# Show that...

• May 29th 2012, 07:47 PM
FRMST
Show that...
I have no idea how to do this......
Any help will be apreaciated

Let $\{ Xn\} _{n \ge 1}^{}$ be a sequence of iid random variabled such that $E({X_1}) = \mu$ and $V({X_1}) = {\sigma ^{^2}}$.
Consider the sequence of real numbers: ${\{ {a_n}\} _{n \ge 1}}$

Proof that:

$P(\frac{{{S_n} - E({S_n})}}{{V({S_n})}} \le {a_n}) - \Phi ({a_n}) \to 0$

Where:

${S_n} = \sum\limits_{i = 1}^n {{X_i}}$

Edit: I know I have to use Central limit theorem, but I'm doubtful cuz of that a_n thing inside PHI
• May 29th 2012, 11:50 PM
Siron
Re: Show that...
The Central limit theorem says:
$\frac{\mbox{S}_n - n\mu}{\sqrt{n}\sigma} \to Z$ with $Z \sim N(0,1)$ for $n \to +\infty$
in other words:
$\forall x \in \mathbb{R}: \lim_{n \to +\infty} P\left(\frac{\mbox{S}_n-n\mu}{\sqrt{n}\sigma}\leq x\right)=P(Z \leq x)=\Phi(x)$
• May 30th 2012, 05:57 AM
FRMST
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.