Convergence of sum of binomial random variables and central limit theorem

Question

I have a doubt regarding the following problem:

The distribution of $\sum _{ i=1 }^{ n }{ { X }_{ i } } $ is a binomial of mean $np$ and variance $np(1-p)$.

The variables ${ X }_{ i }$ are independent.

We suppose that the variance ${ \sigma }^{ 2 }=1/4 $

When $n \rightarrow \infty$, $\sqrt { n } \sum _{ i=1 }^{ n }{ ({ X }_{ i }-p) } $ converges toward ?

I have the following choices:

• $N(p,(1/(4n)))$
• $N(0,(1/4)$
• $N(0,1/2)$
• $N(p,(1/(4n)))$
• $N(p,(1/(2{ n }^{ 1/2 }))$

where $N$ is a normal distribution of parameter $m$ and variance $r$.

My thoughts on this:

• We have to use the Central Limit Theorem.
• $m$ cannot be equal to 0, because we have $({ X }_{ i }-p)$ and not $({ X }_{ i }-np)$
• For the variance, we can use the properties of the variance, ie $V(cX)={ c }^{ 2 }V(X)$ So, we would have, $V(N)=n\times { \sqrt { n } }^{ 2 }\times \frac { 1 }{ 4 } =\frac { { n }^{ 2 } }{ 4 } $

So, if I am correct, the answer should be: $N(p,\frac { { n }^{ 2 } }{ 4 })$

But I see it nowhere in the list of available options.

Thanks in advance.

Regards,

Answer 1

Notice the $X_{i}$ have $E(X_{i})=p$ since $np=E(\sum X_{i})=nE(X)$ similarly to this argument you have $Var(X_{i})=p(1-p)$ we know that as $n\rightarrow\infty$ a binomial distribution is approximately normal but Central Limit Theorem as you have thus we just have to find expected value and variance so we have for the $\mu$: $$\mu=E\left(\sqrt{n}\sum(X_{i}-p)\right)=\sqrt{n}\left(\sum(E(X_{i})-p)\right)=0$$ and as for variance, $\sigma^{2}$ we have $$\sigma^{2}=Var\left(\sqrt{n}\sum(X_{i}-p)\right)=n\left(\sum Var(X_{i})\right)=n\left(n(np(1-p))\right)$$ which you said we assumed we had $np(1-p)=\frac{1}{4}$ so we would have $\sigma^{2}_{n}=\dfrac{n^{2}}{4}$ so as $n\rightarrow \infty$ we have $\sigma^{2}_{n}\rightarrow\infty$ thus as citronrose points out may be a typo, most likely $\frac{1}{\sqrt{n}}$ where $\sigma^{2}_{n}=\frac{1}{4}$ (and here it seems like we didn't use the fact $n\rightarrow\infty$, but was used to establish that this binomial converges to normal as $n\rightarrow\infty$

Answer 2

The central limit theorem is indeed the way to go. The $X_{i}$ all have an expectation $p$. So each of the $X_i - p$ has an expectation zero so their sum is also zero, and multiplying by $\sqrt{n}$ will not change this expectation.

For the variance, I agree with you that it should increase with $n$ so the choices aren't right. But maybe there is a typo in the formal, and it is $1/\sqrt{n}$ instead of $\sqrt{n}$, in which case 2 would be right.