Markov inequality for random variables with negative values.











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I'm given the maximum value of a random variable $X$ (for example $50$) and its mean, $mathbb E(X)=20$. How do I find the upper bound to $P(Xle -10)$?






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  • 1




    I think that you need to make the transformation $Y = X + 11$ and work from there.
    – Ekesh
    Nov 14 at 8:37















up vote
2
down vote

favorite












I'm given the maximum value of a random variable $X$ (for example $50$) and its mean, $mathbb E(X)=20$. How do I find the upper bound to $P(Xle -10)$?






random-variables







share|cite|improve this question




















  • 1




    I think that you need to make the transformation $Y = X + 11$ and work from there.
    – Ekesh
    Nov 14 at 8:37













up vote
2
down vote

favorite









up vote
2
down vote

favorite











I'm given the maximum value of a random variable $X$ (for example $50$) and its mean, $mathbb E(X)=20$. How do I find the upper bound to $P(Xle -10)$?






random-variables







share|cite|improve this question















I'm given the maximum value of a random variable $X$ (for example $50$) and its mean, $mathbb E(X)=20$. How do I find the upper bound to $P(Xle -10)$?






random-variables




random-variables






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share|cite|improve this question













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edited Nov 14 at 8:32









Jimmy R.

32.9k42157




32.9k42157










asked Nov 14 at 6:21









puffles

669




669








  • 1




    I think that you need to make the transformation $Y = X + 11$ and work from there.
    – Ekesh
    Nov 14 at 8:37














  • 1




    I think that you need to make the transformation $Y = X + 11$ and work from there.
    – Ekesh
    Nov 14 at 8:37








1




1




I think that you need to make the transformation $Y = X + 11$ and work from there.
– Ekesh
Nov 14 at 8:37




I think that you need to make the transformation $Y = X + 11$ and work from there.
– Ekesh
Nov 14 at 8:37










2 Answers
2






active

oldest

votes

















up vote
0
down vote



accepted










Define new RV: S = 50 - R



E(S) = 30



P(R<=-10) = P(S>=60)



P(S>=60) = 30/60
= 1/2



You commented: I needed to clarify one last thing. P(50-X >= 60) <= 1/3



It should be 1/2 not 1/3.






share|cite|improve this answer





















  • How is the expectation 30. Can you please explain?
    – puffles
    Nov 17 at 20:53










  • S = 50 -R so E(S) = 50 -E(R) . Since E(R) = 20. We get E(S) = 30
    – helloworld
    Nov 17 at 21:24




















up vote
1
down vote













Hint:




  • $50-X$ is a nonnegative random variable since $50$ is an upperbound.


  • Express your inequality in the form of $Pr(50-X ge c)$.







share|cite|improve this answer





















  • So it should be like this : P(50-X >= -10) ?
    – puffles
    Nov 14 at 11:23










  • $P(X le -10) = P(-X ge 10) = P(50-X ge 60)$, now apply Markov on $50-X$.
    – Siong Thye Goh
    Nov 14 at 11:24










  • Makes sense. Thanks a bunch!
    – puffles
    Nov 14 at 11:25










  • I needed to clarify one last thing. P(50-X >= 60) <= 1/3. Is this correct considering the typical formula of Markov inequality P(X >= a)? What I am trying to ask is that if a constant is added or subtracted in an interval such as in the above case, would it affect the final probability.
    – puffles
    Nov 15 at 15:26






  • 1




    yes, it is correct. if it makes you more comfortable, let $Y=50-X$ and check that $Y$ is nonnegative. If you perform some operations and prove that the two conditions are equivalent, then the probability stays the same.
    – Siong Thye Goh
    Nov 15 at 15:29











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










Define new RV: S = 50 - R



E(S) = 30



P(R<=-10) = P(S>=60)



P(S>=60) = 30/60
= 1/2



You commented: I needed to clarify one last thing. P(50-X >= 60) <= 1/3



It should be 1/2 not 1/3.






share|cite|improve this answer





















  • How is the expectation 30. Can you please explain?
    – puffles
    Nov 17 at 20:53










  • S = 50 -R so E(S) = 50 -E(R) . Since E(R) = 20. We get E(S) = 30
    – helloworld
    Nov 17 at 21:24

















up vote
0
down vote



accepted










Define new RV: S = 50 - R



E(S) = 30



P(R<=-10) = P(S>=60)



