Almost sure convergence exercise
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I have to do this exercise:
Let $(X_n)$ be some sequence of random variables and let $X$ be some random variable such that $X_n to X$ almost surely.
Show that, given $epsilon > 0$, there is a set $A$ such that $P(A)leq epsilon$ and $X_n to X$ uniformly on $Omega setminus A$.
I think that I can use these sets $E(n_k) = bigcup_{mgeq n}{omega in Ω: mid X_m(omega) − X(omega)| geq frac1k}$ and then obtain $A$ from there.
But I don't know how to start and furthermore what $X_n to X$ uniformly means.
probability convergence random-variables
edited Nov 11 '18 at 18:35
Namaste
1
asked Nov 11 '18 at 16:50
Francesca BallatoreFrancesca Ballatore
426
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add a comment |
$begingroup$
I have to do this exercise:
Let $(X_n)$ be some sequence of random variables and let $X$ be some random variable such that $X_n to X$ almost surely.
Show that, given $epsilon > 0$, there is a set $A$ such that $P(A)leq epsilon$ and $X_n to X$ uniformly on $Omega setminus A$.
I think that I can use these sets $E(n_k) = bigcup_{mgeq n}{omega in Ω: mid X_m(omega) − X(omega)| geq frac1k}$ and then obtain $A$ from there.
But I don't know how to start and furthermore what $X_n to X$ uniformly means.
probability convergence random-variables
edited Nov 11 '18 at 18:35
Namaste
1
asked Nov 11 '18 at 16:50
Francesca BallatoreFrancesca Ballatore
426
$endgroup$
2
$begingroup$
This is essentially a celebrated theorem by Egoroff. Look up Egoroff's Thm.
$endgroup$
– TheOscillator
Nov 11 '18 at 17:11
add a comment |
$begingroup$
I have to do this exercise:
Let $(X_n)$ be some sequence of random variables and let $X$ be some random variable such that $X_n to X$ almost surely.
Show that, given $epsilon > 0$, there is a set $A$ such that $P(A)leq epsilon$ and $X_n to X$ uniformly on $Omega setminus A$.
I think that I can use these sets $E(n_k) = bigcup_{mgeq n}{omega in Ω: mid X_m(omega) − X(omega)| geq frac1k}$ and then obtain $A$ from there.
But I don't know how to start and furthermore what $X_n to X$ uniformly means.
probability convergence random-variables
edited Nov 11 '18 at 18:35
Namaste
1
asked Nov 11 '18 at 16:50
Francesca BallatoreFrancesca Ballatore
426
$endgroup$
I have to do this exercise:
Let $(X_n)$ be some sequence of random variables and let $X$ be some random variable such that $X_n to X$ almost surely.
Show that, given $epsilon > 0$, there is a set $A$ such that $P(A)leq epsilon$ and $X_n to X$ uniformly on $Omega setminus A$.
I think that I can use these sets $E(n_k) = bigcup_{mgeq n}{omega in Ω: mid X_m(omega) − X(omega)| geq frac1k}$ and then obtain $A$ from there.
But I don't know how to start and furthermore what $X_n to X$ uniformly means.
probability convergence random-variables
probability convergence random-variables
edited Nov 11 '18 at 18:35
Namaste
1
asked Nov 11 '18 at 16:50
Francesca BallatoreFrancesca Ballatore
426
edited Nov 11 '18 at 18:35
Namaste
1
asked Nov 11 '18 at 16:50
Francesca BallatoreFrancesca Ballatore
426
edited Nov 11 '18 at 18:35
Namaste
1
edited Nov 11 '18 at 18:35
Namaste
1
edited Nov 11 '18 at 18:35
Namaste
1
1
asked Nov 11 '18 at 16:50
Francesca BallatoreFrancesca Ballatore
426
asked Nov 11 '18 at 16:50
Francesca BallatoreFrancesca Ballatore
426
asked Nov 11 '18 at 16:50
Francesca BallatoreFrancesca Ballatore
426
426
2
$begingroup$
This is essentially a celebrated theorem by Egoroff. Look up Egoroff's Thm.
$endgroup$
– TheOscillator
Nov 11 '18 at 17:11
add a comment |
2
$begingroup$
This is essentially a celebrated theorem by Egoroff. Look up Egoroff's Thm.
$endgroup$
– TheOscillator
Nov 11 '18 at 17:11
2
2
$begingroup$
This is essentially a celebrated theorem by Egoroff. Look up Egoroff's Thm.
$endgroup$
– TheOscillator
Nov 11 '18 at 17:11
$begingroup$
This is essentially a celebrated theorem by Egoroff. Look up Egoroff's Thm.
$endgroup$
– TheOscillator
Nov 11 '18 at 17:11
add a comment |
1 Answer
1
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$begingroup$
We say that $X_nto X$ uniformly on a set $E$ if $sup_{omegain E}leftlvert X_n(omega)-X(omega)rightrvertto 0$.
Here are some steps:
For simplicity, assume that $X=0$ and $X_ngeqslant 0$ (consider $leftlvert X_n-Xrightrvert$ instead of $X_n$).
