Suppose $\epsilon_{it}$ is an i.i.d. random variable with $E(\epsilon_{it})=0$ and $E(\epsilon_{it}^2)=1$ for i=1,2,...N and t=1,2,...,T. For large N and small T, from Law of Large Numbers, $\underset{N\to\infty}{plim}\frac{1}{N}\sum_i\left(\frac{\sum_t \epsilon_{it}}{\sqrt{T}} \right)^3=E\left[\left(\frac{\sum_t \epsilon_{it}}{\sqrt{T}} \right)^3\right]=O(T^{-\frac{1}{2}})$.
My question is how large N should be relative to T for the above to hold. However, for small N and large T, from Central Limit Theorem $\frac{\sum_t \epsilon_{it}}{\sqrt{T}}\overset{d}{\rightarrow} N(0,1)$. Hence
$\frac{1}{N}\sum_i\left(\frac{\sum_t \epsilon_{it}}{\sqrt{T}} \right)^3=O_p(N^{-\frac{1}{2}})$.
So how small should N be relative to T for the above to hold? What about the cases in between?
Could anyone help me? Many thanks.
I think I can now work it out by myself. Suppose the cummulants of $\epsilon_{it}$ of different orders exist. Note that $x_{i}=\frac{\sum_t \epsilon_{it}}{\sqrt{T}}$ is a standardized sum with cumulants: $\kappa^1=0$, $\kappa^2=1$, $\kappa^3=O(T^{-\frac{1}{2}})$,...
One can then write down the cumulant generating function (c.g.f.) of $z_{i}=\left(\frac{\sum_t \epsilon_{it}}{\sqrt{T}} \right)^3$ as
$\xi\kappa_{z}^1+\frac{\xi^2\kappa_{z}^2}{2}+\frac{\xi^3\kappa_{z}^3}{3!}+...$ with $\kappa_{z}^1=\kappa^3=O(T^{-\frac{1}{2}})$, $\kappa_{z}^2=O(1)$...
Note that the odd/even order cumulant is of order $O(T^{-\frac{1}{2}})$/$O(1)$. For example, $\kappa_{z}^2=\kappa(x_ix_ix_i,x_ix_ix_i)$ (generalized cumulant), the largest term of which is $\kappa^2\kappa^2\kappa^2=O(1)$. Other terms will either be $0$ or tend to $0$ as $T->\infty$. Any natural number can be represented by $n\cdot 1+m\cdot 2+l\cdot 3$ for n, m and l being natural numbers or 0. To find the largest term, one should choose $n=0$ and minimize $l$ for numbers equal to 6,9,12,...
The c.g.f. for $\frac{1}{N}\sum_i z_{i}$ is therefore
$\xi\kappa_{z}^1+\frac{1}{N}\frac{\xi^2\kappa_{z}^2}{2}+\frac{1}{N^2}\frac{\xi^3\kappa_{z}^3}{3!}+...$
which implies $\underset{N\to\infty}{plim}\frac{1}{N}\sum_i z_{i}=\kappa_{z}^1=O(T^{-\frac{1}{2}})$ regardless of how $T$ behaves as long as $N$ approaches infinite. For fixed $N$ and infinte $T$, one can have $\frac{1}{N}\sum_i\left(\frac{\sum_t \epsilon_{it}}{\sqrt{T}} \right)^3=O_p(N^{-\frac{1}{2}})$.
Also note that the c.g.f. for $\frac{\sum_i z_{i}}{\sqrt{N}}$ is $\sqrt{N}\xi\kappa_{z}^1+\frac{\xi^2\kappa_{z}^2}{2}+N^{-\frac{1}{2}}\frac{\xi^3\kappa_{z}^3}{3!}+\frac{1}{N}\frac{\xi^4\kappa_{z}^4}{4!}...$
If $N$ is fixed and $T$ is infinite, $\frac{\sum_i z_{i}}{\sqrt{N}}$ will converge to a non-normal random variable with mean equal to 0. If $N$ and $T$ grows to infinity at the same rate, $\frac{\sum_i z_{i}}{\sqrt{N}}=O(\sqrt{\frac{N}{T}})+O_p(1)$ will converge to a normal random variable with non-zero mean. To remove the bias, one has to subtract $\sqrt{N}\kappa_{z}^1$ from $\frac{\sum_i z_{i}}{\sqrt{N}}$.
Hence in general, $\sum_i\left(\frac{\sum_t \epsilon_{it}}{\sqrt{T}} \right)^k=\begin{cases} O(N)+O_p(N^{\frac{1}{2}}),&k&is&even\\O(NT^{-\frac{1}{2}})+O_p(N^{\frac{1}{2}}),&k>1,&and\,\,k\,\,is&odd\\O_p(N^{\frac{1}{2}}),&k&=&1\end{cases}$