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$X_1, X_2,\ldots$ are independent, exponentially distributed r. variables with $m_k:=EX_k=\sqrt{2k}$, $v_k:=\operatorname{Var} X_k=2k$.

I want to analyse the weak convergence of $Y_n=\dfrac{\sum_{i=1}^n (X_i−m_i)}{n}$.


CLT: When the Lindeberg Condition

$b^{−2}_n\sum_{i=1}^n E(|X_i−m_i|^2⋅1_{∣Xi−mi∣>ϵ⋅b_n}) ⟶0$ , for $n→∞\text{ and }∀ϵ>0$ is fullfilled, I can state

$\frac{\sum_{i=1}^n (X_i−m_i)}{b_n}⇒N(0,1)$. It converges weakly to a normal distributed r.v.

Here it is $b^2_n=\sum_{i=1}^n \operatorname{Var} X_i=\sum_{i=1}^n 2i=n(n+1)$ So if the condition would be fullfilled it is

$$Y_n=\frac{\sum_{i=1}^n (X_i−m_i)}{n}=\sqrt{\frac{n+1}{n}}\frac{\sum_{i=1}^n (X_i−m_i)}{b_n},$$

where

$$\frac{\sum_{i=1}^{n}(X_i−m_i)}{b_n}⇒N(0,1)\text{ and }\sqrt{\frac{n+1}{n}}=\sqrt{1+\frac{1}{n}}→1,$$

hence

$$Y_n⇒N(0,1).$$

Now the problem is to prove the Lindeberg condition.

$$E(|X_i−m_i|^2⋅1_{∣Xi−mi∣>ϵ⋅b_n})=∫_{∣x−m_i∣>ϵ⋅b_n}(x−m_i)^2\sqrt{2i}e^{−\frac{x}{\sqrt{2i}}}\;dx.$$

But thats where it ends. Am I on the right way for solving this? Can I switch ∑ and ∫ in the condition?

  • 0
    *Lindefeller condition* is a nice creation.2012-02-18
  • 0
    Got something out of the answer?2012-03-08

1 Answers 1