Uniform transformation of a quantile

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Let $x_alpha = inf x inmathbbR: F_X(x) geq alpha$, $U sim Uniform(0,1)$ and $Z=x_U$. I need to prove that Z has the same distribution as X. Obviously this is true as can easily be shown with a numerical example and the intuition behind it is clear. However I cannot seem to formulate a formal mathematical proof. Could anyone provide me with a hint/paper of how to do this?

probability transformation uniform-distribution quantile

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asked Jul 24 at 9:06

Jeannot van den Berg

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up vote
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favorite

Let $x_alpha = inf x inmathbbR: F_X(x) geq alpha$, $U sim Uniform(0,1)$ and $Z=x_U$. I need to prove that Z has the same distribution as X. Obviously this is true as can easily be shown with a numerical example and the intuition behind it is clear. However I cannot seem to formulate a formal mathematical proof. Could anyone provide me with a hint/paper of how to do this?

probability transformation uniform-distribution quantile

share|cite|improve this question

asked Jul 24 at 9:06

Jeannot van den Berg

111

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let $x_alpha = inf x inmathbbR: F_X(x) geq alpha$, $U sim Uniform(0,1)$ and $Z=x_U$. I need to prove that Z has the same distribution as X. Obviously this is true as can easily be shown with a numerical example and the intuition behind it is clear. However I cannot seem to formulate a formal mathematical proof. Could anyone provide me with a hint/paper of how to do this?

probability transformation uniform-distribution quantile

share|cite|improve this question

asked Jul 24 at 9:06

Jeannot van den Berg

111

Let $x_alpha = inf x inmathbbR: F_X(x) geq alpha$, $U sim Uniform(0,1)$ and $Z=x_U$. I need to prove that Z has the same distribution as X. Obviously this is true as can easily be shown with a numerical example and the intuition behind it is clear. However I cannot seem to formulate a formal mathematical proof. Could anyone provide me with a hint/paper of how to do this?

probability transformation uniform-distribution quantile

share|cite|improve this question

asked Jul 24 at 9:06

Jeannot van den Berg

111

share|cite|improve this question

share|cite|improve this question

asked Jul 24 at 9:06

Jeannot van den Berg

111

asked Jul 24 at 9:06

Jeannot van den Berg

111

asked Jul 24 at 9:06

Jeannot van den Berg

111

111

add a comment | 

add a comment | 

1 Answer
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accepted

Verify that $Z leq t$ if and only if $U leq F_X(t)$. You then get $PZleq t =PUleq F_X(t)=F_X(t)$ Hints for the first part: it is immediate from definiton that $Z leq t$ if $U leq F_X(t)$. Suppse $Zleq t$ but $U>F_X(t)$. There exists $a>0$ such that $U>F_X(t+a)$. Then $F_X(s) < U$ for all $s leq t+a$ so $Z geq t+a$, a contradiction.

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edited Jul 24 at 9:17

answered Jul 24 at 9:11

Kavi Rama Murthy

20.2k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I have used right continuity of $F$ to conclude that $F_X(t+a) <U$ for some $a>0$. My proof is valid for all distributions. I did not use the fact that $F_X$ is strictly incraesing.
  – Kavi Rama Murthy
  Jul 26 at 7:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JeannotvandenBerg If you don't want to study my argument you can visit en.wikipedia.org/wiki/Cumulative_distribution_function where the you can find the same statement. What I have proved in my answer if folklore and Probabilists use it routinely.
  – Kavi Rama Murthy
  Jul 26 at 8:16
  
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

Verify that $Z leq t$ if and only if $U leq F_X(t)$. You then get $PZleq t =PUleq F_X(t)=F_X(t)$ Hints for the first part: it is immediate from definiton that $Z leq t$ if $U leq F_X(t)$. Suppse $Zleq t$ but $U>F_X(t)$. There exists $a>0$ such that $U>F_X(t+a)$. Then $F_X(s) < U$ for all $s leq t+a$ so $Z geq t+a$, a contradiction.

share|cite|improve this answer

edited Jul 24 at 9:17

answered Jul 24 at 9:11

Kavi Rama Murthy

20.2k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I have used right continuity of $F$ to conclude that $F_X(t+a) <U$ for some $a>0$. My proof is valid for all distributions. I did not use the fact that $F_X$ is strictly incraesing.
  – Kavi Rama Murthy
  Jul 26 at 7:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JeannotvandenBerg If you don't want to study my argument you can visit en.wikipedia.org/wiki/Cumulative_distribution_function where the you can find the same statement. What I have proved in my answer if folklore and Probabilists use it routinely.
  – Kavi Rama Murthy
  Jul 26 at 8:16
  
