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Expectation of almost impossible events

-1

$begingroup$

In several questions posted here, a satisfying answer, such as 1, may be found on difference between events with zero probability and impossible events. To wrap my mind around this, I usually invoke the interval $I = [0,1]$ as model of explanation - if one picks a countably infinite subset of $I$, with uniformly distributed probability, how many rational numbers are expected in it? Its pointed out to me that zero would be right answer, even some picks may contain rational numbers.

To recollect myself, I'd like to know is that true for any zero-measure subset such as Cantors dust?

probability measure-theory

share|cite|improve this question

edited Mar 25 at 10:25

Stipe Galić

asked Mar 25 at 9:49

Stipe GalićStipe Galić

1143

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Not sure what you are asking. Expectation is countably additive (at least for non-negative random variables, like indicator variables), so since each of your sequence has $0$ probability of being rational, the expected number of rational numbers in the sequence is $0$. Beyond that, I can't really follow what you write.
  $endgroup$
  – lulu
  Mar 25 at 10:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, I wrote that really bad - I'll edit. Basically, I didn't mean that the entire sequence is rational, but only finite number of its members to be rational.
  $endgroup$
  – Stipe Galić
  Mar 25 at 10:12
  

  
  
  
  


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  $begingroup$
  Also the event that $a_ninmathbb Q$ is expected to appear exactly zero times. In that sense there is no distinction with an impossible event. So your model of explanation is not sound.
  $endgroup$
  – drhab
  Mar 25 at 10:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @StipeGalić Once again: the expected number of rational elements in your sequence is $0$. Not "finite". I think you are confused about what "expectation" means.
  $endgroup$
  – lulu
  Mar 25 at 10:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The difference between a possible event and an impossible event is that the former can occur and the latter can not. It is perfectly possible that you choose the number $frac 12$ from $[0,1]$, though it has probability $0$. In this way, continuous probability deviates from our intuition (and from basic properties of discrete probability). But in such cases we either need to surrender our intuition or search for a new theory in which our intuition holds up.
  $endgroup$
  – lulu
  Mar 25 at 10:17
  
  

  
  
  
  

 | 
show 3 more comments

-1

$begingroup$

In several questions posted here, a satisfying answer, such as 1, may be found on difference between events with zero probability and impossible events. To wrap my mind around this, I usually invoke the interval $I = [0,1]$ as model of explanation - if one picks a countably infinite subset of $I$, with uniformly distributed probability, how many rational numbers are expected in it? Its pointed out to me that zero would be right answer, even some picks may contain rational numbers.

To recollect myself, I'd like to know is that true for any zero-measure subset such as Cantors dust?

probability measure-theory

share|cite|improve this question

edited Mar 25 at 10:25

Stipe Galić

asked Mar 25 at 9:49

Stipe GalićStipe Galić

1143

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Not sure what you are asking. Expectation is countably additive (at least for non-negative random variables, like indicator variables), so since each of your sequence has $0$ probability of being rational, the expected number of rational numbers in the sequence is $0$. Beyond that, I can't really follow what you write.
  $endgroup$
  – lulu
  Mar 25 at 10:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, I wrote that really bad - I'll edit. Basically, I didn't mean that the entire sequence is rational, but only finite number of its members to be rational.
  $endgroup$
  – Stipe Galić
  Mar 25 at 10:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also the event that $a_ninmathbb Q$ is expected to appear exactly zero times. In that sense there is no distinction with an impossible event. So your model of explanation is not sound.
  $endgroup$
  – drhab
  Mar 25 at 10:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @StipeGalić Once again: the expected number of rational elements in your sequence is $0$. Not "finite". I think you are confused about what "expectation" means.
  $endgroup$
  – lulu
  Mar 25 at 10:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The difference between a possible event and an impossible event is that the former can occur and the latter can not. It is perfectly possible that you choose the number $frac 12$ from $[0,1]$, though it has probability $0$. In this way, continuous probability deviates from our intuition (and from basic properties of discrete probability). But in such cases we either need to surrender our intuition or search for a new theory in which our intuition holds up.
  $endgroup$
  – lulu
  Mar 25 at 10:17
  
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

In several questions posted here, a satisfying answer, such as 1, may be found on difference between events with zero probability and impossible events. To wrap my mind around this, I usually invoke the interval $I = [0,1]$ as model of explanation - if one picks a countably infinite subset of $I$, with uniformly distributed probability, how many rational numbers are expected in it? Its pointed out to me that zero would be right answer, even some picks may contain rational numbers.

