A simple proof of the Portmanteau Theorem
$begingroup$
A premise. Let $X_n$ be a sequence of random variables and let $X$ be a random variable. Call
$$
F_n(x)=mathbb{P}[X_nleq x]
$$
and
$$
F(x)=mathbb{P}[Xleq x].
$$
I say that $X_n$ converges in distribution to $X$ and I write
$$
X_nstackrel{d}{to} X
$$
if and only if
$$
F_{n}(x)to F(x)
$$
for all the $x$ in which $ F(x)$ is continuous.
My problem. I am looking for a simple proof of this result (which is part of the more general Portmanteau theorem).
Theorem. A sequence of random variables $X_n$ converges in distribution to $X$ if and only if
$$
lim_{nrightarrowinfty}mathbb{E}left[fleft(X_nright)right]=E[f(X)]
$$
for any bounded continuous function $f$.
Any suggestion of textbooks/lectures?
continuity random-variables weak-convergence expected-value
asked Dec 22 '18 at 11:49
AlmostSureUserAlmostSureUser
326418
$endgroup$
add a comment |
$begingroup$
A premise. Let $X_n$ be a sequence of random variables and let $X$ be a random variable. Call
$$
F_n(x)=mathbb{P}[X_nleq x]
$$
and
$$
F(x)=mathbb{P}[Xleq x].
$$
I say that $X_n$ converges in distribution to $X$ and I write
$$
X_nstackrel{d}{to} X
$$
if and only if
$$
F_{n}(x)to F(x)
$$
for all the $x$ in which $ F(x)$ is continuous.
My problem. I am looking for a simple proof of this result (which is part of the more general Portmanteau theorem).
Theorem. A sequence of random variables $X_n$ converges in distribution to $X$ if and only if
$$
lim_{nrightarrowinfty}mathbb{E}left[fleft(X_nright)right]=E[f(X)]
$$
for any bounded continuous function $f$.
Any suggestion of textbooks/lectures?
continuity random-variables weak-convergence expected-value
asked Dec 22 '18 at 11:49
AlmostSureUserAlmostSureUser
326418
$endgroup$
$begingroup$
Call $F_n(x)=mathbb{P}[X_nleq x]$ and $F(x)=mathbb{P}[Xleq x]$. I mean this definition $$ X_nstackrel{d}{to} XLeftrightarrow F_{n}(x)to F(x) $$ for all the $x$ in which $ F(x)$ is continuous.
$endgroup$
– AlmostSureUser
Dec 22 '18 at 13:47
$begingroup$
Now it should be ok.
$endgroup$
– AlmostSureUser
Dec 22 '18 at 13:53
$begingroup$
What if you take the function $$h(x) = left{ begin{array}{ll} 1 &mbox{ if $x leq a$} \ 0 & mbox{ otherwise} end{array} right.$$ and find $E[h(X_n)]$ and $E[h(X)]$. Now $h$ is not continuous, can you approximate it by a continuous function?
$endgroup$
– Michael
Dec 22 '18 at 14:06
$begingroup$
Ok, and concerning the other implication?
$endgroup$
– AlmostSureUser
Dec 22 '18 at 14:15
$begingroup$
There seems to be a proof given here, I don't know if it is the "simplest possible": theanalysisofdata.com/probability/8_5.html
$endgroup$
– Michael
Dec 22 '18 at 14:23
add a comment |
$begingroup$
A premise. Let $X_n$ be a sequence of random variables and let $X$ be a random variable. Call
$$
F_n(x)=mathbb{P}[X_nleq x]
$$
and
$$
F(x)=mathbb{P}[Xleq x].
$$
I say that $X_n$ converges in distribution to $X$ and I write
$$
X_nstackrel{d}{to} X
$$
if and only if
$$
F_{n}(x)to F(x)
$$
for all the $x$ in which $ F(x)$ is continuous.
My problem. I am looking for a simple proof of this result (which is part of the more general Portmanteau theorem).
Theorem. A sequence of random variables $X_n$ converges in distribution to $X$ if and only if
$$
lim_{nrightarrowinfty}mathbb{E}left[fleft(X_nright)right]=E[f(X)]
$$
for any bounded continuous function $f$.
