Let $p \in(0,1)$. What is the distribution of the sum of $n$ independent Bernoulli random variables with parameter $p$? Let $0 \leq a < b \leq 1$. Use approprtiate limit theorems to determine how the following depends on $a$ and $b$: $$\lim_{n\to\infty} \sum_{r=\lfloor an \rfloor}^{\lfloor bn \rfloor} {n \choose r } p^r (1-p)^{n-r}\ $$

I know that the sum of $n$ Bernoulli random variables with parameter $p$ has binomial distribution $Bin(n,p)$.

I see that the above is equal to $P(an \leq X \leq bn) $ where $X$ is the sum, and I can see how to use the weak law of large numbers in the cases where $p \neq a$ and $p \neq b$ to find the appropriate limits, but I can't see what I have to do when $p=a$ or $p=b$.

I assume I have to apply the central limit theorem, but I really don't see how to do that. Any help you could give me would be really appreciated.


As you observed, we can express $X=X_1+\dots+X_n$ where the $X_i$ are i.i.d Bernoulli random variables with parameter $p$.

The central limit theorem tells you that $\sqrt{n} \left(\frac{1}{n} X - p\right)$ is distributed as $\mathcal{N}(0, p(1-p))$ when $n$ tends to $\infty$ because $p$ and $p(1-p)$ are respectively the mean and variance of $X_i$ for any $i$. This is true if your variable satisfy the standard conditions for the CLT, which it does but try to check it.

I think that you should be able to conclude now.

EDIT : as you mentioned you are looking for (as $n$ tends to $\infty$) \begin{align*} \mathbb{P}(an\leq X\leq bn) &= \mathbb{P}\left(\sqrt{n}(a-p)\leq \sqrt{n}\left(\frac{1}{n} X - p\right)\leq \sqrt{n}(b-p)\right)\\ &= Q(\sqrt{n}(a-p)) - Q(\sqrt{n}(b-p)) \end{align*}

So now if $a=p$ or $b=p$, one of these two is actually exactly $1/2$, You can just apply the different cases you mentioned to this one.

EDIT 2 : For completeness let me specify the cases. We use the property $Q(0)=1/2$, $\lim_{x\to\infty} Q(x) = 0$ and $\lim_{x\to-\infty} Q(x) = 1$

  • $p\notin [a,b]$ : Then the sign of $a-p$ and $b-p$ are the same and cannot be $0$, this means that for $n\to\infty$, $Q(\sqrt{n}(a-p)) = Q(\sqrt{n}(b-p)) \in \lbrace 0,1\rbrace$ and so $\mathbb{P}(an\leq X\leq bn)=0$
  • $p\in \lbrace a,b \rbrace$ : then the two possible values for $Q(\sqrt{n}(a-p))$ and $Q(\sqrt{n}(b-p))$ are respectively $1/2$, $0$ or $1$, $1/2$. In any case $\mathbb{P}(an\leq X\leq bn)=1/2$
  • $p\in ]a,b[$ : Then $Q(\sqrt{n}(a-p))=1$ and $Q(\sqrt{n}(b-p))=0$ and $\mathbb{P}(an\leq X\leq bn)=1$.
  • $\begingroup$ I see how $\sqrt{n} \left(\frac{1}{n} X - p\right)$ tends towards $\mathcal{N}(0, p(1-p))$, but I'm struggling to work out how I'm supposed to use that. If it's not too much trouble, would you be able to walk me through it in a bit more depth? $\endgroup$
    – bobphimist
    Sep 11 '18 at 18:21
  • $\begingroup$ no problem at all, see my edit, and if you need anything more, feel free to ask. $\endgroup$
    – P. Quinton
    Sep 11 '18 at 18:38
  • $\begingroup$ Thank you so much! $\endgroup$
    – bobphimist
    Sep 16 '18 at 17:21

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