Sets constructed by random variable series limit
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Let $(Omega$, $mathcal{F}$, $P)$ be a probability space, $X_n$, $X$ random variables in that space. Show that:
$(a)$ ${omegainOmega| lim_{nrightarrowinfty} X_n(omega)$ $exists}inmathcal{F}$,
$(b)$ ${omegainOmega| lim_{nrightarrowinfty} X_n(omega)=X(omega)}inmathcal{F}$.
Edit:
I found quite simply solution.
${omegainOmega| lim_{nrightarrowinfty} X_n(omega)exists}=\={omegainOmega| forall_{kinmathbb{N}}exists_{Ninmathbb{N}}forall_{m,n>N}|X_n(omega)-X_m(omega)|<frac1k}= \=bigcaplimits_{kinmathbb{N}}bigcuplimits_{Ninmathbb{N}}bigcaplimits_{n,m>N}{|X_n(omega)-X_m(omega)|<frac1k}=\=bigcaplimits_{kinmathbb{N}}bigcuplimits_{Ninmathbb{N}}bigcaplimits_{n,m>N}big({X_n(omega)-X_m(omega)<frac1k}cap{X_n(omega)-X_m(omega)>-frac1k}big)inmathcal{F}$
probability probability-theory random-variables
asked Oct 7 at 19:13
Matt53
1227
add a comment |
up vote
0
down vote
favorite
Let $(Omega$, $mathcal{F}$, $P)$ be a probability space, $X_n$, $X$ random variables in that space. Show that:
$(a)$ ${omegainOmega| lim_{nrightarrowinfty} X_n(omega)$ $exists}inmathcal{F}$,
$(b)$ ${omegainOmega| lim_{nrightarrowinfty} X_n(omega)=X(omega)}inmathcal{F}$.
Edit:
I found quite simply solution.
${omegainOmega| lim_{nrightarrowinfty} X_n(omega)exists}=\={omegainOmega| forall_{kinmathbb{N}}exists_{Ninmathbb{N}}forall_{m,n>N}|X_n(omega)-X_m(omega)|<frac1k}= \=bigcaplimits_{kinmathbb{N}}bigcuplimits_{Ninmathbb{N}}bigcaplimits_{n,m>N}{|X_n(omega)-X_m(omega)|<frac1k}=\=bigcaplimits_{kinmathbb{N}}bigcuplimits_{Ninmathbb{N}}bigcaplimits_{n,m>N}big({X_n(omega)-X_m(omega)<frac1k}cap{X_n(omega)-X_m(omega)>-frac1k}big)inmathcal{F}$
probability probability-theory random-variables
asked Oct 7 at 19:13
Matt53
1227
2
Can you show that $limsup_n X_n$ is measurable (and $liminf$)? If so, the set of points at which the limit exists can be written as the set of points satisfying $limsup_n X_n(omega) = liminf_n X_n(omega)$.
– copper.hat
Oct 7 at 19:17
Is ${omega|limsup_nX_n(omega)>a}=bigcuplimits_{n=1}^{infty}{omega|X_n(omega)>a}$ enough to show that $limsup_nX_n$ measureable?
– Matt53
Oct 7 at 19:29
1
That is not a correct equality.
– copper.hat
Oct 7 at 19:32
It should be ${omega|sup_{kgeq n}X_k(omega)>a}=bigcuplimits_{k=n}^{infty}{omega|X_k(omega)>a}$ right? And then by $limsup_nX_n=lim_nsup_{kgeq n}X_k$ we got that $limsup_nX_n$ is measureable.
– Matt53
Oct 7 at 20:55
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $(Omega$, $mathcal{F}$, $P)$ be a probability space, $X_n$, $X$ random variables in that space. Show that:
$(a)$ ${omegainOmega| lim_{nrightarrowinfty} X_n(omega)$ $exists}inmathcal{F}$,
$(b)$ ${omegainOmega| lim_{nrightarrowinfty} X_n(omega)=X(omega)}inmathcal{F}$.
Edit:
I found quite simply solution.
${omegainOmega| lim_{nrightarrowinfty} X_n(omega)exists}=\={omegainOmega| forall_{kinmathbb{N}}exists_{Ninmathbb{N}}forall_{m,n>N}|X_n(omega)-X_m(omega)|<frac1k}= \=bigcaplimits_{kinmathbb{N}}bigcuplimits_{Ninmathbb{N}}bigcaplimits_{n,m>N}{|X_n(omega)-X_m(omega)|<frac1k}=\=bigcaplimits_{kinmathbb{N}}bigcuplimits_{Ninmathbb{N}}bigcaplimits_{n,m>N}big({X_n(omega)-X_m(omega)<frac1k}cap{X_n(omega)-X_m(omega)>-frac1k}big)inmathcal{F}$
probability probability-theory random-variables
asked Oct 7 at 19:13
Matt53
1227
Let $(Omega$, $mathcal{F}$, $P)$ be a probability space, $X_n$, $X$ random variables in that space. Show that:
$(a)$ ${omegainOmega| lim_{nrightarrowinfty} X_n(omega)$ $exists}inmathcal{F}$,
$(b)$ ${omegainOmega| lim_{nrightarrowinfty} X_n(omega)=X(omega)}inmathcal{F}$.
Edit:
I found quite simply solution.
