Probability limits of random variable sums I have $X_1, X_2, X_3, \cdots$ which are independent random variables with the same non-zero mean ($\mu\ne0$) and same variance $\sigma^2$.
I would like to compute $$\lim_{n\to\infty} P[\frac{1}n\sum^n_{i=1}X_i < \frac{\mu}{2}]$$ for $\mu<0$ and $\mu>0$.
My initial thought was to use the central limit theorem but it indicates the variables must be identically distributed which I dont have here -- only have first and second order moments are similar.
Any thoughts on how to start tackling this?
 A: Chebychev's inequality may help.

 For $\mu > 0$, we have $$P\left(\frac{1}{n} \sum_{i=1}^n X_i < \frac{\mu}{2}\right) \le P\left(\left|\frac{1}{n} \sum_{i=1}^n X_i - \mu \right| > \frac{\mu}{2}\right) \le \frac{\sigma^2/n}{\mu^2/4} \to 0.$$    For $\mu < 0$, we have $$P\left(\frac{1}{n} \sum_{i=1}^n X_i \ge \frac{\mu}{2}\right) \le P\left(\left|\frac{1}{n} \sum_{i=1}^n X_i - \mu \right| \ge \frac{\mu}{2}\right) \le \frac{\sigma^2/n}{\mu^2/4} \to 0$$ so $P\left(\frac{1}{n} \sum_{i=1}^n X_i < \frac{\mu}{2}\right) \to 1$.

A: Define $M_n=\frac{1}{n}\sum_{i-1}^nX_i$.
$E[M_n]=\mu$, $Var[M_n]=\frac{\sigma_X^2}{n}$
For $\mu<0$:
$$P[M_n<\frac{\mu}{2}]$$
$$P[M_n-\mu<\frac{\mu}{2}-\mu]$$ (Subtracting $\mu$ from both sides)
$$P[M_n-\mu<\frac{\mu}{2}-\mu]$$
$$P[M_n-\mu< -\frac{\mu}{2}]$$
(using Chebyshev inequality)
$$P[|M_n-\mu| < -\frac{\mu}{2}]$$ 
$$P[|M_n-\mu| < -\frac{\mu}{2}] \geq 1 - \frac{\sigma_X^2/n}{\mu^2/4}$$ which goes to zero when n goes to infinity, so the Probability is $1-0=1$. 
For $\mu > 0$:
$$P[M_n<\frac{\mu}{2}]$$
$$P[M_n-\mu<\frac{\mu}{2}-\mu]$$ (Subtracting $\mu$ from both sides)
$$P[M_n-\mu<\frac{\mu}{2}-\mu]$$
$$P[M_n-\mu< -\frac{\mu}{2}]$$
(flip inequality due to division of both sides by -1)
$$P[-M_n+\mu > \frac{\mu}{2}]$$
(using Chebyshev inequality and fact of $|A|=|-A|$)
$$P[|M_n-\mu| > \frac{\mu}{2}]$$ 
$$P[|M_n-\mu| < \frac{\mu}{2}] \leq \frac{\sigma_X^2/n}{\mu^2/4}$$ which goes to zero when n goes to infinity. So, the probability is 0.
