Convergence in probability implies convergence in distribution
up vote
2
down vote
favorite
A sequence of random variables ${X_n}$ converges to $X$ in probability if for any $varepsilon > 0$,
$$P(|X_n-X| geq varepsilon) rightarrow 0$$
They converge in distribution if
$$F_{X_n} rightarrow F_X$$
at points where $F_X$ is continuous.
(There is another equivalent definition of converge in distribution in terms of weak convergence.)
It seems like a very simple result, but I cannot think of a clever proof.
probability probability-theory weak-convergence
edited Jun 18 at 18:52
Davide Giraudo
124k16150259
asked Nov 14 '12 at 4:08
Hawii
666514
add a comment |
up vote
2
down vote
favorite
A sequence of random variables ${X_n}$ converges to $X$ in probability if for any $varepsilon > 0$,
$$P(|X_n-X| geq varepsilon) rightarrow 0$$
They converge in distribution if
$$F_{X_n} rightarrow F_X$$
at points where $F_X$ is continuous.
(There is another equivalent definition of converge in distribution in terms of weak convergence.)
It seems like a very simple result, but I cannot think of a clever proof.
probability probability-theory weak-convergence
edited Jun 18 at 18:52
Davide Giraudo
124k16150259
asked Nov 14 '12 at 4:08
Hawii
666514
1
Have you tried the wikipedia article: en.wikipedia.org/wiki/… ? Most books on probability theory include a proof.
– Gautam Shenoy
Nov 14 '12 at 4:54
Oh, how come I didn't find it! It looks like something I have in mind. Thank you so much!
– Hawii
Nov 14 '12 at 5:34
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
A sequence of random variables ${X_n}$ converges to $X$ in probability if for any $varepsilon > 0$,
$$P(|X_n-X| geq varepsilon) rightarrow 0$$
They converge in distribution if
$$F_{X_n} rightarrow F_X$$
at points where $F_X$ is continuous.
(There is another equivalent definition of converge in distribution in terms of weak convergence.)
It seems like a very simple result, but I cannot think of a clever proof.
probability probability-theory weak-convergence
edited Jun 18 at 18:52
Davide Giraudo
124k16150259
asked Nov 14 '12 at 4:08
Hawii
666514
A sequence of random variables ${X_n}$ converges to $X$ in probability if for any $varepsilon > 0$,
$$P(|X_n-X| geq varepsilon) rightarrow 0$$
They converge in distribution if
$$F_{X_n} rightarrow F_X$$
at points where $F_X$ is continuous.
(There is another equivalent definition of converge in distribution in terms of weak convergence.)
It seems like a very simple result, but I cannot think of a clever proof.
probability probability-theory weak-convergence
probability probability-theory weak-convergence
edited Jun 18 at 18:52
Davide Giraudo
124k16150259
asked Nov 14 '12 at 4:08
Hawii
666514
edited Jun 18 at 18:52
Davide Giraudo
124k16150259
asked Nov 14 '12 at 4:08
Hawii
666514
edited Jun 18 at 18:52
Davide Giraudo
124k16150259
edited Jun 18 at 18:52
Davide Giraudo
124k16150259
edited Jun 18 at 18:52
Davide Giraudo
124k16150259
124k16150259
asked Nov 14 '12 at 4:08
Hawii
666514
asked Nov 14 '12 at 4:08
Hawii
666514
asked Nov 14 '12 at 4:08
Hawii
666514
666514
1
Have you tried the wikipedia article: en.wikipedia.org/wiki/… ? Most books on probability theory include a proof.
– Gautam Shenoy
Nov 14 '12 at 4:54
Oh, how come I didn't find it! It looks like something I have in mind. Thank you so much!
– Hawii
Nov 14 '12 at 5:34
add a comment |
1
Have you tried the wikipedia article: en.wikipedia.org/wiki/… ? Most books on probability theory include a proof.
– Gautam Shenoy
Nov 14 '12 at 4:54
Oh, how come I didn't find it! It looks like something I have in mind. Thank you so much!
– Hawii
Nov 14 '12 at 5:34
1
1
Have you tried the wikipedia article: en.wikipedia.org/wiki/… ? Most books on probability theory include a proof.
– Gautam Shenoy
Nov 14 '12 at 4:54
Have you tried the wikipedia article: en.wikipedia.org/wiki/… ? Most books on probability theory include a proof.
– Gautam Shenoy
Nov 14 '12 at 4:54
Oh, how come I didn't find it! It looks like something I have in mind. Thank you so much!
– Hawii
Nov 14 '12 at 5:34
Oh, how come I didn't find it! It looks like something I have in mind. Thank you so much!
