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I have a question considering Chebyshev's inequality.

$$ P[|X-E[X]| > a] \leq \frac{Var[X]}{a^2} $$

Let's say we have 5 possible events, probability is uniformly distributed, so every event has $$ PR[event] = \frac{1}5$$

Now we want to know the upper bound of the probability of a specific event happening at least 10 times, if we have 50 trials. This would lead to $$E[X] = 10$$

If we then use Chebyshev's inequality:

$$PR[X > 10] = PR[X -10 > 0] = PR[|X-10|>0] \leq \frac{Var[X]}{a^2} $$

Which is a problem. But there surely must be a way of getting a upper bound for this case with Chebyshev's inequality right? Or did I miss something?

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The Chebyshev's inequality is useful only when $a^2> Var[X]$

Otherwise you get some unuseful bounds such as $P(Something)<1000$!

P.S.

In the common proof of chebyshev you get this:

$a^2 P(|X-EX|\ge a)\le Var[X]$

So there is not any algebraic problem of dividing by zero