Proving $\frac{1}{n}(\sum_{k=1}^{n}a_k\xi_{k}+\sum_{k=1}^{n}\eta_{k})\stackrel{P}\to 0 \iff \frac{1}{n^2}\sum_{k=1}^{n}a_k^2\to0$

Question

A problem related to the weak law of large numbers，$\{\xi_k\}$ and $\{\eta_k\}$ are independent of each other and $\{\xi_k\}$ and $\{\eta_k\}$ are all independent sequence. They all obey the standard normal distribution. Prove $$ \frac{1}{n}(\sum_{k=1}^{n}a_k\xi_{k}+\sum_{k=1}^{n}\eta_{k})\rightarrow 0 \quad \text{in probability} \\ \iff \\ \frac{1}{n^2}\sum_{k=1}^{n}a_k^2\rightarrow0 $$ I don’t understand why such conditions are needed to ensure convergence in probability. Why can't these conditions guarantee the establishment of the law of large numbers?

Answer 1

Based on the assumptions, the vector $(\xi_1,\dots,\xi_n,\eta_1,\dots,\eta_n)$ is Gaussian centered and consists of independent random variables. As a consequence, $\frac{1}{n}\left(\sum_{k=1}^na_k\xi_k+\sum_{k=1}^n\eta_k\right)$ has a centered normal distribution with variance $\frac1{n^2}\sum_{k=1}^na_k^2+\frac 1n$. To conclude, it remains to notice that if $Y_n$ has a normal distribution with mean $0$ and variance $\sigma_n^2$, then convergence in probability of $(Y_n)$ to $0$ is equivalent to $\sigma_n\to 0$.