– Hawii
Nov 14 '12 at 5:34
add a comment |
3 Answers
3
active
oldest
votes
up vote
10
down vote
A slicker proof (and more importantly one that generalizes) than the one in the wikipedia article is to observe that $X_n Longrightarrow X$ if and only if for all bounded continuous functions $f$ we have $E f(X_n) to E f(X)$. If you have convergence in probability then you can apply the dominated convergence theorem (recalling that $f$ is bounded and that for continuous functions $X_n to X$ in probability implies $f(X_n) to f(X)$ in probability) to conclude that $E |f(X_n) - f(X)| to 0$, which implies the result.
answered Nov 14 '12 at 20:55
Chris Janjigian
4,89341835
add a comment |
up vote
1
down vote
Here is an answer that does not rely on dominated convergence.
To prove convergence in distribution, we only need to establish that $E[f(X_n)]$ converges to $E[f(X)]$ for bounded continuous functions $f$.
By definition of the limit, we need to prove that for any $epsilon>0$, there some $n_0=n_0(epsilon)$ such that for all $n>n_0$ the inequality $| E[f(X_n)] - E[f(X)]| < epsilon $ holds.
- As suggested in another answer, the first step is to show that if $X_n$ converge to $X$ in probability then $f(X_n)$ also converges in probability to $f(X)$ for any continuous $f$.
- Let $f$ be any continuous function bounded by $K$. Take any $epsilon>0$ and show that
$$| E[f(X_n)] - E[f(X)]| le E[|f(X_n)] - E[f(X)|] le (epsilon/2) ; P(A_n^c) + K ; P(A_n)$$
where $A_n$ is the event ${ |f(X_n)] - E[f(X)| > epsilon /2 }$. - It remains to show that $P(A_n^c)le 1$ (obvious) and that for $n$ large enough, one has $P(A_n^c)le epsilon/(2 K)$ thanks to the convergence in probability established in 1.
answered Sep 26 '17 at 4:11
jlewk
765
add a comment |
up vote
0
down vote
Chris J.'s answer more or less is correct, but you require almost sure convergence to be able to apply dominated convergence. Fortunately, convergence in probability implies almost sure convergence along a subsequence, and the proof more or less can proceed as desired.
For more details, Kallenberg's Foundations of Modern Probability, First Edition, Lemma 3.7 is useful.
answered Feb 16 '14 at 3:14
Roy D.
404211
Or you could apply the bounded convergence theorem.
– Calculon
Mar 15 '15 at 11:33
2
Dominated convergence theorem also applies with convergence in probability.
– perlman
Oct 29 '17 at 0:26
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
A slicker proof (and more importantly one that generalizes) than the one in the wikipedia article is to observe that $X_n Longrightarrow X$ if and only if for all bounded continuous functions $f$ we have $E f(X_n) to E f(X)$. If you have convergence in probability then you can apply the dominated convergence theorem (recalling that $f$ is bounded and that for continuous functions $X_n to X$ in probability implies $f(X_n) to f(X)$ in probability) to conclude that $E |f(X_n) - f(X)| to 0$, which implies the result.
answered Nov 14 '12 at 20:55
Chris Janjigian
4,89341835
add a comment |
up vote
10
down vote
A slicker proof (and more importantly one that generalizes) than the one in the wikipedia article is to observe that $X_n Longrightarrow X$ if and only if for all bounded continuous functions $f$ we have $E f(X_n) to E f(X)$. If you have convergence in probability then you can apply the dominated convergence theorem (recalling that $f$ is bounded and that for continuous functions $X_n to X$ in probability implies $f(X_n) to f(X)$ in probability) to conclude that $E |f(X_n) - f(X)| to 0$, which implies the result.
answered Nov 14 '12 at 20:55
Chris Janjigian
4,89341835
add a comment |
up vote
10
down vote
up vote
10
down vote
A slicker proof (and more importantly one that generalizes) than the one in the wikipedia article is to observe that $X_n Longrightarrow X$ if and only if for all bounded continuous functions $f$ we have $E f(X_n) to E f(X)$. If you have convergence in probability then you can apply the dominated convergence theorem (recalling that $f$ is bounded and that for continuous functions $X_n to X$ in probability implies $f(X_n) to f(X)$ in probability) to conclude that $E |f(X_n) - f(X)| to 0$, which implies the result.
answered Nov 14 '12 at 20:55
Chris Janjigian
4,89341835
A slicker proof (and more importantly one that generalizes) than the one in the wikipedia article is to observe that $X_n Longrightarrow X$ if and only if for all bounded continuous functions $f$ we have $E f(X_n) to E f(X)$. If you have convergence in probability then you can apply the dominated convergence theorem (recalling that $f$ is bounded and that for continuous functions $X_n to X$ in probability implies $f(X_n) to f(X)$ in probability) to conclude that $E |f(X_n) - f(X)| to 0$, which implies the result.
answered Nov 14 '12 at 20:55
Chris Janjigian
4,89341835
answered Nov 14 '12 at 20:55
Chris Janjigian
4,89341835
answered Nov 14 '12 at 20:55
Chris Janjigian
4,89341835
answered Nov 14 '12 at 20:55
Chris Janjigian
4,89341835
4,89341835
add a comment |
add a comment |
up vote
1
down vote
Here is an answer that does not rely on dominated convergence.
