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*}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.