Does the Central Limit Theorem Imply the strong Law of Large Numbers?

Question

Assume that $(X_{k})_{k\geq 0}$ is a stationary (or weakly stationary) process defined on the same probability space $(\Omega,\mathcal{F},\mathbb{P})$. Can we assert from the convergence in distribution $$\frac{1}{\sqrt{n}}\sum_{k=0}^{n-1}X_{k}\Rightarrow_{n} N(0,1)$$ that $$\lim_{n}\frac{1}{n}\sum_{k=0}^{n-1}X_{k}=0,$$ $\mathbb{P}-$a.s.?

Answer 1

I will treat the strictly stationary case.

• Assume that $\mathbb E\left|X_0\right|\lt\infty$ and that $\frac{1}{\sqrt{n}}\sum_{k=0}^{n-1}X_{k}\Rightarrow_{n} \mathcal N(0,1)$. Then $\frac{1}{n}\sum_{k=0}^{n-1}X_{k}\to 0$ in probability. Using Birkhoff ergodic theorem, we find that $\frac{1}{n}\sum_{k=0}^{n-1}X_{k}\to \mathbb E\left[X_0\mid\mathcal I\right]$ in $\mathbb L^1$ and almost surely, where $\mathcal I$ denotes the $\sigma$-algebra of shift-invariant sets of the sequence $\left(X_k\right)_{k\geqslant 0}$. Consequently, by identifying the limits, we get that $\mathbb E\left[X_0\mid\mathcal I\right]=0$ almost surely, hence $\frac{1}{n}\sum_{k=0}^{n-1}X_{k}\to 0$ in the almost sure sense.

• If we do not assume a finite first moment, we may not have the strong law of large numbers. Actually, we can construct a $1$-dependent sequence $\left(X_k\right)_{k\geqslant 0}$ which satisfies the central limit theorem but not the strong law of large numbers. Take $X_k=\varepsilon_k+g_k-g_{k+1}$ where $\left(\varepsilon_k\right)_{k\geqslant 0}$ and $\left(g_k\right)_{k\geqslant 0}$ are independent i.i.d. sequences and $\varepsilon_0$ is centered with unit variance, and $g_0$ is not integrable. Since $\sum_{k=0}^{n-1}X_k=\sum_{k=0}^{n-1}\varepsilon_k+g_0-g_n$ and $(g_0-g_n)/\sqrt n\to 0$ in probability, we have $\sum_{k=0}^{n-1}X_k/\sqrt n\to \mathcal N(0,1)$. On the other hand, the sequence $\left(g_n/n\right)_{n\geqslant 1}$ does not converge to $0$ almost surely. This is due to the divergence of the series $\sum_{n\geqslant 1} \mathbb P\left\{g_n\geqslant n\right\}$ and the second Borel-Cantelli lemma.