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Apex Framework / library for consuming REST services

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Proof that the probability $P(X

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I'm struggling to prove analytically that for two iid (independent and identically distributed) random variables $ P(X<Y) = 0.5$ for any distribution. I'm able to do it by direct integration on specific distributions, but not in the general usecase.

probability-theory probability-distributions

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edited Mar 28 at 20:18

Max

9851319

asked Mar 28 at 19:43

abcdaireabcdaire

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abcdaire is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

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  It's not true for arbitrary distributions. If there are point masses, such as for any discrete distribution, the nonzero probability that $X=Y$ throws this off. Perhaps there should be additional restrictions?
  $endgroup$
  – jmerry
  Mar 28 at 19:56
  

  
  
  
  


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  Sorry its in the case of continuous distribution, basically it just come from another thread (that I lost) in the specific case of a continous uniform, and people started arguing that its valid for any distribution (which makes sense by intuiton, symmetry)
  $endgroup$
  – abcdaire
  Mar 28 at 20:05
  

  
  
  
  


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  Hint: First show that $P(X = Y) = 0$. To accomplish then, first show that $X,Y$ being iid with a pdf implies that $(X,Y)$ has a pdf. (You can explicitly write down what the pdf is in terms of the pdf of the distribution of $X$.)
  $endgroup$
  – Lorenzo
  Mar 28 at 20:25
  
  

  
  
  
  


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  $(x,y) to (y,x)$ is a probability-preserving transformation of the sample space. Thus $P(X>Y) = P(Y>X)$. And if $P(X=Y)=0$ then each must be $1/2$.
  $endgroup$
  – Robert Israel
  Mar 28 at 20:30
  

  
  
  
  

add a comment | 

0

$begingroup$

I'm struggling to prove analytically that for two iid (independent and identically distributed) random variables $ P(X<Y) = 0.5$ for any distribution. I'm able to do it by direct integration on specific distributions, but not in the general usecase.

probability-theory probability-distributions

share|cite|improve this question

edited Mar 28 at 20:18

Max

9851319

asked Mar 28 at 19:43

abcdaireabcdaire

101

New contributor

abcdaire is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's not true for arbitrary distributions. If there are point masses, such as for any discrete distribution, the nonzero probability that $X=Y$ throws this off. Perhaps there should be additional restrictions?
  $endgroup$
  – jmerry
  Mar 28 at 19:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry its in the case of continuous distribution, basically it just come from another thread (that I lost) in the specific case of a continous uniform, and people started arguing that its valid for any distribution (which makes sense by intuiton, symmetry)
  $endgroup$
  – abcdaire
  Mar 28 at 20:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: First show that $P(X = Y) = 0$. To accomplish then, first show that $X,Y$ being iid with a pdf implies that $(X,Y)$ has a pdf. (You can explicitly write down what the pdf is in terms of the pdf of the distribution of $X$.)
  $endgroup$
  – Lorenzo
  Mar 28 at 20:25
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $(x,y) to (y,x)$ is a probability-preserving transformation of the sample space. Thus $P(X>Y) = P(Y>X)$. And if $P(X=Y)=0$ then each must be $1/2$.
  $endgroup$
  – Robert Israel
  Mar 28 at 20:30
  

  
  
  
  

add a comment | 

$begingroup$

I'm struggling to prove analytically that for two iid (independent and identically distributed) random variables $ P(X<Y) = 0.5$ for any distribution. I'm able to do it by direct integration on specific distributions, but not in the general usecase.

probability-theory probability-distributions

share|cite|improve this question

edited Mar 28 at 20:18

Max

9851319

asked Mar 28 at 19:43

abcdaireabcdaire

101

New contributor

abcdaire is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 28 at 20:18

Max

9851319

asked Mar 28 at 19:43

abcdaireabcdaire

101

New contributor

abcdaire is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 28 at 20:18

Max

9851319

asked Mar 28 at 19:43

abcdaireabcdaire

101

New contributor

abcdaire is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Mar 28 at 20:18

Max

9851319

edited Mar 28 at 20:18

Max

9851319

asked Mar 28 at 19:43

abcdaireabcdaire

101

New contributor

abcdaire is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 28 at 19:43

abcdaireabcdaire

101