Probabilities inequalities

1. Jun 7, 2013

ParisSpart

the random value X has the inequality , -10<=X<=10 and E(X)=2, what is the minimum upper bound
of the probability P(X>=5) ?

my first thought was to find this P(X>=t)<=E(X)/t which is 2/5 from Markov but its not correct, any ideas?

2. Jun 7, 2013

verty

My very crude idea is to set all the weights to zero and then choose two weights only to adjust. I won't say more.

3. Jun 7, 2013

ParisSpart

weights? what do you mean?

4. Jun 7, 2013

Ray Vickson

Why do you say 2/5 is wrong? It is not wrong, because for any number p (0 ≤ p ≤ 2/5) we can find a random variable X that satisfies the conditions and gives P(X ≥ 5) = p. So, no number less than 2/5 can possibly be an UPPER bound, because if we take a number p with p < 2/5 we can always find a suitable X having P(X ≥ 5) > p (but ≤ 2/5); in fact, we can find infinitely many suitable X having P(X ≥ 5) = 2/5 exactly. On the other hand, 2/5 is certainly an upper bound, because no suitable X can have a probability P(X ≥ 5) that exceeds 2/5---that's what Markov's inequality is all about.

I will leave it to you to verify the statements I have made; you do need to verify them, since otherwise your "solution" would be that 'somebody said so', and that is not a proof.

5. Jun 7, 2013

ParisSpart

yea but why X is -10<X<10 ? this made me to think that i must find other bounds upper...

6. Jun 7, 2013

Staff: Mentor

Consider distributions which can take two values only.
This is the same as "the weight is non-zero for two values only", verty's idea.

7. Jun 7, 2013

8. Jun 7, 2013

haruspex

Isn't that predicated on X>=0?

9. Jun 7, 2013

Ray Vickson

Yes, sorry. Disregard my silly posting; I've been dizzy and feverish for most of today and I should have stayed in bed.

10. Jun 7, 2013

haruspex

Sorry to hear that - hope you feel better soon.

ParisSpart, in order to make use of the Markov inequality you will need to map X to a random variable which is always >= 0.

11. Jun 7, 2013

Ray Vickson

Better now.

12. Jun 8, 2013

ParisSpart

how i will find X>=0 , i can find that abs(X)=<10 what i can take from this?

13. Jun 8, 2013

haruspex

No, X can be < 0, so I'm suggesting creating a new random variable Y as a function of X (the simpler the better) which satisfies Y >= 0. P[X>=5] will equal P[Y>=c] for some c. You can apply LMVT to Y.

14. Jun 8, 2013

ParisSpart

What do you mean with LMVT?

15. Jun 8, 2013

haruspex

Sorry, mixing up threads. I meant Markov's inequality.