To prove convergence in distribution, we only need to establish that $E[f(X_n)]$ converges to $E[f(X)]$ for bounded continuous functions $f$.
By definition of the limit, we need to prove that for any $epsilon>0$, there some $n_0=n_0(epsilon)$ such that for all $n>n_0$ the inequality $| E[f(X_n)] - E[f(X)]| < epsilon $ holds.
- As suggested in another answer, the first step is to show that if $X_n$ converge to $X$ in probability then $f(X_n)$ also converges in probability to $f(X)$ for any continuous $f$.
- Let $f$ be any continuous function bounded by $K$. Take any $epsilon>0$ and show that
$$| E[f(X_n)] - E[f(X)]| le E[|f(X_n)] - E[f(X)|] le (epsilon/2) ; P(A_n^c) + K ; P(A_n)$$
where $A_n$ is the event ${ |f(X_n)] - E[f(X)| > epsilon /2 }$. - It remains to show that $P(A_n^c)le 1$ (obvious) and that for $n$ large enough, one has $P(A_n^c)le epsilon/(2 K)$ thanks to the convergence in probability established in 1.
answered Sep 26 '17 at 4:11
jlewk
765
add a comment |
up vote
1
down vote
Here is an answer that does not rely on dominated convergence.
To prove convergence in distribution, we only need to establish that $E[f(X_n)]$ converges to $E[f(X)]$ for bounded continuous functions $f$.
By definition of the limit, we need to prove that for any $epsilon>0$, there some $n_0=n_0(epsilon)$ such that for all $n>n_0$ the inequality $| E[f(X_n)] - E[f(X)]| < epsilon $ holds.
- As suggested in another answer, the first step is to show that if $X_n$ converge to $X$ in probability then $f(X_n)$ also converges in probability to $f(X)$ for any continuous $f$.
- Let $f$ be any continuous function bounded by $K$. Take any $epsilon>0$ and show that
$$| E[f(X_n)] - E[f(X)]| le E[|f(X_n)] - E[f(X)|] le (epsilon/2) ; P(A_n^c) + K ; P(A_n)$$
where $A_n$ is the event ${ |f(X_n)] - E[f(X)| > epsilon /2 }$. - It remains to show that $P(A_n^c)le 1$ (obvious) and that for $n$ large enough, one has $P(A_n^c)le epsilon/(2 K)$ thanks to the convergence in probability established in 1.
answered Sep 26 '17 at 4:11
jlewk
765
add a comment |
up vote
1
down vote
up vote
1
down vote
Here is an answer that does not rely on dominated convergence.
To prove convergence in distribution, we only need to establish that $E[f(X_n)]$ converges to $E[f(X)]$ for bounded continuous functions $f$.
By definition of the limit, we need to prove that for any $epsilon>0$, there some $n_0=n_0(epsilon)$ such that for all $n>n_0$ the inequality $| E[f(X_n)] - E[f(X)]| < epsilon $ holds.
- As suggested in another answer, the first step is to show that if $X_n$ converge to $X$ in probability then $f(X_n)$ also converges in probability to $f(X)$ for any continuous $f$.
- Let $f$ be any continuous function bounded by $K$. Take any $epsilon>0$ and show that
$$| E[f(X_n)] - E[f(X)]| le E[|f(X_n)] - E[f(X)|] le (epsilon/2) ; P(A_n^c) + K ; P(A_n)$$
where $A_n$ is the event ${ |f(X_n)] - E[f(X)| > epsilon /2 }$. - It remains to show that $P(A_n^c)le 1$ (obvious) and that for $n$ large enough, one has $P(A_n^c)le epsilon/(2 K)$ thanks to the convergence in probability established in 1.
answered Sep 26 '17 at 4:11
jlewk
765
Here is an answer that does not rely on dominated convergence.
To prove convergence in distribution, we only need to establish that $E[f(X_n)]$ converges to $E[f(X)]$ for bounded continuous functions $f$.
