An independent sequence of square-integrable random variables with convergent sum of variances converges stochastically

Question

Let $(X_i)_{i\in\mathbb{N}}$ be a sequence of independent and square-integrable random variables with $\operatorname{E}\left[X_i\right]=0$ and $$\sum_{i\in\mathbb{N}}\operatorname{Var}\left[X_i\right]<\infty$$ Let $$X^{(n)}:=\sum_{i=1}^nX_i\;\;\;\text{and}\;\;\;D^{(m,n)}:=X^{(n)}-X^{(m)}$$ I want to show that $\left(X^{(n)}\right)_{n\in\mathbb{N}}$ is a stochastic Cauchy sequence, i.e. $$\lim_{m,n\to\infty}\Pr\left[\left|D^{(m,n)}\right|>\varepsilon\right]=0\;\;\;\text{for all }\varepsilon>0\tag{0}$$

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I've tried the following:

• Since $(X_i)_{i\in\mathbb{N}}$ is independent, the Bienaymé equation yields $$V_n:=\operatorname{Var}\left[X^{(n)}\right]=\sum_{i=1}^n\operatorname{Var}\left[X_i\right]$$
• Thus, $$\lim_{n\to\infty}V_n=\sum_{i\in\mathbb{N}}\operatorname{Var}\left[X_i\right]<\infty$$ implies $$\sup_{n\in\mathbb{N}}V_n<\infty\tag{1}$$
• Moreover, $$\operatorname{E}\left[X^{(n)}\right]=\sum_{i=1}^n\operatorname{E}\left[X_i\right]=0\tag{2}$$
• $(1)$ and $(2)$ imply the uniform integrability of $\left(X^{(n)}\right)_{n\in\mathbb{N}}$

So, if I could show $(0)$ - which implies that $\left(X^{(n)}\right)_{n\in\mathbb{N}}$ converges stochastically - I could conclude that $\left(X^{(n)}\right)_{n\in\mathbb{N}}$ converges in $L^2$. However, I wasn't able to find an easy proof for $(0)$.

Answer 1

Fix $\delta>0$. Since $\sum_{i \in \mathbb{N}} \text{var} \, (X_i)<\infty$, we can choose $N \in \mathbb{N}$ such that

$$\text{var} \, (D^{(m,n)}) = \sum_{i=m+1}^n \text{var}(X_i) \leq \delta$$

for all $n \geq m \geq N$. By Markov's inequality,

$$\mathbb{P}(|D^{(m,n)}| \geq \epsilon) \leq \frac{1}{\epsilon^2} \text{var} \, (D^{(m,n)}) = \frac{\delta}{\epsilon^2}.$$

Since $\delta>0$ is arbitrary, this finishes the proof.