$Convergence in Lp$

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$Convergence in Lp$

Let $\{X_n\}$ be a sequence of random variables and $S_n:=\sum_{j=1}^n X_j$. Proof that if $X_n\overset{\mathcal{L}_p}{\to}0$ for some $p\geq 1$, then $n^{-1}S_n\overset{\mathcal{L}_p}{\to} 0$ but the converse is not true in general.

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For the converse let $X_n$ simply be the constant random variable $(-1)^n$, then $X_n$ does not converge to $0$ in $L^p$, but $S_n = \frac{1+(-1)^n}n$ certainly does.

Now if $X_n\to 0$ in $L^p$ then $\|X_n\|_{L^p}\to0$ and we suppose $\|X_n\|\neq\infty$ for all $n$. Now for any $\epsilon$ there must be some $N$ where one has if $n≥N$ then $\|X_n\|<\epsilon$. It follows that for $n>N$ that then:

$$\|S_n\|≤\frac1n \sum_{j=1}^N\|X_j\|+\frac1n\sum_{j=N+1}^n\|X_j\|<\frac1n\sum_{j=1}^N\|X_j\|+\frac{n\epsilon}{n}$$
Now clearly you may find an $M\in\Bbb N$ with $M>\frac{\sum_{j=1}^N\|X_j\|}{\epsilon}$, whence for any $n>\max(N,M)$ you have
$$\|S_n\|≤ \frac\epsilon{\sum_{j=1}^N\|X_j\|}\sum_{j=1}^N\|X_j\|+\epsilon = 2\epsilon$$

Since $\epsilon$ was arbitrary this shows that $S_n\to0$ in $L^p$.

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$Convergence in Lp$

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