Let $(X_n)_{n\geq 1} $ a sequence of independent random variables. We assume that $X_n \sim \mathcal{N}(0,\sigma_n^2)$ and that $(\sigma_n)_{n\geq 1}$ is a vanishing sequence of positive numbers. Let $\delta_0$ be the Dirac law with mass at $0$. Using Lévy's continuity theorem, it is easy to see that $(X_n)_{n\geq 1} $ weakly converges to $\delta_0$.
My question is : does $(X_n)_{n\geq 1} $ almost surely converges to $0$ ?
A natural way to prove the (potential) almost sure convergence would be to use the Borel-Cantelli lemma. It tells us that, if $\sum_{n\geq 1}\mathbb{P}(|X_n|\geq\varepsilon)<\infty$ for all $\varepsilon >0$, then $X_n \overset{a.s}{\longrightarrow} 0$.
Using a variant of Chebyshev's inequality, we can see that, for all integer $p\geq 2$, $$\mathbb{P}(|X_n|\geq\varepsilon )\leq\frac{\mathbb E |X_n|^p}{\varepsilon^p}=\frac{\mu_p}{\varepsilon^p}\sigma_n^p$$where $\mu_p$ is the $p$-th moment of the standard Gaussian.
which leads to $$\sum_{n\geq 1}\sigma_n^p<\infty \; \text{for some} \;p \Rightarrow (X_n \overset{a.s}{\longrightarrow} 0). $$
Is is possible to obtain a more general result? Or to find a necessary and sufficient condition on the speed of convergence of $(\sigma_n)_{n\geq 1}$ for $(X_n \overset{a.s}{\longrightarrow} 0)$ to be true?
Best Answer
Since the sequence $(X_n)_{n\geqslant 1}$ is independent, an application of the Borel-Cantelli lemma shows that $X_n\to 0$ almost surely is equivalent to the convergence of the series $\sum_{n\geqslant 1}\mathbb P(|X_n|>\varepsilon)$ for each fixed $\varepsilon$. Thus, $X_n\to 0$ almost surely is and only if $\sum_{n=1}^\infty\mathbb P(\sigma_n|N|>\varepsilon)$ converges for each positive $\varepsilon$, where $N$ is a random variable with standard normal distribution.
To get a more tractable condition, one can use asymptotic equivalents for the normal distribution, like $$\left(\frac{1}{x}-\frac{1}{x^3}\right)\exp\left(-\frac{x^2}{2}\right)\leqslant\mathbb P(|N|>x)\leqslant \frac{1}{x}\exp\left(-\frac{x^2}{2}\right).$$