Proof of a.s. uniqueness of probability convergence limit [closed]
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
asked Jan 4 at 19:07
J. DoeJ. Doe
1286
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.
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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
asked Jan 4 at 19:07
J. DoeJ. Doe
1286
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 |
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
asked Jan 4 at 19:07
J. DoeJ. Doe
1286
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
asked Jan 4 at 19:07
J. DoeJ. Doe
1286
asked Jan 4 at 19:07
J. DoeJ. Doe
1286
edited Jan 5 at 11:02
J. Doe
asked Jan 4 at 19:07
J. DoeJ. Doe
1286
asked Jan 4 at 19:07
J. DoeJ. Doe
1286
asked Jan 4 at 19:07
J. DoeJ. Doe
1286
1286
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.
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1 Answer
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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,
$$
\ mathbb{P}(|X-Y|>0)=0Rightarrow mathbb{P}(X=Y)=1
$$
answered Jan 4 at 19:07
J. DoeJ. Doe
1286
1
yes, as with many standard uniqueness proofs, it is the triangle inequality... this proof is correct
– 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
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,
$$
\ mathbb{P}(|X-Y|>0)=0Rightarrow mathbb{P}(X=Y)=1
$$
answered Jan 4 at 19:07
J. DoeJ. Doe
1286
1
yes, as with many standard uniqueness proofs, it is the triangle inequality... this proof is correct
– Lost1
Jan 4 at 23:58
add a comment |
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,
$$
\ mathbb{P}(|X-Y|>0)=0Rightarrow mathbb{P}(X=Y)=1
$$
answered Jan 4 at 19:07
J. DoeJ. Doe
1286
1
yes, as with many standard uniqueness proofs, it is the triangle inequality... this proof is correct
– Lost1
Jan 4 at 23:58
add a comment |
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,
$$
\ mathbb{P}(|X-Y|>0)=0Rightarrow mathbb{P}(X=Y)=1
$$
answered Jan 4 at 19:07
J. DoeJ. Doe
1286
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,
$$
\ mathbb{P}(|X-Y|>0)=0Rightarrow mathbb{P}(X=Y)=1
$$
answered Jan 4 at 19:07
J. DoeJ. Doe
1286
answered Jan 4 at 19:07
J. DoeJ. Doe
1286
answered Jan 4 at 19:07
J. DoeJ. Doe
1286
answered Jan 4 at 19:07
J. DoeJ. Doe
1286
1286
1
yes, as with many standard uniqueness proofs, it is the triangle inequality... this proof is correct
– Lost1
Jan 4 at 23:58
add a comment |
1
yes, as with many standard uniqueness proofs, it is the triangle inequality... this proof is correct
– Lost1
Jan 4 at 23:58
1
1
yes, as with many standard uniqueness proofs, it is the triangle inequality... this proof is correct
– Lost1
Jan 4 at 23:58
yes, as with many standard uniqueness proofs, it is the triangle inequality... this proof is correct
– Lost1
Jan 4 at 23:58
add a comment |