P(S>=60) = 30/60
= 1/2



You commented: I needed to clarify one last thing. P(50-X >= 60) <= 1/3



It should be 1/2 not 1/3.






share|cite|improve this answer





















  • How is the expectation 30. Can you please explain?
    – puffles
    Nov 17 at 20:53










  • S = 50 -R so E(S) = 50 -E(R) . Since E(R) = 20. We get E(S) = 30
    – helloworld
    Nov 17 at 21:24















up vote
0
down vote



accepted







up vote
0
down vote



accepted






Define new RV: S = 50 - R



E(S) = 30



P(R<=-10) = P(S>=60)



P(S>=60) = 30/60
= 1/2



You commented: I needed to clarify one last thing. P(50-X >= 60) <= 1/3



It should be 1/2 not 1/3.






share|cite|improve this answer












Define new RV: S = 50 - R



E(S) = 30



P(R<=-10) = P(S>=60)



P(S>=60) = 30/60
= 1/2



You commented: I needed to clarify one last thing. P(50-X >= 60) <= 1/3



It should be 1/2 not 1/3.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 17 at 20:17









helloworld

477




477












  • How is the expectation 30. Can you please explain?
    – puffles
    Nov 17 at 20:53










  • S = 50 -R so E(S) = 50 -E(R) . Since E(R) = 20. We get E(S) = 30
    – helloworld
    Nov 17 at 21:24




















  • How is the expectation 30. Can you please explain?
    – puffles
    Nov 17 at 20:53










  • S = 50 -R so E(S) = 50 -E(R) . Since E(R) = 20. We get E(S) = 30
    – helloworld
    Nov 17 at 21:24


















How is the expectation 30. Can you please explain?
– puffles
Nov 17 at 20:53




How is the expectation 30. Can you please explain?
– puffles
Nov 17 at 20:53












S = 50 -R so E(S) = 50 -E(R) . Since E(R) = 20. We get E(S) = 30
– helloworld
Nov 17 at 21:24






S = 50 -R so E(S) = 50 -E(R) . Since E(R) = 20. We get E(S) = 30
– helloworld
Nov 17 at 21:24












up vote
1
down vote













Hint:




  • $50-X$ is a nonnegative random variable since $50$ is an upperbound.


  • Express your inequality in the form of $Pr(50-X ge c)$.







share|cite|improve this answer





















  • So it should be like this : P(50-X >= -10) ?
    – puffles
    Nov 14 at 11:23










  • $P(X le -10) = P(-X ge 10) = P(50-X ge 60)$, now apply Markov on $50-X$.
    – Siong Thye Goh
    Nov 14 at 11:24










  • Makes sense. Thanks a bunch!
    – puffles
    Nov 14 at 11:25










  • I needed to clarify one last thing. P(50-X >= 60) <= 1/3. Is this correct considering the typical formula of Markov inequality P(X >= a)? What I am trying to ask is that if a constant is added or subtracted in an interval such as in the above case, would it affect the final probability.
    – puffles
    Nov 15 at 15:26






  • 1




    yes, it is correct. if it makes you more comfortable, let $Y=50-X$ and check that $Y$ is nonnegative. If you perform some operations and prove that the two conditions are equivalent, then the probability stays the same.
    – Siong Thye Goh
    Nov 15 at 15:29















up vote
1
down vote













Hint:




  • $50-X$ is a nonnegative random variable since $50$ is an upperbound.


  • Express your inequality in the form of $Pr(50-X ge c)$.







share|cite|improve this answer





















  • So it should be like this : P(50-X >= -10) ?
    – puffles
    Nov 14 at 11:23










  • $P(X le -10) = P(-X ge 10) = P(50-X ge 60)$, now apply Markov on $50-X$.
    – Siong Thye Goh
    Nov 14 at 11:24










  • Makes sense. Thanks a bunch!
    – puffles
    Nov 14 at 11:25










  • I needed to clarify one last thing. P(50-X >= 60) <= 1/3. Is this correct considering the typical formula of Markov inequality P(X >= a)? What I am trying to ask is that if a constant is added or subtracted in an interval such as in the above case, would it affect the final probability.
    – puffles
    Nov 15 at 15:26






  • 1




    yes, it is correct. if it makes you more comfortable, let $Y=50-X$ and check that $Y$ is nonnegative. If you perform some operations and prove that the two conditions are equivalent, then the probability stays the same.
    – Siong Thye Goh
    Nov 15 at 15:29













up vote
1
down vote










up vote
1
down vote









Hint:




  • $50-X$ is a nonnegative random variable since $50$ is an upperbound.


