What Is the Minimum Upper Bound of P(X>=5) Given -10<=X<=10 and E(X)=2?

[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?

 

[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.

 

[ParisSpart]

weights? what do you mean?

 

[Ray Vickson]

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?

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.

 

[ParisSpart]

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

 

[mfb]

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.

 

[Ray Vickson]

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

Which post are your responding to? Use the proper 'reply' buttons; otherwise, it is impossible to tell what part of the thread your post addresses.

 

[haruspex]

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.

Isn't that predicated on X>=0?

 

[Ray Vickson]

Isn't that predicated on X>=0?

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

 

[haruspex]

I've been dizzy and feverish for most of today and I should have stayed in bed.

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.

 

[Ray Vickson]

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.

Better now.

 

[ParisSpart]

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

 

[haruspex]

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

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.

 

[ParisSpart]

What do you mean with LMVT?

 

[haruspex]

What do you mean with LMVT?

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