1
$\begingroup$

Say I have a sequence of i.i.d random variables $(X_n)$ with mean $\mu< \infty $ and variance $\sigma^2 < \infty$.

Using the CLT, I can state that:

$\lim_{n\rightarrow\infty} \mathbb{P}\bigg( \frac{\sum_{i=1}^n{X_i} - n\mu}{\sigma n^{1/2}} \leqslant x\bigg) = \Phi(x)$

where $\Phi$ is the CDF of a standard normal distribution.

I'm wondering if there is a simple manipulation of this identify to obtain the value of the following limit in terms of the function $\Phi$:

$\lim_{n\rightarrow\infty} \mathbb{P}\bigg( \frac{\sum_{i=1}^n{X_i} - n\mu}{\sigma n^{\alpha}} \leqslant x\bigg).$

Note that the only difference between the CLT and the second equation is that we are now dividing by $n^\alpha$ instead of $n^{1/2}$. In addition, we can assume that $\alpha >0$.

  • 0
    That's easy, assuming the CLT applies, if $a\not = 1/2$, then you either got zero in the limit or the whole thing explodes, and it does not converge to any random variable. (Or if you want, it goes to infinity)2012-11-27

1 Answers 1

1

Let $Y_n^{\alpha}:=\frac{\sum_{i=1}^nX_i-n\mu}{\sigma n^{\alpha}}$.

  • If $\alpha>1/2$ and $x>0$, then $P(Y_n^\alpha\leqslant x)=P(Y^{1/2}_n\leqslant n^{\alpha-1/2}x);$ for a fixed $A$, we have for $n$ large enough that $n^{\alpha-1/2}\geqslant A$ so for these $N$, we have $P(Y_n^\alpha\leqslant x)\geqslant P(Y_n^{1/2}\leqslant xA),$ so for each $A$, $\liminf_{n\to +\infty}P(Y_n^\alpha\leqslant x)\geqslant \Phi(Ax).$ Letting $A\to +\infty$, we get the wanted result. If $x=0$ the convergence is to $\Phi(0)=1/2$. If $x<0$, the limit is $0$.