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 $\alpha >0$.

$\endgroup$
  • $\begingroup$ 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) $\endgroup$ – kjetil b halvorsen Nov 27 '12 at 4:28
1
$\begingroup$

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 $0$.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.