Proof of a.s. uniqueness of probability convergence limit [closed]

1

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Prove that the limit of a random variable sequence (converges in probability) is almost surely uniuque.

Let $X, Y$ random variables and let a random variable sequence $(X_n)_n$ such that $X_n$ converges in probability to $X$ and $X_n$ converges in probability to $Y$ (i.e. $X_n overset{P}{rightarrow} X, X_noverset{P}{rightarrow} Y$). Prove that $mathbb{P}(X=Y)=1$.

probability limits proof-verification

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edited Jan 5 at 11:02

J. Doe

asked Jan 4 at 19:07

J. DoeJ. Doe

1737

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closed as off-topic by Davide Giraudo, amWhy, Lee David Chung Lin, jgon, KReiser Jan 5 at 0:23

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If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

1

$begingroup$

Prove that the limit of a random variable sequence (converges in probability) is almost surely uniuque.

Let $X, Y$ random variables and let a random variable sequence $(X_n)_n$ such that $X_n$ converges in probability to $X$ and $X_n$ converges in probability to $Y$ (i.e. $X_n overset{P}{rightarrow} X, X_noverset{P}{rightarrow} Y$). Prove that $mathbb{P}(X=Y)=1$.

probability limits proof-verification

share|cite|improve this question

edited Jan 5 at 11:02

J. Doe

asked Jan 4 at 19:07

J. DoeJ. Doe

1737

$endgroup$

closed as off-topic by Davide Giraudo, amWhy, Lee David Chung Lin, jgon, KReiser Jan 5 at 0:23

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Davide Giraudo, amWhy, Lee David Chung Lin, jgon, KReiser

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

1

1

1

0

$begingroup$

Prove that the limit of a random variable sequence (converges in probability) is almost surely uniuque.

Let $X, Y$ random variables and let a random variable sequence $(X_n)_n$ such that $X_n$ converges in probability to $X$ and $X_n$ converges in probability to $Y$ (i.e. $X_n overset{P}{rightarrow} X, X_noverset{P}{rightarrow} Y$). Prove that $mathbb{P}(X=Y)=1$.

probability limits proof-verification

share|cite|improve this question

edited Jan 5 at 11:02

J. Doe

asked Jan 4 at 19:07

J. DoeJ. Doe

1737

$endgroup$

Prove that the limit of a random variable sequence (converges in probability) is almost surely uniuque.

Let $X, Y$ random variables and let a random variable sequence $(X_n)_n$ such that $X_n$ converges in probability to $X$ and $X_n$ converges in probability to $Y$ (i.e. $X_n overset{P}{rightarrow} X, X_noverset{P}{rightarrow} Y$). Prove that $mathbb{P}(X=Y)=1$.

probability limits proof-verification

probability limits proof-verification

share|cite|improve this question

edited Jan 5 at 11:02

J. Doe

asked Jan 4 at 19:07

J. DoeJ. Doe

1737

share|cite|improve this question

edited Jan 5 at 11:02

J. Doe

asked Jan 4 at 19:07

J. DoeJ. Doe

1737

share|cite|improve this question

share|cite|improve this question

edited Jan 5 at 11:02

J. Doe

edited Jan 5 at 11:02

J. Doe

edited Jan 5 at 11:02

J. Doe

asked Jan 4 at 19:07

J. DoeJ. Doe

1737

asked Jan 4 at 19:07

J. DoeJ. Doe

1737

asked Jan 4 at 19:07

J. DoeJ. Doe

1737

1737

closed as off-topic by Davide Giraudo, amWhy, Lee David Chung Lin, jgon, KReiser Jan 5 at 0:23

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Davide Giraudo, amWhy, Lee David Chung Lin, jgon, KReiser

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Davide Giraudo, amWhy, Lee David Chung Lin, jgon, KReiser Jan 5 at 0:23

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Davide Giraudo, amWhy, Lee David Chung Lin, jgon, KReiser

