2
$\begingroup$

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).

$\endgroup$
  • 1
    $\begingroup$ The two statements are equivalent, just divide by $\mu$. A sequence $(u_n)$ tends to $1$ iff $|1-u_n| \to 0$. $\endgroup$ – Ewan Delanoy Nov 5 '13 at 6:02
  • $\begingroup$ Ok, thank you. The explanation with sequence makes sense... $\endgroup$ – HellRazor Nov 5 '13 at 6:06
2
$\begingroup$

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 $\mu<0$,

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

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

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.