By definition of the limit, we need to prove that for any $epsilon>0$, there some $n_0=n_0(epsilon)$ such that for all $n>n_0$ the inequality $| E[f(X_n)] - E[f(X)]| < epsilon $ holds.
- As suggested in another answer, the first step is to show that if $X_n$ converge to $X$ in probability then $f(X_n)$ also converges in probability to $f(X)$ for any continuous $f$.
- Let $f$ be any continuous function bounded by $K$. Take any $epsilon>0$ and show that
$$| E[f(X_n)] - E[f(X)]| le E[|f(X_n)] - E[f(X)|] le (epsilon/2) ; P(A_n^c) + K ; P(A_n)$$
where $A_n$ is the event ${ |f(X_n)] - E[f(X)| > epsilon /2 }$. - It remains to show that $P(A_n^c)le 1$ (obvious) and that for $n$ large enough, one has $P(A_n^c)le epsilon/(2 K)$ thanks to the convergence in probability established in 1.
answered Sep 26 '17 at 4:11
jlewk
765
edited Nov 17 '17 at 18:21
answered Sep 26 '17 at 4:11
jlewk
765
answered Sep 26 '17 at 4:11
jlewk
765
answered Sep 26 '17 at 4:11
jlewk
765
765
add a comment |
add a comment |
up vote
0
down vote
Chris J.'s answer more or less is correct, but you require almost sure convergence to be able to apply dominated convergence. Fortunately, convergence in probability implies almost sure convergence along a subsequence, and the proof more or less can proceed as desired.
For more details, Kallenberg's Foundations of Modern Probability, First Edition, Lemma 3.7 is useful.
answered Feb 16 '14 at 3:14
Roy D.
404211
Or you could apply the bounded convergence theorem.
– Calculon
Mar 15 '15 at 11:33
2
Dominated convergence theorem also applies with convergence in probability.
– perlman
Oct 29 '17 at 0:26
add a comment |
up vote
0
down vote
Chris J.'s answer more or less is correct, but you require almost sure convergence to be able to apply dominated convergence. Fortunately, convergence in probability implies almost sure convergence along a subsequence, and the proof more or less can proceed as desired.
For more details, Kallenberg's Foundations of Modern Probability, First Edition, Lemma 3.7 is useful.
answered Feb 16 '14 at 3:14
Roy D.
404211
Or you could apply the bounded convergence theorem.
– Calculon
Mar 15 '15 at 11:33
2
Dominated convergence theorem also applies with convergence in probability.
– perlman
Oct 29 '17 at 0:26
add a comment |
up vote
0
down vote
up vote
0
down vote
Chris J.'s answer more or less is correct, but you require almost sure convergence to be able to apply dominated convergence. Fortunately, convergence in probability implies almost sure convergence along a subsequence, and the proof more or less can proceed as desired.
For more details, Kallenberg's Foundations of Modern Probability, First Edition, Lemma 3.7 is useful.
answered Feb 16 '14 at 3:14
Roy D.
404211
Chris J.'s answer more or less is correct, but you require almost sure convergence to be able to apply dominated convergence. Fortunately, convergence in probability implies almost sure convergence along a subsequence, and the proof more or less can proceed as desired.
For more details, Kallenberg's Foundations of Modern Probability, First Edition, Lemma 3.7 is useful.
answered Feb 16 '14 at 3:14
Roy D.
404211
answered Feb 16 '14 at 3:14
Roy D.
404211
answered Feb 16 '14 at 3:14
Roy D.
404211
answered Feb 16 '14 at 3:14
Roy D.
404211
404211
Or you could apply the bounded convergence theorem.
– Calculon
Mar 15 '15 at 11:33
2
Dominated convergence theorem also applies with convergence in probability.
– perlman
Oct 29 '17 at 0:26
add a comment |
Or you could apply the bounded convergence theorem.
– Calculon
Mar 15 '15 at 11:33
2
Dominated convergence theorem also applies with convergence in probability.
– perlman
Oct 29 '17 at 0:26
Or you could apply the bounded convergence theorem.
– Calculon
Mar 15 '15 at 11:33
Or you could apply the bounded convergence theorem.
– Calculon
Mar 15 '15 at 11:33
2
2
Dominated convergence theorem also applies with convergence in probability.
– perlman
Oct 29 '17 at 0:26
Dominated convergence theorem also applies with convergence in probability.
– perlman
Oct 29 '17 at 0:26
add a comment |
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1
Have you tried the wikipedia article: en.wikipedia.org/wiki/… ? Most books on probability theory include a proof.
– Gautam Shenoy
Nov 14 '12 at 4:54
Oh, how come I didn't find it! It looks like something I have in mind. Thank you so much!
– Hawii
Nov 14 '12 at 5:34