$$\lim _{n\to \:\infty }\mathbb{P}\left(\sum _{n=1}^{\infty }\:X_n>\frac{pn^2}{2}\right)=?$$
When $p\in (0,1)$ , $\left\{X_n\right\}^{\infty }_{_{_{n=1}}}$, $X_n~Bin\left(n,p\right)$, as they are independant of course.
What I tried is using as usual, like I do with normal, poi, and other stuff, the CLT.
There are two ways to do the solution:
- divide the binomials to bernoli so we have $n\cdot n$ indexes and then use CLT with $n^2$.
- keep the binominals, but we know the sum of binominal is a sum of their number of indexes.
By trying both ways, I could not reach the answer.
The way to answer those questions, is basically reach this ( it has always worked to me ):
$$\lim _{n\to \infty }\left(\mathbb{P}\left(\frac{\frac{X_1+X_2+...+X_n}{n}-E\left[X_i\right]}{\frac{\sigma }{\sqrt{n}}}<=a\right)\right)=\:\Phi \left(a\right)$$ As the expression inside is basically $N(0,1)$. while trying to do, let us say the first way with $n^2$ indexes, I reach this:
While $X_{i_j}~Ber(p)$ $$\lim _{n\to \infty }\mathbb{P}\left(\sum _{n=1}^{\infty }X_n>\frac{pn^2}{2}\right)=1-\lim _{n\to \infty }\mathbb{P}\left(\sum _{n=1}^{\infty }X_n\le \frac{pn^2}{2}\right)$$=$$1-\lim \:_{n\to \infty \:}\mathbb{P}\left(\frac{X_{11}+..+X_{1n}+..X_{n1}+..+X_{nn}}{n^2}\le \:\frac{p}{2}\right)$$=$$1-\lim \:_{n\to \infty \:}\mathbb{P}\left(\frac{X_{11}+..+X_{1n}+..X_{n1}+..+X_{nn}}{n^2}-p\le \:-\frac{p}{2}\right)$$=$$1-\lim \:_{n\to \infty \:}\mathbb{P}\left(\frac{\frac{X_{11}+..+X_{1n}+..X_{n1}+..+X_{nn}}{n^2}-p}{\frac{p\left(1-p\right)}{n}}\le \:\frac{-\frac{p}{2}}{\frac{p\left(1-p\right)}{n}}\right)$$ =$$1-\lim \:_{n\to \infty \:}\mathbb{P}\left(\frac{\frac{X_{11}+..+X_{1n}+..X_{n1}+..+X_{nn}}{n^2}-p}{\frac{p\left(1-p\right)}{n}}\le \:-\frac{np}{2p\left(1-p\right)}\right)$$ and here is the problem, The answer should be $$\Phi \left(-\sqrt{\frac{p}{2\left(1-p\right)}}\right)$$
and if I do it by way 2 ( with $n$ indexes and not $n^2$, I receive ):
$$1-\lim \:_{n\to \infty \:}\mathbb{P}\left(\frac{\frac{X_1+X_2+...+X_n}{n}-pn}{\frac{np\left(1-p\right)}{\sqrt{n}}}\le -\frac{p\sqrt{n}}{2\left(1-p\right)}\:\right)\:$$
Which is not the correct answer also.
What can be the problem? why is it not working?
Let $Y_n = \sum _{i=1}^{n }\:X_i\sim B(\frac{n(n+1)}{2},p)$, we have $$\mathop{{}\mathbb{E}}(Y_n)=\frac{n(n+1)p}{2},\text{Var}(Y_n) = \frac{n(n+1)p(1-p)}{2}$$
The question you are asking is $$\lim _{n\to \:\infty }\mathbb{P}\left(\sum _{i=1}^{n }\:X_i>\frac{pn^2}{2}\right) =\lim _{n\to \:\infty }\mathbb{P}\left(Y_n>\frac{pn^2}{2}\right) \\ =\lim _{n\to \:\infty }\mathbb{P}\left(Y_n-\frac{n(n+1)p}{2}>\frac{pn^2}{2}-\frac{n(n+1)p}{2}\right) \\ =\lim _{n\to \:\infty }\mathbb{P}\left(\frac{Y_n-\frac{n(n+1)p}{2}}{\sqrt{\frac{n(n+1)p(1-p)}{2}}}>\frac{-\frac{np}{2}}{\sqrt{\frac{n(n+1)p(1-p)}{2}}}\right) \\ =1-\Phi\left(\lim _{n\to \:\infty }\frac{-\frac{np}{2}}{\sqrt{\frac{n(n+1)p(1-p)}{2}}}\right) =1-\Phi \left(-\sqrt{\frac{p}{2\left(1-p\right)}}\right)$$