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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 $\alpha >0$.

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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) –  kjetil b halvorsen Nov 27 '12 at 4:28
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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 $0$.
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