
Show that...
I have no idea how to do this......
Any help will be apreaciated
Let $\displaystyle \{ Xn\} _{n \ge 1}^{}$ be a sequence of iid random variabled such that $\displaystyle E({X_1}) = \mu $ and $\displaystyle V({X_1}) = {\sigma ^{^2}}$.
Consider the sequence of real numbers: $\displaystyle {\{ {a_n}\} _{n \ge 1}}$
Proof that:
$\displaystyle P(\frac{{{S_n}  E({S_n})}}{{V({S_n})}} \le {a_n})  \Phi ({a_n}) \to 0$
Where:
$\displaystyle {S_n} = \sum\limits_{i = 1}^n {{X_i}} $
Edit: I know I have to use Central limit theorem, but I'm doubtful cuz of that a_n thing inside PHI

Re: Show that...
The Central limit theorem says:
$\displaystyle \frac{\mbox{S}_n  n\mu}{\sqrt{n}\sigma} \to Z $ with $\displaystyle Z \sim N(0,1)$ for $\displaystyle n \to +\infty$
in other words:
$\displaystyle \forall x \in \mathbb{R}: \lim_{n \to +\infty} P\left(\frac{\mbox{S}_nn\mu}{\sqrt{n}\sigma}\leq x\right)=P(Z \leq x)=\Phi(x)$

Re: Show that...
Thank you, but what happens when I have a sequence...
The sequence is going to infinity... so, I don't know if I can use straight the central limit theorem
EDIT:
Oh... I got it...
So, the sequence in this case would be:
( x  n*u )/( sqrt(n)*Sigma)
Thank you buddy.