Let $X_1, X_2, ...$ be a sequence of independent and identically distributed random variables, such that $X_i\sim Bern(p)$. Now, let $Y_{1,n}, Y_{2,n}, ...$ be random variables, such that $Y_{i,n}\sim Bern(p_{i,n})$, with $\lim_{n \to \infty} Cov(Y_{i,n}, Y_{j,n})=0$, for $i\neq j$, and $Y_{i,n} \overset{P}{\to} X_i, \forall i$.
I have to prove that
$\frac{Y_{1,n}+...+Y_{n,n}}{n}\overset{P}{\to}p$.
Is it possible? Otherwise, what additional assumptions should be made?
Thank you for your attention.
This would be true if $Cov(Y_{i,n},Y_{j,n})=0$ for all $i\neq j$.
It is false otherwise. Here is a counter-example: Consider $\{X_i\}_{i=1}^{\infty}$ i.i.d. $Bern(1/2)$. For $i,n \in \{1, 2, 3, ...\}$ define $$Y_{i,n} = \left\{\begin{array}{cc} X_i & \mbox{if $i\leq n/2$} \\ X_1 & \mbox{else} \end{array}\right.$$ Then $Y_{i,n} \sim Bern(1/2)$ for all $i,n \in \{1, 2, 3, ...\}$. Also, for any fixed positive integer $i$, we have $Y_{i,n}=X_i$ whenever $n\geq 2i$. So $Y_{i,n}\rightarrow X_i$ surely (as $n\rightarrow \infty$).
Also, for any fixed $i,j$ with $i \neq j$, we have for all $n> 2\max\{i,j\}$: $$Cov(Y_{i,n}, Y_{j,n}) = Cov(X_i,X_j)=0$$ and so $$ \lim_{n\rightarrow\infty} Cov(Y_{i,n}, Y_{j,n})=0$$
However for positive integers $n$ we have $$ \frac{1}{2n}\sum_{i=1}^{2n}Y_{i,2n}= \frac{1}{2n}\sum_{i=1}^nX_i + \frac{X_1}{2}$$ and this does not coverge to $1/2$ in probability. In fact it converges to $\frac{1}{4} + \frac{X_1}{2}$ almost surely.