To recollect myself, I'd like to know is that true for any zero-measure subset such as Cantors dust?

probability measure-theory

share|cite|improve this question

edited Mar 25 at 10:25

Stipe Galić

asked Mar 25 at 9:49

Stipe GalićStipe Galić

1143

$endgroup$

share|cite|improve this question

edited Mar 25 at 10:25

Stipe Galić

asked Mar 25 at 9:49

Stipe GalićStipe Galić

1143

share|cite|improve this question

edited Mar 25 at 10:25

Stipe Galić

asked Mar 25 at 9:49

Stipe GalićStipe Galić

1143

asked Mar 25 at 9:49

Stipe GalićStipe Galić

1143

asked Mar 25 at 9:49

Stipe GalićStipe Galić

1143

1 Answer
1

active

oldest

votes

0

$begingroup$

I may have grossly misunderstood your question but I think one instance where I felt equally confused initially was with continuous probability density function. Say, you want to pick a random point on the circumference of a circle. In polar co-ordinates, the point can be $(r,theta)$ where $theta$ can be drawn from a uniform distribution in the interval of $[0,2pi]$. In this entire interval, the p.d.f has a value $1/2pi$. So you might think that the probability that $theta$ is $pi/6$ is $1/2pi$. Nope. It is actually zero.

In fact, the probability that any particular value of $theta$ being drawn is zero and yet when a point is chosen it has a definite value. The expectation of $theta$ is a non zero value. So I think that is a good example of zero probability that isn't impossible.

share|cite|improve this answer

answered Mar 25 at 12:02

Balakrishnan RajanBalakrishnan Rajan

15210

$endgroup$

add a comment | 

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0

$begingroup$

I may have grossly misunderstood your question but I think one instance where I felt equally confused initially was with continuous probability density function. Say, you want to pick a random point on the circumference of a circle. In polar co-ordinates, the point can be $(r,theta)$ where $theta$ can be drawn from a uniform distribution in the interval of $[0,2pi]$. In this entire interval, the p.d.f has a value $1/2pi$. So you might think that the probability that $theta$ is $pi/6$ is $1/2pi$. Nope. It is actually zero.

In fact, the probability that any particular value of $theta$ being drawn is zero and yet when a point is chosen it has a definite value. The expectation of $theta$ is a non zero value. So I think that is a good example of zero probability that isn't impossible.

share|cite|improve this answer

answered Mar 25 at 12:02

Balakrishnan RajanBalakrishnan Rajan

15210

$endgroup$

add a comment | 

0

$begingroup$

I may have grossly misunderstood your question but I think one instance where I felt equally confused initially was with continuous probability density function. Say, you want to pick a random point on the circumference of a circle. In polar co-ordinates, the point can be $(r,theta)$ where $theta$ can be drawn from a uniform distribution in the interval of $[0,2pi]$. In this entire interval, the p.d.f has a value $1/2pi$. So you might think that the probability that $theta$ is $pi/6$ is $1/2pi$. Nope. It is actually zero.

In fact, the probability that any particular value of $theta$ being drawn is zero and yet when a point is chosen it has a definite value. The expectation of $theta$ is a non zero value. So I think that is a good example of zero probability that isn't impossible.

share|cite|improve this answer

answered Mar 25 at 12:02

Balakrishnan RajanBalakrishnan Rajan

15210

$endgroup$

add a comment | 

$begingroup$

I may have grossly misunderstood your question but I think one instance where I felt equally confused initially was with continuous probability density function. Say, you want to pick a random point on the circumference of a circle. In polar co-ordinates, the point can be $(r,theta)$ where $theta$ can be drawn from a uniform distribution in the interval of $[0,2pi]$. In this entire interval, the p.d.f has a value $1/2pi$. So you might think that the probability that $theta$ is $pi/6$ is $1/2pi$. Nope. It is actually zero.

In fact, the probability that any particular value of $theta$ being drawn is zero and yet when a point is chosen it has a definite value. The expectation of $theta$ is a non zero value. So I think that is a good example of zero probability that isn't impossible.

share|cite|improve this answer

answered Mar 25 at 12:02

Balakrishnan RajanBalakrishnan Rajan

15210

$endgroup$

share|cite|improve this answer

answered Mar 25 at 12:02

Balakrishnan RajanBalakrishnan Rajan

15210

answered Mar 25 at 12:02

Balakrishnan RajanBalakrishnan Rajan

15210

answered Mar 25 at 12:02

Balakrishnan RajanBalakrishnan Rajan

15210