Does the weak law of large numbers imply that the ratio $\frac{\frac{1}{n}\sum_{i=1}^nX_i}{\mu}\rightarrow 1$ in probability?

Question

Suppose there is a sequence of $n$ bounded i.i.d. random variables $X_1,\ldots,X_n$, i.e. for all $i$, $a<X_i<b$ with $a$ and $b$ real constants. Denote the mean of these random variables by $\mu$. Weak law of large numbers (WLLN) states that $\frac{1}{n}\sum_{i=1}^nX_i\rightarrow\mu$ in probability, i.e. $\lim P(|\mu-\frac{1}{n}\sum_{i=1}^nX_i|\geq\epsilon)=0$ for any $\epsilon>0$.

The statement of the WLLN uses the absolute difference. I am wondering whether showing that $\lim P(|\mu-\frac{1}{n}\sum_{i=1}^nX_i|\geq\epsilon)=0$ for any $\epsilon>0$ also implies the following:

$$\frac{\frac{1}{n}\sum_{i=1}^nX_i}{\mu}\rightarrow 1$$

in probability (in my particular case with bounded r.v.'s or in general).

Answer 1

For $\mu>0$,

$$\lim P(|\mu-\frac{1}{n}\sum_{i=1}^nX_i|\geq\epsilon)=0 \Rightarrow \lim P\left(\mu\left|1-\frac{\frac{1}{n}\sum_{i=1}^nX_i}{\mu}\right|\geq\epsilon\right) = \lim P\left(\left|1-\frac{\frac{1}{n}\sum_{i=1}^nX_i}{\mu}\right|\geq\epsilon\mu^{-1}\right) = \lim P\left(\left|1-\frac{\frac{1}{n}\sum_{i=1}^nX_i}{\mu}\right|\geq\epsilon^*\right) = 0$$

For $\mu<0$,

$$\lim P(|\mu-\frac{1}{n}\sum_{i=1}^nX_i|\geq\epsilon)=0 \Rightarrow \lim P\left(\left|-\left((-\mu)+\frac {1}{n}\sum_{i=1}^nX_i\right)\right|\geq\epsilon\right) = \lim P\left((-\mu)\left|1+\frac{\frac{1}{n}\sum_{i=1}^nX_i}{-\mu}\right|\geq\epsilon\right)=\lim P\left(\left|1-\frac{\frac{1}{n}\sum_{i=1}^nX_i}{\mu}\right|\geq\epsilon(-\mu)^{-1}\right) = \lim P\left(\left|1-\frac{\frac{1}{n}\sum_{i=1}^nX_i}{\mu}\right|\geq\epsilon^{**}\right) = 0$$

Answer 2

If $\mu\neq 0$, just use the definition of convergence in probability of $\frac1n\sum_{i=1}^nX_i$ with $\mu\varepsilon$ instead of $\varepsilon$.