${omegainOmega| lim_{nrightarrowinfty} X_n(omega)exists}=\={omegainOmega| forall_{kinmathbb{N}}exists_{Ninmathbb{N}}forall_{m,n>N}|X_n(omega)-X_m(omega)|<frac1k}= \=bigcaplimits_{kinmathbb{N}}bigcuplimits_{Ninmathbb{N}}bigcaplimits_{n,m>N}{|X_n(omega)-X_m(omega)|<frac1k}=\=bigcaplimits_{kinmathbb{N}}bigcuplimits_{Ninmathbb{N}}bigcaplimits_{n,m>N}big({X_n(omega)-X_m(omega)<frac1k}cap{X_n(omega)-X_m(omega)>-frac1k}big)inmathcal{F}$
probability probability-theory random-variables
probability probability-theory random-variables
asked Oct 7 at 19:13
Matt53
1227
asked Oct 7 at 19:13
Matt53
1227
edited Nov 25 at 10:54
asked Oct 7 at 19:13
Matt53
1227
asked Oct 7 at 19:13
Matt53
1227
asked Oct 7 at 19:13
Matt53
1227
1227
2
Can you show that $limsup_n X_n$ is measurable (and $liminf$)? If so, the set of points at which the limit exists can be written as the set of points satisfying $limsup_n X_n(omega) = liminf_n X_n(omega)$.
– copper.hat
Oct 7 at 19:17
Is ${omega|limsup_nX_n(omega)>a}=bigcuplimits_{n=1}^{infty}{omega|X_n(omega)>a}$ enough to show that $limsup_nX_n$ measureable?
– Matt53
Oct 7 at 19:29
1
That is not a correct equality.
– copper.hat
Oct 7 at 19:32
It should be ${omega|sup_{kgeq n}X_k(omega)>a}=bigcuplimits_{k=n}^{infty}{omega|X_k(omega)>a}$ right? And then by $limsup_nX_n=lim_nsup_{kgeq n}X_k$ we got that $limsup_nX_n$ is measureable.
– Matt53
Oct 7 at 20:55
add a comment |
2
Can you show that $limsup_n X_n$ is measurable (and $liminf$)? If so, the set of points at which the limit exists can be written as the set of points satisfying $limsup_n X_n(omega) = liminf_n X_n(omega)$.
– copper.hat
Oct 7 at 19:17
Is ${omega|limsup_nX_n(omega)>a}=bigcuplimits_{n=1}^{infty}{omega|X_n(omega)>a}$ enough to show that $limsup_nX_n$ measureable?
– Matt53
Oct 7 at 19:29
1
That is not a correct equality.
– copper.hat
Oct 7 at 19:32
It should be ${omega|sup_{kgeq n}X_k(omega)>a}=bigcuplimits_{k=n}^{infty}{omega|X_k(omega)>a}$ right? And then by $limsup_nX_n=lim_nsup_{kgeq n}X_k$ we got that $limsup_nX_n$ is measureable.
– Matt53
Oct 7 at 20:55
2
2
Can you show that $limsup_n X_n$ is measurable (and $liminf$)? If so, the set of points at which the limit exists can be written as the set of points satisfying $limsup_n X_n(omega) = liminf_n X_n(omega)$.
– copper.hat
Oct 7 at 19:17
Can you show that $limsup_n X_n$ is measurable (and $liminf$)? If so, the set of points at which the limit exists can be written as the set of points satisfying $limsup_n X_n(omega) = liminf_n X_n(omega)$.
– copper.hat
Oct 7 at 19:17
Is ${omega|limsup_nX_n(omega)>a}=bigcuplimits_{n=1}^{infty}{omega|X_n(omega)>a}$ enough to show that $limsup_nX_n$ measureable?
– Matt53
Oct 7 at 19:29
Is ${omega|limsup_nX_n(omega)>a}=bigcuplimits_{n=1}^{infty}{omega|X_n(omega)>a}$ enough to show that $limsup_nX_n$ measureable?
– Matt53
Oct 7 at 19:29
1
1
That is not a correct equality.
– copper.hat
Oct 7 at 19:32
That is not a correct equality.
– copper.hat
Oct 7 at 19:32
It should be ${omega|sup_{kgeq n}X_k(omega)>a}=bigcuplimits_{k=n}^{infty}{omega|X_k(omega)>a}$ right? And then by $limsup_nX_n=lim_nsup_{kgeq n}X_k$ we got that $limsup_nX_n$ is measureable.
– Matt53
Oct 7 at 20:55
It should be ${omega|sup_{kgeq n}X_k(omega)>a}=bigcuplimits_{k=n}^{infty}{omega|X_k(omega)>a}$ right? And then by $limsup_nX_n=lim_nsup_{kgeq n}X_k$ we got that $limsup_nX_n$ is measureable.
– Matt53
Oct 7 at 20:55
add a comment |
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2
Can you show that $limsup_n X_n$ is measurable (and $liminf$)? If so, the set of points at which the limit exists can be written as the set of points satisfying $limsup_n X_n(omega) = liminf_n X_n(omega)$.
– copper.hat
Oct 7 at 19:17
Is ${omega|limsup_nX_n(omega)>a}=bigcuplimits_{n=1}^{infty}{omega|X_n(omega)>a}$ enough to show that $limsup_nX_n$ measureable?
– Matt53
Oct 7 at 19:29
1
That is not a correct equality.
– copper.hat
Oct 7 at 19:32
It should be ${omega|sup_{kgeq n}X_k(omega)>a}=bigcuplimits_{k=n}^{infty}{omega|X_k(omega)>a}$ right? And then by $limsup_nX_n=lim_nsup_{kgeq n}X_k$ we got that $limsup_nX_n$ is measureable.
– Matt53
Oct 7 at 20:55