Let $(V^i_N)_{i\ge 1}$ be a sequence of i.i.d. real random variables such that $|V^i_N|\le c$ for some $c$ independent of $N$. Suppose that each $V^i_N$ converges in law to some random variable $V^i$ with $E[V^i]=m$. Then
$
\lim\limits_{N\to \infty} \frac 1N \sum\limits_{i=1}^N V^i_N=m
$
in law.
It seems like it should be solved by the law of large numbers, but they don't have the same mean, so I don't know how to use it here.
Thanks for your answers!
Best Answer
Setup
For each $N \in \{1, 2 ,3 ,...\}$ let $\{V_N^i\}_{i=1}^{\infty}$ be a sequence of i.i.d. random variables. The variables across different sequences (for different $N$) need not be independent nor identically distributed. Assume there is a finite constant $c$ such that with prob 1: $$ |V_N^i|\leq c \quad \forall N, i \in \{1, 2, 3, ...\}$$ Suppose that there are additional random variables $\{V^i\}_{i=1}^{\infty}$ such that $E[V^i]=m$ for each $i \in \{1, 2, 3,...\}$ (for some $m \in \mathbb{R}$) and $V_N^i$ converges in distribution to $V^i$ as $N\rightarrow\infty$. That is, for all points $t \in \mathbb{R}$ at which the CDF of $V^i$ is continuous we have: $$ \lim_{N\rightarrow\infty} P[V_N^i>t] = P[V^i>t]$$
Define \begin{align} m_N &= E[V_N^1] \quad \forall N \in \{1, 2, 3, ...\}\\ A_N &= \frac{1}{N}\sum_{i=1}^N V_N^i\quad \forall N \in \{1, 2, 3, ...\} \end{align}
Observe that $m_N = E[V_N^i]$ for all $i, N \in \{1, 2, 3, ...\}$.
Claim 1 (Convergence of the means):
$$\lim_{N\rightarrow\infty} m_N = m$$
Claim 2 (Convergence in probability):
For all $\epsilon>0$ we have $\lim_{N\rightarrow\infty} P[|A_N-m|\geq \epsilon] = 0$.
Proof steps:
Assume Claim 1 holds. To prove Claim 2 you can fix $\epsilon>0$, argue that for all sufficiently large $N$ we have $P[|A_N-m|\geq \epsilon] \leq P[|A_N-m_N|\geq \epsilon/2]$, and use the Markov/Chebyshev inequality.
To prove Claim 1 you can use $V_N^i+c \in [0, 2c]$ with prob 1 and $E[V_N^i+c] = \int_0^{\infty} P[V_N^i+c>t] dt$.