Let $$E(n,k) = bigcup_{mgeqslant n}left{omega in Ω: X_m(omega) geqslant frac1kright}.$$
The assumption that $X_mto 0$ implies that for all fixed $k$, $Prleft(bigcap_{ngeqslant 1}E_{n,k}right)=0$.The sequence $left(E_{n,k}right)_{ngeqslant 1}$ is non-increasing for each $k$ hence there exists $n_k$ such that $Prleft( E_{n_k,k}right)lt varepsilon 2^{-k}$
Define $A:=bigcup_{kgeqslant 1}E_{n_k,k}$ and check that $A$ does the job.
answered Dec 10 '18 at 11:32
Davide GiraudoDavide Giraudo
127k17154268
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add a comment |
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1 Answer
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$begingroup$
We say that $X_nto X$ uniformly on a set $E$ if $sup_{omegain E}leftlvert X_n(omega)-X(omega)rightrvertto 0$.
Here are some steps:
For simplicity, assume that $X=0$ and $X_ngeqslant 0$ (consider $leftlvert X_n-Xrightrvert$ instead of $X_n$).
Let $$E(n,k) = bigcup_{mgeqslant n}left{omega in Ω: X_m(omega) geqslant frac1kright}.$$
The assumption that $X_mto 0$ implies that for all fixed $k$, $Prleft(bigcap_{ngeqslant 1}E_{n,k}right)=0$.The sequence $left(E_{n,k}right)_{ngeqslant 1}$ is non-increasing for each $k$ hence there exists $n_k$ such that $Prleft( E_{n_k,k}right)lt varepsilon 2^{-k}$
Define $A:=bigcup_{kgeqslant 1}E_{n_k,k}$ and check that $A$ does the job.
answered Dec 10 '18 at 11:32
Davide GiraudoDavide Giraudo
127k17154268
$endgroup$
add a comment |
$begingroup$
We say that $X_nto X$ uniformly on a set $E$ if $sup_{omegain E}leftlvert X_n(omega)-X(omega)rightrvertto 0$.
Here are some steps:
For simplicity, assume that $X=0$ and $X_ngeqslant 0$ (consider $leftlvert X_n-Xrightrvert$ instead of $X_n$).
Let $$E(n,k) = bigcup_{mgeqslant n}left{omega in Ω: X_m(omega) geqslant frac1kright}.$$
The assumption that $X_mto 0$ implies that for all fixed $k$, $Prleft(bigcap_{ngeqslant 1}E_{n,k}right)=0$.The sequence $left(E_{n,k}right)_{ngeqslant 1}$ is non-increasing for each $k$ hence there exists $n_k$ such that $Prleft( E_{n_k,k}right)lt varepsilon 2^{-k}$
Define $A:=bigcup_{kgeqslant 1}E_{n_k,k}$ and check that $A$ does the job.
answered Dec 10 '18 at 11:32
Davide GiraudoDavide Giraudo
127k17154268
$endgroup$
add a comment |
$begingroup$
We say that $X_nto X$ uniformly on a set $E$ if $sup_{omegain E}leftlvert X_n(omega)-X(omega)rightrvertto 0$.
Here are some steps:
For simplicity, assume that $X=0$ and $X_ngeqslant 0$ (consider $leftlvert X_n-Xrightrvert$ instead of $X_n$).
Let $$E(n,k) = bigcup_{mgeqslant n}left{omega in Ω: X_m(omega) geqslant frac1kright}.$$
The assumption that $X_mto 0$ implies that for all fixed $k$, $Prleft(bigcap_{ngeqslant 1}E_{n,k}right)=0$.The sequence $left(E_{n,k}right)_{ngeqslant 1}$ is non-increasing for each $k$ hence there exists $n_k$ such that $Prleft( E_{n_k,k}right)lt varepsilon 2^{-k}$
Define $A:=bigcup_{kgeqslant 1}E_{n_k,k}$ and check that $A$ does the job.
answered Dec 10 '18 at 11:32
Davide GiraudoDavide Giraudo
127k17154268
$endgroup$
We say that $X_nto X$ uniformly on a set $E$ if $sup_{omegain E}leftlvert X_n(omega)-X(omega)rightrvertto 0$.
Here are some steps:
For simplicity, assume that $X=0$ and $X_ngeqslant 0$ (consider $leftlvert X_n-Xrightrvert$ instead of $X_n$).
Let $$E(n,k) = bigcup_{mgeqslant n}left{omega in Ω: X_m(omega) geqslant frac1kright}.$$
The assumption that $X_mto 0$ implies that for all fixed $k$, $Prleft(bigcap_{ngeqslant 1}E_{n,k}right)=0$.The sequence $left(E_{n,k}right)_{ngeqslant 1}$ is non-increasing for each $k$ hence there exists $n_k$ such that $Prleft( E_{n_k,k}right)lt varepsilon 2^{-k}$
Define $A:=bigcup_{kgeqslant 1}E_{n_k,k}$ and check that $A$ does the job.
answered Dec 10 '18 at 11:32
Davide GiraudoDavide Giraudo
127k17154268
answered Dec 10 '18 at 11:32
Davide GiraudoDavide Giraudo
127k17154268
answered Dec 10 '18 at 11:32
Davide GiraudoDavide Giraudo
127k17154268
answered Dec 10 '18 at 11:32
Davide GiraudoDavide Giraudo
127k17154268
127k17154268
add a comment |
add a comment |
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$begingroup$
This is essentially a celebrated theorem by Egoroff. Look up Egoroff's Thm.
$endgroup$
– TheOscillator
Nov 11 '18 at 17:11