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I am showing that Lyapunov's condition $$ \exists\delta>0:\sum_{m=1}^ns_n^{-(2+\delta)}\mathbb{E}[(X_m)^{2+\delta}]\to0 $$ implies the Lindeberg's condition $$ \forall\epsilon>0:\sum_{m=1}^n \mathbb{E}\left[\left(\frac{X_m}{s_n}\right)^2\mathbb{1}_{\{X_m\geq\epsilon s_n\}}\right]\to0. $$

However, I do not see why the $\delta>0$ is needed. If we have $$ \sum_{m=1}^ns_n^{-2}\mathbb{E}[(X_m)^{2}]\to0 $$ then apparently Lindeberg's condition holds since $\left(\frac{X_m}{s_n}\right)^2>\left(\frac{X_m}{s_n}\right)^2\mathbb{1}_{\{X_m\geq\epsilon s_n\}}\geq0,\forall\epsilon>0$. I cannot identify the problem in the argument above.

I am aware of a duplicate of this question (here) but I fail to see why the answerer suggests that "The $\dfrac{1}{c\delta}$ with $\delta>0$ is needed to make this go to 0" and the counterexample (it doesn't satisfy the Lyapunov's condition in first place). May someone help me identify the errors I made in my arguments and understand how $\delta>0$ is needed? Any input is appreciated.

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  • $\begingroup$ I can also not see why $\delta=0$ is a problem, and I agree with you that the counterexample provided in the answer to the other post is a bit irrelevant since Lyapunov's condition isn't satisfied (even for $\delta=0$). My guess is that the condition has $\delta>0$ because it was needed for the original proof of Lyapunov's CLT. $\endgroup$
    – jakobdt
    Commented Apr 29, 2022 at 13:32
  • $\begingroup$ @jakobdt Thank you for your comment! Perhaps it's because the condition $\delta=0$ "overlaps" with Lindeberg's condition, hence is not suitable to be stated as a separate criterion. $\endgroup$
    – barbatos233
    Commented Apr 29, 2022 at 13:43
  • $\begingroup$ Lyapunov's condition when $\delta = 0$ is much stronger than Lindeberg's condition, and Lyapunov's condition for $\delta > 0$ much less stronger than Lindeberg's. Stronger means here that it imposes mpre restrictions (and therefeore less cases will satisfy it). Therefore, assuming Lyapunov's condition for $\delta > 0$ is assuming Lindeberg's and much more, which is kinda redundant. $\endgroup$
    – William M.
    Commented Apr 29, 2022 at 16:50

1 Answer 1

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For $\delta=0$, the expression $\sum_{m=1}^ns_n^{-(2+\delta)}\mathbb{E}[(X_m)^{2+\delta}]$ becomes $\sum_{m=1}^ns_n^{-2}\mathbb{E}[(X_m)^{2}]=1$ hence a variation of Lyapunov's condition for $\delta=0$ cannot take place.

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