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

Verify that $Z leq t$ if and only if $U leq F_X(t)$. You then get $PZleq t =PUleq F_X(t)=F_X(t)$ Hints for the first part: it is immediate from definiton that $Z leq t$ if $U leq F_X(t)$. Suppse $Zleq t$ but $U>F_X(t)$. There exists $a>0$ such that $U>F_X(t+a)$. Then $F_X(s) < U$ for all $s leq t+a$ so $Z geq t+a$, a contradiction.

share|cite|improve this answer

edited Jul 24 at 9:17

answered Jul 24 at 9:11

Kavi Rama Murthy

20.2k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I have used right continuity of $F$ to conclude that $F_X(t+a) <U$ for some $a>0$. My proof is valid for all distributions. I did not use the fact that $F_X$ is strictly incraesing.
  – Kavi Rama Murthy
  Jul 26 at 7:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JeannotvandenBerg If you don't want to study my argument you can visit en.wikipedia.org/wiki/Cumulative_distribution_function where the you can find the same statement. What I have proved in my answer if folklore and Probabilists use it routinely.
  – Kavi Rama Murthy
  Jul 26 at 8:16
  
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

Verify that $Z leq t$ if and only if $U leq F_X(t)$. You then get $PZleq t =PUleq F_X(t)=F_X(t)$ Hints for the first part: it is immediate from definiton that $Z leq t$ if $U leq F_X(t)$. Suppse $Zleq t$ but $U>F_X(t)$. There exists $a>0$ such that $U>F_X(t+a)$. Then $F_X(s) < U$ for all $s leq t+a$ so $Z geq t+a$, a contradiction.

share|cite|improve this answer

edited Jul 24 at 9:17

answered Jul 24 at 9:11

Kavi Rama Murthy

20.2k2829

Verify that $Z leq t$ if and only if $U leq F_X(t)$. You then get $PZleq t =PUleq F_X(t)=F_X(t)$ Hints for the first part: it is immediate from definiton that $Z leq t$ if $U leq F_X(t)$. Suppse $Zleq t$ but $U>F_X(t)$. There exists $a>0$ such that $U>F_X(t+a)$. Then $F_X(s) < U$ for all $s leq t+a$ so $Z geq t+a$, a contradiction.

share|cite|improve this answer

edited Jul 24 at 9:17

answered Jul 24 at 9:11

Kavi Rama Murthy

20.2k2829

share|cite|improve this answer

share|cite|improve this answer

edited Jul 24 at 9:17

edited Jul 24 at 9:17

edited Jul 24 at 9:17

answered Jul 24 at 9:11

Kavi Rama Murthy

20.2k2829

answered Jul 24 at 9:11

Kavi Rama Murthy

20.2k2829

answered Jul 24 at 9:11

Kavi Rama Murthy

20.2k2829

20.2k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I have used right continuity of $F$ to conclude that $F_X(t+a) <U$ for some $a>0$. My proof is valid for all distributions. I did not use the fact that $F_X$ is strictly incraesing.
  – Kavi Rama Murthy
  Jul 26 at 7:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JeannotvandenBerg If you don't want to study my argument you can visit en.wikipedia.org/wiki/Cumulative_distribution_function where the you can find the same statement. What I have proved in my answer if folklore and Probabilists use it routinely.
  – Kavi Rama Murthy
  Jul 26 at 8:16
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I have used right continuity of $F$ to conclude that $F_X(t+a) <U$ for some $a>0$. My proof is valid for all distributions. I did not use the fact that $F_X$ is strictly incraesing.
  – Kavi Rama Murthy
  Jul 26 at 7:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JeannotvandenBerg If you don't want to study my argument you can visit en.wikipedia.org/wiki/Cumulative_distribution_function where the you can find the same statement. What I have proved in my answer if folklore and Probabilists use it routinely.
  – Kavi Rama Murthy
  Jul 26 at 8:16
  
  

  
  
  
  

I have used right continuity of $F$ to conclude that $F_X(t+a) <U$ for some $a>0$. My proof is valid for all distributions. I did not use the fact that $F_X$ is strictly incraesing.
– Kavi Rama Murthy
Jul 26 at 7:51

I have used right continuity of $F$ to conclude that $F_X(t+a) <U$ for some $a>0$. My proof is valid for all distributions. I did not use the fact that $F_X$ is strictly incraesing.
– Kavi Rama Murthy
Jul 26 at 7:51

@JeannotvandenBerg If you don't want to study my argument you can visit en.wikipedia.org/wiki/Cumulative_distribution_function where the you can find the same statement. What I have proved in my answer if folklore and Probabilists use it routinely.
– Kavi Rama Murthy
Jul 26 at 8:16

@JeannotvandenBerg If you don't want to study my argument you can visit en.wikipedia.org/wiki/Cumulative_distribution_function where the you can find the same statement. What I have proved in my answer if folklore and Probabilists use it routinely.
– Kavi Rama Murthy
Jul 26 at 8:16

add a comment | 

 

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