Any suggestion of textbooks/lectures?
continuity random-variables weak-convergence expected-value
asked Dec 22 '18 at 11:49
AlmostSureUserAlmostSureUser
326418
$endgroup$
A premise. Let $X_n$ be a sequence of random variables and let $X$ be a random variable. Call
$$
F_n(x)=mathbb{P}[X_nleq x]
$$
and
$$
F(x)=mathbb{P}[Xleq x].
$$
I say that $X_n$ converges in distribution to $X$ and I write
$$
X_nstackrel{d}{to} X
$$
if and only if
$$
F_{n}(x)to F(x)
$$
for all the $x$ in which $ F(x)$ is continuous.
My problem. I am looking for a simple proof of this result (which is part of the more general Portmanteau theorem).
Theorem. A sequence of random variables $X_n$ converges in distribution to $X$ if and only if
$$
lim_{nrightarrowinfty}mathbb{E}left[fleft(X_nright)right]=E[f(X)]
$$
for any bounded continuous function $f$.
Any suggestion of textbooks/lectures?
continuity random-variables weak-convergence expected-value
continuity random-variables weak-convergence expected-value
asked Dec 22 '18 at 11:49
AlmostSureUserAlmostSureUser
326418
asked Dec 22 '18 at 11:49
AlmostSureUserAlmostSureUser
326418
edited Dec 22 '18 at 13:52
AlmostSureUser
asked Dec 22 '18 at 11:49
AlmostSureUserAlmostSureUser
326418
asked Dec 22 '18 at 11:49
AlmostSureUserAlmostSureUser
326418
asked Dec 22 '18 at 11:49
AlmostSureUserAlmostSureUser
326418
326418
$begingroup$
Call $F_n(x)=mathbb{P}[X_nleq x]$ and $F(x)=mathbb{P}[Xleq x]$. I mean this definition $$ X_nstackrel{d}{to} XLeftrightarrow F_{n}(x)to F(x) $$ for all the $x$ in which $ F(x)$ is continuous.
$endgroup$
– AlmostSureUser
Dec 22 '18 at 13:47
$begingroup$
Now it should be ok.
$endgroup$
– AlmostSureUser
Dec 22 '18 at 13:53
$begingroup$
What if you take the function $$h(x) = left{ begin{array}{ll} 1 &mbox{ if $x leq a$} \ 0 & mbox{ otherwise} end{array} right.$$ and find $E[h(X_n)]$ and $E[h(X)]$. Now $h$ is not continuous, can you approximate it by a continuous function?
$endgroup$
– Michael
Dec 22 '18 at 14:06
$begingroup$
Ok, and concerning the other implication?
$endgroup$
– AlmostSureUser
Dec 22 '18 at 14:15
$begingroup$
There seems to be a proof given here, I don't know if it is the "simplest possible": theanalysisofdata.com/probability/8_5.html
$endgroup$
– Michael
Dec 22 '18 at 14:23
add a comment |
$begingroup$
Call $F_n(x)=mathbb{P}[X_nleq x]$ and $F(x)=mathbb{P}[Xleq x]$. I mean this definition $$ X_nstackrel{d}{to} XLeftrightarrow F_{n}(x)to F(x) $$ for all the $x$ in which $ F(x)$ is continuous.
$endgroup$
– AlmostSureUser
Dec 22 '18 at 13:47
$begingroup$
Now it should be ok.
$endgroup$
– AlmostSureUser
Dec 22 '18 at 13:53
$begingroup$
What if you take the function $$h(x) = left{ begin{array}{ll} 1 &mbox{ if $x leq a$} \ 0 & mbox{ otherwise} end{array} right.$$ and find $E[h(X_n)]$ and $E[h(X)]$. Now $h$ is not continuous, can you approximate it by a continuous function?
$endgroup$
– Michael
Dec 22 '18 at 14:06
$begingroup$
Ok, and concerning the other implication?
