Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is a question that I happen to think of when looking at the Chebyshev's Inequality. In the inequality, it has this: $$ P(\left| X-\mu \right| \ge k\sigma )\le \frac { 1 }{ { k }^{ 2 } } $$

Suppose I have an expectation of $67.5$ and standard deviation of $7.5$. I want to find what probability of $P(30\le X\le 105)$.

$ P(30\le X\le 105)\quad \le \quad P(\left| X-\mu \right| \le 37.5)\\ \qquad \qquad \qquad \qquad \le \quad P(\left| X-\mu \right| \le 5\sigma )\\ \qquad \qquad \qquad \qquad \le \quad 1-P(\left| X-\mu \right| >5\sigma ) $

At this stage, there is a problem. By using Chebyshev's inequality, I can say $P(\left| X-\mu \right| \ge 5\sigma )\le \frac { 1 }{ { 5 }^{ 2 } } $ but in the above steps that I have done, I reached $1-P(\left| X-\mu \right| >5\sigma )$, which is just "more than", not "more than or equals". If I still use $\frac { 1 }{ { 5 }^{ 2 } } $ as its result, the result will not be accurate because something extra is minus out.

What should I do here?

share|improve this question
    
It seems your problem is not the strictness of the inequality, but that you should be using Chebychev's to put a lower bound on the probability $P(30\leq X \leq 105)$, not an upper bound. –  William DeMeo Feb 22 '12 at 9:24

2 Answers 2

up vote 1 down vote accepted

You have

$$P(30 \leq X \leq 105) = P(|X-\mu| \leq 5 \sigma) = 1 - P(|X-\mu|>5\sigma).$$

By Chebychev's, you have

$$ P(|X-\mu|>5\sigma) \leq P(|X-\mu|\geq5\sigma) \leq 1/5^2.$$

Therefore, $$P(30 \leq X \leq 105) = 1 - P(|X-\mu|>5\sigma) \geq 1 - 1/5^2$$

share|improve this answer

If $P(\left| X-\mu \right| \ge k\sigma )\le \dfrac { 1 }{ { k }^{ 2 } }$ then clearly you have $P(\left| X-\mu \right| \gt k\sigma ) \le P(\left| X-\mu \right| \ge k\sigma )\le \dfrac { 1 }{ { k }^{ 2 } }$. You can go further than this and show that you cannot have equality simultaneously in both cases, leading to

$$P(\left| X-\mu \right| \gt k\sigma )\lt \frac { 1 }{ { k }^{ 2 } }$$

but for $k \ge 1$ you can get arbitrarily close to equality, so without more information you cannot do better than this.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.