The calculations using something like the Chebyshev inequality that give you the LLN, tell you that $\lim_n P(\sum X_i > n/5+n^{\lambda}/5)$ goes to 0 for $\lambda>1/2$ and to 1 for $\lambda<1/2$. For $\lambda=1/2$ you need the more sophisticated computation you find in the CLT.

Since $\lim_n P(\sum X_i > n/5+n^{\lambda}/5)$ isn't 0 or 1 it is at least bounded, and every subsequence must have a convergent subsubsequence converging to some $c\in(0,1)$. Knowing more than that requires the CLT, which tells you that in fact $\lim_n P(\sum X_i > n/5+\sqrt{n}/5)$ has a well-defined limit, and also tells you what it is.


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