$endgroup$
– AlmostSureUser
Dec 22 '18 at 14:15
$begingroup$
There seems to be a proof given here, I don't know if it is the "simplest possible": theanalysisofdata.com/probability/8_5.html
$endgroup$
– Michael
Dec 22 '18 at 14:23
$begingroup$
Call $F_n(x)=mathbb{P}[X_nleq x]$ and $F(x)=mathbb{P}[Xleq x]$. I mean this definition $$ X_nstackrel{d}{to} XLeftrightarrow F_{n}(x)to F(x) $$ for all the $x$ in which $ F(x)$ is continuous.
$endgroup$
– AlmostSureUser
Dec 22 '18 at 13:47
$begingroup$
Call $F_n(x)=mathbb{P}[X_nleq x]$ and $F(x)=mathbb{P}[Xleq x]$. I mean this definition $$ X_nstackrel{d}{to} XLeftrightarrow F_{n}(x)to F(x) $$ for all the $x$ in which $ F(x)$ is continuous.
$endgroup$
– AlmostSureUser
Dec 22 '18 at 13:47
$begingroup$
Now it should be ok.
$endgroup$
– AlmostSureUser
Dec 22 '18 at 13:53
$begingroup$
Now it should be ok.
$endgroup$
– AlmostSureUser
Dec 22 '18 at 13:53
$begingroup$
What if you take the function $$h(x) = left{ begin{array}{ll} 1 &mbox{ if $x leq a$} \ 0 & mbox{ otherwise} end{array} right.$$ and find $E[h(X_n)]$ and $E[h(X)]$. Now $h$ is not continuous, can you approximate it by a continuous function?
$endgroup$
– Michael
Dec 22 '18 at 14:06
$begingroup$
What if you take the function $$h(x) = left{ begin{array}{ll} 1 &mbox{ if $x leq a$} \ 0 & mbox{ otherwise} end{array} right.$$ and find $E[h(X_n)]$ and $E[h(X)]$. Now $h$ is not continuous, can you approximate it by a continuous function?
$endgroup$
– Michael
Dec 22 '18 at 14:06
$begingroup$
Ok, and concerning the other implication?
$endgroup$
– AlmostSureUser
Dec 22 '18 at 14:15
$begingroup$
Ok, and concerning the other implication?
$endgroup$
– AlmostSureUser
Dec 22 '18 at 14:15
$begingroup$
There seems to be a proof given here, I don't know if it is the "simplest possible": theanalysisofdata.com/probability/8_5.html
$endgroup$
– Michael
Dec 22 '18 at 14:23
$begingroup$
There seems to be a proof given here, I don't know if it is the "simplest possible": theanalysisofdata.com/probability/8_5.html
$endgroup$
– Michael
Dec 22 '18 at 14:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here is a sketch.
You need to show that for each $epsilon>0$, the inequality $|Ef(X_n)-Ef(X)|<epsilon$ holds for all $n$ sufficiently large. You can start out by picking $K$ so that $P(|X|>K)<epsilon/(2|f|).$ There is a finite set of intervals covering the compact set $[-K,K]$ such that $f$ is within $epsilon/2$ of a constant on each of them. You can make all the above choices so that all the endpoints are points of continuity of $F$. Now replace $f$ with a piecewise constant function suggested by the above partition of $mathbb R$ into intervals, upper bounding $|Ef(X_n)-Ef(X)|$ by a quantity that converges to something less than $epsilon$.
answered Dec 22 '18 at 15:29
kimchi loverkimchi lover
11k31128
$endgroup$
add a comment |
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$begingroup$
Here is a sketch.
You need to show that for each $epsilon>0$, the inequality $|Ef(X_n)-Ef(X)|<epsilon$ holds for all $n$ sufficiently large. You can start out by picking $K$ so that $P(|X|>K)<epsilon/(2|f|).$ There is a finite set of intervals covering the compact set $[-K,K]$ such that $f$ is within $epsilon/2$ of a constant on each of them. You can make all the above choices so that all the endpoints are points of continuity of $F$. Now replace $f$ with a piecewise constant function suggested by the above partition of $mathbb R$ into intervals, upper bounding $|Ef(X_n)-Ef(X)|$ by a quantity that converges to something less than $epsilon$.
answered Dec 22 '18 at 15:29
kimchi loverkimchi lover
11k31128
$endgroup$
add a comment |
$begingroup$
Here is a sketch.
