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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?

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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 $P\left(\frac{1}{n} \sum_{i=1}^n X_i < \frac{\mu}{2}\right) \to 1$.

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    $\begingroup$ @ClementC. You're right, thanks for catching my mistake. $\endgroup$ – angryavian Dec 21 '18 at 2:56
  • $\begingroup$ Why is the right side of Chebychev's inquality have n and not n^2? $\endgroup$ – Avedis Dec 23 '18 at 1:43
  • $\begingroup$ @Avedis Think carefully about what the variance of $\frac{1}{n} \sum_{i=1}^n X_i$ is. $\endgroup$ – angryavian Dec 23 '18 at 2:08
  • $\begingroup$ Thank you, I just figure out what you mentioned on my own. However, when I complete the problem, I get a different answer then you do for $\mu>0$. Why? $\endgroup$ – Avedis Dec 23 '18 at 3:51
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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.

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  • $\begingroup$ In both cases you should have $(\mu/2)^2=\mu^2/4$ rather than $\mu^2/2$. And it seems our answers agree. $\endgroup$ – angryavian Dec 23 '18 at 4:20
  • $\begingroup$ You are right and I will fix. I misread your answer. I thought the last statement was a summary of the entire problem, not the last part exclusively. Thank you! $\endgroup$ – Avedis Dec 23 '18 at 14:04

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