Suppose we have random variables, $\{Y_{in}\}_{i\leq n, n\in \mathbb{N}}$, such that
$$ Y_{in} \xrightarrow{p} 0, \text{ as $n \to \infty$}, $$
for all $i$. We also have that, for any fixed $n$, the random variables $Y_{1n}, Y_{2n}, \ldots, Y_{nn}$ are identically distributed (but not independent).
Is it true that
$$ \frac{1}{n} \sum_{i=1}^n Y_{in} \xrightarrow{p} 0, \text{ as $n \to \infty$ (?)} $$
I am unable to come up with a counterexample or proof.
Best Answer
Let $\Omega=[0,\,1)]$ with Borel $\sigma$-algebra and $\mathbb P(B)=\lambda(B)$ where $\lambda$ is the Lebesgue measure. Set $$ Y_{in}(\omega) = \begin{cases}n, & \frac{i-1}{n}\leq \omega <\frac{i}{n},\cr 0 & \text{otherwise}. \end{cases} $$ Then for any $i\leq n$, $$ \mathbb P(Y_{in}=n)=\frac1n=1-\mathbb P(Y_{in}=0) $$ and $Y_{in}\xrightarrow{p}0$ as $n\to\infty$ for any $i$.
But $\sum_{i=1}^n Y_{in}=n$ for any $\omega\in\Omega$ and $$ \frac1n\sum_{i=1}^n Y_{in} \to 1. $$ Moreover, this ratio equals to $1$ for any $n$, for any $\omega\in\Omega$.