# Inequality regarding sample mean

I was looking at the book "Asymptotic Theory of statistics and probability, DasGupta A., 2008" and in one point of a proof they use an inequality which I have not been able to understand.

Given that $$X_i, i \in \{1,...,n\}$$ are independent and identically distributed random variables, with mean $$\mu$$, and with sample mean $$\overline{X}_n$$, they state that

$$\sum_{i=1}^n |X_i - \overline{X}_n|^3 \leq 2³\left( \sum_{i=1}^n |X_i - \mu|^3 + n|\mu - \overline{X}_n|^3 \right)$$

I don't know how the 2³ term appears. I've tried adding and substracting $$\mu$$ but I have not been able to proof the inequality. It would be perfect if you could lend me a hand.

Thanks

• which page does it appear on? – user321627 Apr 12 at 19:01
• Page 466, chapter 29 – RJ-mac Apr 12 at 23:33

## 2 Answers

For $$a$$ and $$b$$ non-negative you have $$(a+b)^3 \le (2\max(a,b))^3 = 2^3\max(a,b)^3 \le 2^3 (a^3+b^3)$$

So here you can say $$\sum\limits_{i=1}^n \bigg|X_i - \overline{X}_n\bigg|^3 \\ = \sum\limits_{i=1}^n \bigg|(X_i -\mu) + (\mu-\overline{X}_n)\bigg|^3 \\ \le \sum\limits_{i=1}^n \bigg( |X_i -\mu| + |\mu-\overline{X}_n|\bigg)^3 \\ \le \sum\limits_{i=1}^n 2^3\bigg( |X_i -\mu|^3 + |\mu-\overline{X}_n|^3\bigg) \\ = 2^3\bigg(\bigg( \sum\limits_{i=1}^n |X_i -\mu|^3\bigg) + \bigg( \sum\limits_{i=1}^n|\mu-\overline{X}_n|^3\bigg)\bigg)\\ = 2^3\bigg(\bigg( \sum\limits_{i=1}^n |X_i -\mu|^3\bigg) + n|\mu-\overline{X}_n|^3\bigg)$$

• Ok perfect! Thank you – RJ-mac Apr 12 at 22:13

The inequality is ugly but it is true. First call $$X_i-\mu=a_i$$ and $$\mu-\bar X_n=b$$. If $$b=0$$ it is trivial. Otherwise divide by $$|b|^3$$ both sides. Your inequality now reads $$\sum|a_i/b+1|^3\le 8(\sum|a_i/b|^3+n)$$ It is enough to prove that $$|t+1|^3\le 8(|t|^3+1)$$ For all $$t$$ (You can then plug in $$t=a_i/b$$ and add them up). Consider the cases one by one to remove the absolute value. If $$t>0$$ your inequality reduces to $$7t^2+6t+7\ge 0$$ Which is clearly true for all $$t$$. If $$-1\le t\le 0$$ it reduces to $$9t^3+3t^2+3t-7\le 0$$ Which is true even without the $$-7$$, and if $$t\le -1$$ it reduces to $$7t^2+10t+7\ge 0,$$ Which is true for all $$t$$ as well.

• Nice! Thank you – RJ-mac Apr 12 at 22:14