  • Express your inequality in the form of $Pr(50-X ge c)$.







share|cite|improve this answer












Hint:




  • $50-X$ is a nonnegative random variable since $50$ is an upperbound.


  • Express your inequality in the form of $Pr(50-X ge c)$.








share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 14 at 8:38









Siong Thye Goh

94k1462114




94k1462114












  • So it should be like this : P(50-X >= -10) ?
    – puffles
    Nov 14 at 11:23










  • $P(X le -10) = P(-X ge 10) = P(50-X ge 60)$, now apply Markov on $50-X$.
    – Siong Thye Goh
    Nov 14 at 11:24










  • Makes sense. Thanks a bunch!
    – puffles
    Nov 14 at 11:25










  • I needed to clarify one last thing. P(50-X >= 60) <= 1/3. Is this correct considering the typical formula of Markov inequality P(X >= a)? What I am trying to ask is that if a constant is added or subtracted in an interval such as in the above case, would it affect the final probability.
    – puffles
    Nov 15 at 15:26






  • 1




    yes, it is correct. if it makes you more comfortable, let $Y=50-X$ and check that $Y$ is nonnegative. If you perform some operations and prove that the two conditions are equivalent, then the probability stays the same.
    – Siong Thye Goh
    Nov 15 at 15:29


















  • So it should be like this : P(50-X >= -10) ?
    – puffles
    Nov 14 at 11:23










  • $P(X le -10) = P(-X ge 10) = P(50-X ge 60)$, now apply Markov on $50-X$.
    – Siong Thye Goh
    Nov 14 at 11:24










  • Makes sense. Thanks a bunch!
    – puffles
    Nov 14 at 11:25










  • I needed to clarify one last thing. P(50-X >= 60) <= 1/3. Is this correct considering the typical formula of Markov inequality P(X >= a)? What I am trying to ask is that if a constant is added or subtracted in an interval such as in the above case, would it affect the final probability.
    – puffles
    Nov 15 at 15:26






  • 1




    yes, it is correct. if it makes you more comfortable, let $Y=50-X$ and check that $Y$ is nonnegative. If you perform some operations and prove that the two conditions are equivalent, then the probability stays the same.
    – Siong Thye Goh
    Nov 15 at 15:29
















So it should be like this : P(50-X >= -10) ?
– puffles
Nov 14 at 11:23




So it should be like this : P(50-X >= -10) ?
– puffles
Nov 14 at 11:23












$P(X le -10) = P(-X ge 10) = P(50-X ge 60)$, now apply Markov on $50-X$.
– Siong Thye Goh
Nov 14 at 11:24




$P(X le -10) = P(-X ge 10) = P(50-X ge 60)$, now apply Markov on $50-X$.
– Siong Thye Goh
Nov 14 at 11:24












Makes sense. Thanks a bunch!
– puffles
Nov 14 at 11:25




Makes sense. Thanks a bunch!
– puffles
Nov 14 at 11:25












I needed to clarify one last thing. P(50-X >= 60) <= 1/3. Is this correct considering the typical formula of Markov inequality P(X >= a)? What I am trying to ask is that if a constant is added or subtracted in an interval such as in the above case, would it affect the final probability.
– puffles
Nov 15 at 15:26




I needed to clarify one last thing. P(50-X >= 60) <= 1/3. Is this correct considering the typical formula of Markov inequality P(X >= a)? What I am trying to ask is that if a constant is added or subtracted in an interval such as in the above case, would it affect the final probability.
– puffles
Nov 15 at 15:26




1




1




yes, it is correct. if it makes you more comfortable, let $Y=50-X$ and check that $Y$ is nonnegative. If you perform some operations and prove that the two conditions are equivalent, then the probability stays the same.
– Siong Thye Goh
Nov 15 at 15:29




yes, it is correct. if it makes you more comfortable, let $Y=50-X$ and check that $Y$ is nonnegative. If you perform some operations and prove that the two conditions are equivalent, then the probability stays the same.
– Siong Thye Goh
Nov 15 at 15:29


















 

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