# Weak Law of Large Numbers -Special Case

For independent and identically distributed random variables $X_1, X_2, \cdots$ with finite mean and variance, by the weak Law of Large Numbers, $\frac{1}{N}\sum_{i=1}^{N} X_i$ converges in probability to $\mathbb{E}[X_i]$: $$\frac{1}{N}\sum_{i=1}^{N} X_i \xrightarrow{p}\mathbb{E}[X_i]$$

If $f(N)$ is a continuous function such that $\lim_{N \to \infty} f(N)=c \in \mathbb{R}^{+}$, then is the following true? $$\frac{f(N)}{N}\sum_{i=1}^{N} X_i \xrightarrow{p}c \mathbb{E}[X_i]$$

We know that $\lim\limits_{N \to \infty} \mathbb{P}\left(|\frac{1}{N}\sum_{i=1}^{N} X_i - \mathbb{E}[X_i]|>\epsilon\right)=0$ and thus

$$\lim\limits_{N \to \infty} \mathbb{P}\left(|\frac{f(N)}{N}\sum_{i=1}^{N} X_i - f(N)\mathbb{E}[X_i]|>\epsilon f(N)\right)=0$$

How can we continue form here?

We can assume that $|f(N)|\le M$, for all $N$ and certain $M>0$. Moreover, there exist $N_0$ such that, for $N>N_0$, $|f(N)-c|< \varepsilon/(2 E(X_1))$. Since \begin{align*} \Big|\frac{f(N)}{N}\sum_{i=1}^N X_i - cE(X_1) \Big| \le \Big|\frac{f(N)}{N}\sum_{i=1}^N X_i - f(N)E(X_1) \Big| + \big|f(N)-c\big|\, E(X_1), \end{align*} we can have that \begin{align*} \Big|\frac{1}{N}\sum_{i=1}^N X_i - E(X_1) \Big| &\ge \frac{\Big|\frac{f(N)}{N}\sum_{i=1}^N X_i - cE(X_1) \Big| - \big|f(N)-c\big|\,E(X_1)}{f(N)}\\ &\ge \frac{\Big|\frac{f(N)}{N}\sum_{i=1}^N X_i - cE(X_1) \Big| - \frac{\varepsilon}{2}}{M}. \end{align*} Then, for $N> N_0$, \begin{align*} \left\{\omega: \Big|\frac{f(N)}{N}\sum_{i=1}^N X_i - cE(X_1) \Big|\ge \varepsilon\right\} \subset \left\{\omega: \Big|\frac{1}{N}\sum_{i=1}^N X_i - E(X_1) \Big|\ge \frac{\varepsilon}{2M}\right\}. \end{align*} That is, \begin{align*} \lim_{N\rightarrow\infty}\mathbb{P}\left(\Big|\frac{f(N)}{N}\sum_{i=1}^N X_i - cE(X_1) \Big|\ge \varepsilon\right) = 0. \end{align*}