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

1

$begingroup$

Let $varepsilon>0$. Let $omegainOmega$ (if there exists such $omega$) such that $|X^{-1}(omega)-Y^{-1}(omega)|>varepsilon$. Thus, for all $ninmathbb{N}$, definitely $|X^{-1}(omega)-X_n^{-1}(omega)|>varepsilon$ or $|X_n^{-1}(omega)-Y^{-1}(omega)|>varepsilon$. Thus,
$$
\ {|X-Y|>varepsilon}subseteq({|X_n-X|> {varepsilonover 3}}cup{|X_n-Y|>{varepsilonover 3}})
$$
Thus,
$$
\ mathbb{P}(|X-Y|>{1over n})leqmathbb{P}(|X_n-X|>{varepsilonover 3}cup|X_n-Y|leq{varepsilonover 3})
\ leqmathbb{P}(|X_n-X|>{varepsilonover 3})+mathbb{P}(|X_n-Y|>{varepsilonover 3})underset{ntoinfty}{rightarrow} 0
$$
Thus,
$$
\ mathbb{P}(|X-Y|>0)=0Rightarrow mathbb{P}(X=Y)=1
$$

share|cite|improve this answer

answered Jan 4 at 19:07

J. DoeJ. Doe

1737

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  yes, as with many standard uniqueness proofs, it is the triangle inequality... this proof is correct
  $endgroup$
  – Lost1
  Jan 4 at 23:58
  
  
  

  
  
  
  

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

Let $varepsilon>0$. Let $omegainOmega$ (if there exists such $omega$) such that $|X^{-1}(omega)-Y^{-1}(omega)|>varepsilon$. Thus, for all $ninmathbb{N}$, definitely $|X^{-1}(omega)-X_n^{-1}(omega)|>varepsilon$ or $|X_n^{-1}(omega)-Y^{-1}(omega)|>varepsilon$. Thus,
$$
\ {|X-Y|>varepsilon}subseteq({|X_n-X|> {varepsilonover 3}}cup{|X_n-Y|>{varepsilonover 3}})
$$
Thus,
$$
\ mathbb{P}(|X-Y|>{1over n})leqmathbb{P}(|X_n-X|>{varepsilonover 3}cup|X_n-Y|leq{varepsilonover 3})
\ leqmathbb{P}(|X_n-X|>{varepsilonover 3})+mathbb{P}(|X_n-Y|>{varepsilonover 3})underset{ntoinfty}{rightarrow} 0
$$
Thus,
$$
\ mathbb{P}(|X-Y|>0)=0Rightarrow mathbb{P}(X=Y)=1
$$

share|cite|improve this answer

answered Jan 4 at 19:07

J. DoeJ. Doe

1737

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  yes, as with many standard uniqueness proofs, it is the triangle inequality... this proof is correct
  $endgroup$
  – Lost1
  Jan 4 at 23:58
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $varepsilon>0$. Let $omegainOmega$ (if there exists such $omega$) such that $|X^{-1}(omega)-Y^{-1}(omega)|>varepsilon$. Thus, for all $ninmathbb{N}$, definitely $|X^{-1}(omega)-X_n^{-1}(omega)|>varepsilon$ or $|X_n^{-1}(omega)-Y^{-1}(omega)|>varepsilon$. Thus,
$$
\ {|X-Y|>varepsilon}subseteq({|X_n-X|> {varepsilonover 3}}cup{|X_n-Y|>{varepsilonover 3}})
$$
Thus,
$$
\ mathbb{P}(|X-Y|>{1over n})leqmathbb{P}(|X_n-X|>{varepsilonover 3}cup|X_n-Y|leq{varepsilonover 3})
\ leqmathbb{P}(|X_n-X|>{varepsilonover 3})+mathbb{P}(|X_n-Y|>{varepsilonover 3})underset{ntoinfty}{rightarrow} 0
$$
Thus,
$$
\ mathbb{P}(|X-Y|>0)=0Rightarrow mathbb{P}(X=Y)=1
$$

share|cite|improve this answer

answered Jan 4 at 19:07

J. DoeJ. Doe

1737

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  yes, as with many standard uniqueness proofs, it is the triangle inequality... this proof is correct
  $endgroup$
  – Lost1
  Jan 4 at 23:58
  
