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Lower bound on the probability that a random variable is greater than half of its mean

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1

$begingroup$

Consider a random variable $X$ that takes values between $-1$ and $1$. What is a non-trivial lower bound on the probability that the outcome of $X$ is greater than half of its mean?

---

EDIT:

I am looking for a bound that is a function of the mean (and so would cover the case where the mean is positive).

I.e.

$textPr left[ X > mathbbE[X] over 2 right] geq f(mathbbE[X]).$

probability probability-theory random-variables information-theory

share|cite|improve this question

edited Mar 30 at 9:23

user120404

asked Mar 29 at 22:34

user120404user120404

83118

$endgroup$

add a comment | 

1

$begingroup$

Consider a random variable $X$ that takes values between $-1$ and $1$. What is a non-trivial lower bound on the probability that the outcome of $X$ is greater than half of its mean?

---

EDIT:

I am looking for a bound that is a function of the mean (and so would cover the case where the mean is positive).

I.e.

$textPr left[ X > mathbbE[X] over 2 right] geq f(mathbbE[X]).$

probability probability-theory random-variables information-theory

share|cite|improve this question

edited Mar 30 at 9:23

user120404

asked Mar 29 at 22:34

user120404user120404

83118

$endgroup$

add a comment | 

$begingroup$

Consider a random variable $X$ that takes values between $-1$ and $1$. What is a non-trivial lower bound on the probability that the outcome of $X$ is greater than half of its mean?

---

EDIT:

I am looking for a bound that is a function of the mean (and so would cover the case where the mean is positive).

I.e.

$textPr left[ X > mathbbE[X] over 2 right] geq f(mathbbE[X]).$

probability probability-theory random-variables information-theory

share|cite|improve this question

edited Mar 30 at 9:23

user120404

asked Mar 29 at 22:34

user120404user120404

83118

$endgroup$

share|cite|improve this question

edited Mar 30 at 9:23

user120404

asked Mar 29 at 22:34

user120404user120404

83118

share|cite|improve this question

edited Mar 30 at 9:23

user120404

asked Mar 29 at 22:34

user120404user120404

83118

asked Mar 29 at 22:34

user120404user120404

83118

asked Mar 29 at 22:34

user120404user120404

83118

2 Answers
2

active

oldest

votes

2

$begingroup$

Let $mu=E[X]$ and $Z = mathbb1_X > mu/2 $ (indicator variable), $a=P(Z=1)$

We are seeking for the maximal $g(mu)$ such that $a ge g(mu)$

If $mu < 0 $ then we cannot do better than the trivial $g(mu)=0$.

(That can be seen by considering a Dirac delta on $x=mu$)

For $mu >0$:

$$
mu = E[E[X | Z]]= a E[X| Z=1] + (1-a) E[X|Z=0] tag1
$$

but $E[X| Z=1] le 1$ (because $Xle 1$) and $E[X|Z=0] le mu/ 2$. Hence

$$ mu le a + (1-a) mu /2 implies a ge fracmu2-mu tag2$$

This bound is attained
by two Dirac deltas on $x=mu/2$ and $x=1$ with weights $a$ and $1-a$ resp.
Hence he desired bound is
$$g(mu)=begincases
0 & mule 0\
fracmu2-mu & mu>0
endcasestag3$$

share|cite|improve this answer

edited Mar 30 at 15:49

answered Mar 30 at 12:45

leonbloyleonbloy

42.2k647108

$endgroup$

add a comment | 

2

$begingroup$

For this distribution the mean is near $-1$, so half the mean is $-1/2$, and clearly the probability of finding $x>-1/2$ can be made arbitrarily small.

share|cite|improve this answer

answered Mar 29 at 22:45

David G. StorkDavid G. Stork

12k41735

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes thank you, but is there an expression for the lower bound as a function of the mean (which would cover the cases in which the mean is positive)?
  $endgroup$
  – user120404
  Mar 29 at 22:58
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That wasn't your question. Please edit accordingly.
  $endgroup$
  – David G. Stork
  Mar 29 at 23:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please check the edited question.
  $endgroup$
  – user120404
  Mar 30 at 10:03
  

  
  
  
  

add a comment | 

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2

$begingroup$

Let $mu=E[X]$ and $Z = mathbb1_X > mu/2 $ (indicator variable), $a=P(Z=1)$

We are seeking for the maximal $g(mu)$ such that $a ge g(mu)$

If $mu < 0 $ then we cannot do better than the trivial $g(mu)=0$.

