2
$\begingroup$

I have a pretty simple question about how to use the Chebyshev's inequality in this case:

First of all we know that for expected value $\mu$ and variance $\sigma^2 \leq \infty$ and a random variable $X$, it holds for $k > 0$:

$$P\left[|X - \mu| \geq k \right] \leq \frac{\sigma^2}{k^2}$$

In a case where we have a random variable Y with $\mu = 50$ and $\sigma^2 = 25$ what is the probability of the random variable to have a value between 40 and 70 ?

I know that is possible to find this probability if instead of 70 we had 60 since: $$P[40 < X < 60] = P[-10 < X - 50 < 10] = P[|X - 50| < 10]$$ So we compute $1 - P[|X - 50| \geq 10]$ to find what we search for.

But it's not possible to use the same approach for a value between 40 and 70 right? Cause we would have $P[|X - 55| < 15]$ and we couldn't use the Chebyshev's inequality am I right? Which other ways would we have to solve the problem, if some ?

$\endgroup$
  • $\begingroup$ "...it's possible to find this probability if..." The Chebyshev's inequality does not allow to find the probability, only to bound it (and the bounds are not typically tight). $\endgroup$ – leonbloy Jan 8 '17 at 15:35
2
$\begingroup$

Note $P[|X-55|>15]\leq \dfrac{E(X-55)^2}{15^2}=\dfrac{Var(X)+(E(X)-55)^2}{15^2}$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Oh I didn't knew that, so I could solve this with $\frac{25 + (50-55)^2}{15^2}$? Is there a reference to that equation? Or could you show me how the transofrmations you made to reach that equality? $\endgroup$ – Ergo Jan 8 '17 at 14:59
  • $\begingroup$ Google markov inequality and note $|x-55|>15$ iff $(x-55)^2>15^2$ $\endgroup$ – Landon Carter Jan 8 '17 at 15:47
  • $\begingroup$ Oh thanks I got it :) $\endgroup$ – Ergo Jan 9 '17 at 12:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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