  
  

  
  
  
  

add a comment | 

1

1

1

$begingroup$

Let $varepsilon>0$. Let $omegainOmega$ (if there exists such $omega$) such that $|X^{-1}(omega)-Y^{-1}(omega)|>varepsilon$. Thus, for all $ninmathbb{N}$, definitely $|X^{-1}(omega)-X_n^{-1}(omega)|>varepsilon$ or $|X_n^{-1}(omega)-Y^{-1}(omega)|>varepsilon$. Thus,
$$
\ {|X-Y|>varepsilon}subseteq({|X_n-X|> {varepsilonover 3}}cup{|X_n-Y|>{varepsilonover 3}})
$$
Thus,
$$
\ mathbb{P}(|X-Y|>{1over n})leqmathbb{P}(|X_n-X|>{varepsilonover 3}cup|X_n-Y|leq{varepsilonover 3})
\ leqmathbb{P}(|X_n-X|>{varepsilonover 3})+mathbb{P}(|X_n-Y|>{varepsilonover 3})underset{ntoinfty}{rightarrow} 0
$$
Thus,
$$
\ mathbb{P}(|X-Y|>0)=0Rightarrow mathbb{P}(X=Y)=1
$$

share|cite|improve this answer

answered Jan 4 at 19:07

J. DoeJ. Doe

1737

$endgroup$

Let $varepsilon>0$. Let $omegainOmega$ (if there exists such $omega$) such that $|X^{-1}(omega)-Y^{-1}(omega)|>varepsilon$. Thus, for all $ninmathbb{N}$, definitely $|X^{-1}(omega)-X_n^{-1}(omega)|>varepsilon$ or $|X_n^{-1}(omega)-Y^{-1}(omega)|>varepsilon$. Thus,
$$
\ {|X-Y|>varepsilon}subseteq({|X_n-X|> {varepsilonover 3}}cup{|X_n-Y|>{varepsilonover 3}})
$$
Thus,
$$
\ mathbb{P}(|X-Y|>{1over n})leqmathbb{P}(|X_n-X|>{varepsilonover 3}cup|X_n-Y|leq{varepsilonover 3})
\ leqmathbb{P}(|X_n-X|>{varepsilonover 3})+mathbb{P}(|X_n-Y|>{varepsilonover 3})underset{ntoinfty}{rightarrow} 0
$$
Thus,
$$
\ mathbb{P}(|X-Y|>0)=0Rightarrow mathbb{P}(X=Y)=1
$$

share|cite|improve this answer

answered Jan 4 at 19:07

J. DoeJ. Doe

1737

share|cite|improve this answer

share|cite|improve this answer

answered Jan 4 at 19:07

J. DoeJ. Doe

1737

answered Jan 4 at 19:07

J. DoeJ. Doe

1737

answered Jan 4 at 19:07

J. DoeJ. Doe

1737

1737

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  yes, as with many standard uniqueness proofs, it is the triangle inequality... this proof is correct
  $endgroup$
  – Lost1
  Jan 4 at 23:58
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  yes, as with many standard uniqueness proofs, it is the triangle inequality... this proof is correct
  $endgroup$
  – Lost1
  Jan 4 at 23:58
  
  
  

  
  
  
  

1

1

$begingroup$
yes, as with many standard uniqueness proofs, it is the triangle inequality... this proof is correct
$endgroup$
– Lost1
Jan 4 at 23:58

$begingroup$
yes, as with many standard uniqueness proofs, it is the triangle inequality... this proof is correct
$endgroup$
– Lost1
Jan 4 at 23:58

add a comment | 

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