(That can be seen by considering a Dirac delta on $x=mu$)

For $mu >0$:

$$
mu = E[E[X | Z]]= a E[X| Z=1] + (1-a) E[X|Z=0] tag1
$$

but $E[X| Z=1] le 1$ (because $Xle 1$) and $E[X|Z=0] le mu/ 2$. Hence

$$ mu le a + (1-a) mu /2 implies a ge fracmu2-mu tag2$$

This bound is attained
by two Dirac deltas on $x=mu/2$ and $x=1$ with weights $a$ and $1-a$ resp.
Hence he desired bound is
$$g(mu)=begincases
0 & mule 0\
fracmu2-mu & mu>0
endcasestag3$$

share|cite|improve this answer

edited Mar 30 at 15:49

answered Mar 30 at 12:45

leonbloyleonbloy

42.2k647108

$endgroup$

add a comment | 

2

$begingroup$

Let $mu=E[X]$ and $Z = mathbb1_X > mu/2 $ (indicator variable), $a=P(Z=1)$

We are seeking for the maximal $g(mu)$ such that $a ge g(mu)$

If $mu < 0 $ then we cannot do better than the trivial $g(mu)=0$.

(That can be seen by considering a Dirac delta on $x=mu$)

For $mu >0$:

$$
mu = E[E[X | Z]]= a E[X| Z=1] + (1-a) E[X|Z=0] tag1
$$

but $E[X| Z=1] le 1$ (because $Xle 1$) and $E[X|Z=0] le mu/ 2$. Hence

$$ mu le a + (1-a) mu /2 implies a ge fracmu2-mu tag2$$

This bound is attained
by two Dirac deltas on $x=mu/2$ and $x=1$ with weights $a$ and $1-a$ resp.
Hence he desired bound is
$$g(mu)=begincases
0 & mule 0\
fracmu2-mu & mu>0
endcasestag3$$

share|cite|improve this answer

edited Mar 30 at 15:49

answered Mar 30 at 12:45

leonbloyleonbloy

42.2k647108

$endgroup$

add a comment | 

$begingroup$

Let $mu=E[X]$ and $Z = mathbb1_X > mu/2 $ (indicator variable), $a=P(Z=1)$

We are seeking for the maximal $g(mu)$ such that $a ge g(mu)$

If $mu < 0 $ then we cannot do better than the trivial $g(mu)=0$.

(That can be seen by considering a Dirac delta on $x=mu$)

For $mu >0$:

$$
mu = E[E[X | Z]]= a E[X| Z=1] + (1-a) E[X|Z=0] tag1
$$

but $E[X| Z=1] le 1$ (because $Xle 1$) and $E[X|Z=0] le mu/ 2$. Hence

$$ mu le a + (1-a) mu /2 implies a ge fracmu2-mu tag2$$

This bound is attained
by two Dirac deltas on $x=mu/2$ and $x=1$ with weights $a$ and $1-a$ resp.
Hence he desired bound is
$$g(mu)=begincases
0 & mule 0\
fracmu2-mu & mu>0
endcasestag3$$

share|cite|improve this answer

edited Mar 30 at 15:49

answered Mar 30 at 12:45

leonbloyleonbloy

42.2k647108

$endgroup$

share|cite|improve this answer

edited Mar 30 at 15:49

answered Mar 30 at 12:45

leonbloyleonbloy

42.2k647108

answered Mar 30 at 12:45

leonbloyleonbloy

42.2k647108

answered Mar 30 at 12:45

leonbloyleonbloy

42.2k647108

2

$begingroup$

For this distribution the mean is near $-1$, so half the mean is $-1/2$, and clearly the probability of finding $x>-1/2$ can be made arbitrarily small.

share|cite|improve this answer

answered Mar 29 at 22:45

David G. StorkDavid G. Stork

12k41735

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes thank you, but is there an expression for the lower bound as a function of the mean (which would cover the cases in which the mean is positive)?
  $endgroup$
  – user120404
  Mar 29 at 22:58
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That wasn't your question. Please edit accordingly.
  $endgroup$
  – David G. Stork
  Mar 29 at 23:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please check the edited question.
  $endgroup$
  – user120404
  Mar 30 at 10:03
  

  
  
  
  

add a comment | 

2

$begingroup$

For this distribution the mean is near $-1$, so half the mean is $-1/2$, and clearly the probability of finding $x>-1/2$ can be made arbitrarily small.

share|cite|improve this answer

answered Mar 29 at 22:45

David G. StorkDavid G. Stork

12k41735

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes thank you, but is there an expression for the lower bound as a function of the mean (which would cover the cases in which the mean is positive)?
  $endgroup$
  – user120404
  Mar 29 at 22:58
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That wasn't your question. Please edit accordingly.
  $endgroup$
  – David G. Stork
  Mar 29 at 23:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please check the edited question.
  $endgroup$
  – user120404
  Mar 30 at 10:03
  

  
  
  
  

add a comment | 

$begingroup$

For this distribution the mean is near $-1$, so half the mean is $-1/2$, and clearly the probability of finding $x>-1/2$ can be made arbitrarily small.

share|cite|improve this answer

answered Mar 29 at 22:45

David G. StorkDavid G. Stork

12k41735

$endgroup$

share|cite|improve this answer

answered Mar 29 at 22:45

David G. StorkDavid G. Stork

12k41735

answered Mar 29 at 22:45

David G. StorkDavid G. Stork

12k41735

answered Mar 29 at 22:45

David G. StorkDavid G. Stork

12k41735