# Understanding why do we need $\delta>0$ in Lyapunov's condition?

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.

• 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. Commented Apr 29, 2022 at 13:32
• @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. Commented Apr 29, 2022 at 13:43
• 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. Commented Apr 29, 2022 at 16:50

## 1 Answer

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.