You need to show that for each $epsilon>0$, the inequality $|Ef(X_n)-Ef(X)|<epsilon$ holds for all $n$ sufficiently large. You can start out by picking $K$ so that $P(|X|>K)<epsilon/(2|f|).$ There is a finite set of intervals covering the compact set $[-K,K]$ such that $f$ is within $epsilon/2$ of a constant on each of them. You can make all the above choices so that all the endpoints are points of continuity of $F$. Now replace $f$ with a piecewise constant function suggested by the above partition of $mathbb R$ into intervals, upper bounding $|Ef(X_n)-Ef(X)|$ by a quantity that converges to something less than $epsilon$.
answered Dec 22 '18 at 15:29
kimchi loverkimchi lover
11k31128
$endgroup$
add a comment |
$begingroup$
Here is a sketch.
You need to show that for each $epsilon>0$, the inequality $|Ef(X_n)-Ef(X)|<epsilon$ holds for all $n$ sufficiently large. You can start out by picking $K$ so that $P(|X|>K)<epsilon/(2|f|).$ There is a finite set of intervals covering the compact set $[-K,K]$ such that $f$ is within $epsilon/2$ of a constant on each of them. You can make all the above choices so that all the endpoints are points of continuity of $F$. Now replace $f$ with a piecewise constant function suggested by the above partition of $mathbb R$ into intervals, upper bounding $|Ef(X_n)-Ef(X)|$ by a quantity that converges to something less than $epsilon$.
answered Dec 22 '18 at 15:29
kimchi loverkimchi lover
11k31128
$endgroup$
Here is a sketch.
You need to show that for each $epsilon>0$, the inequality $|Ef(X_n)-Ef(X)|<epsilon$ holds for all $n$ sufficiently large. You can start out by picking $K$ so that $P(|X|>K)<epsilon/(2|f|).$ There is a finite set of intervals covering the compact set $[-K,K]$ such that $f$ is within $epsilon/2$ of a constant on each of them. You can make all the above choices so that all the endpoints are points of continuity of $F$. Now replace $f$ with a piecewise constant function suggested by the above partition of $mathbb R$ into intervals, upper bounding $|Ef(X_n)-Ef(X)|$ by a quantity that converges to something less than $epsilon$.
answered Dec 22 '18 at 15:29
kimchi loverkimchi lover
11k31128
edited Dec 23 '18 at 2:36
answered Dec 22 '18 at 15:29
kimchi loverkimchi lover
11k31128
answered Dec 22 '18 at 15:29
kimchi loverkimchi lover
11k31128
answered Dec 22 '18 at 15:29
kimchi loverkimchi lover
11k31128
11k31128
add a comment |
add a comment |
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$begingroup$
Call $F_n(x)=mathbb{P}[X_nleq x]$ and $F(x)=mathbb{P}[Xleq x]$. I mean this definition $$ X_nstackrel{d}{to} XLeftrightarrow F_{n}(x)to F(x) $$ for all the $x$ in which $ F(x)$ is continuous.
$endgroup$
– AlmostSureUser
Dec 22 '18 at 13:47
$begingroup$
Now it should be ok.
$endgroup$
– AlmostSureUser
Dec 22 '18 at 13:53
$begingroup$
What if you take the function $$h(x) = left{ begin{array}{ll} 1 &mbox{ if $x leq a$} \ 0 & mbox{ otherwise} end{array} right.$$ and find $E[h(X_n)]$ and $E[h(X)]$. Now $h$ is not continuous, can you approximate it by a continuous function?
$endgroup$
– Michael
Dec 22 '18 at 14:06
$begingroup$
Ok, and concerning the other implication?
$endgroup$
– AlmostSureUser
Dec 22 '18 at 14:15
$begingroup$
There seems to be a proof given here, I don't know if it is the "simplest possible": theanalysisofdata.com/probability/8_5.html
$endgroup$
– Michael
Dec 22 '18 at 14:23