limsup of a sequence of random variables (definition)convergence of sequence of random variables and cauchy sequencesConvergence of discrete random variables, show $fracS_nsqrtnto0$ a.s.Prove that limsup and liminf of an independent sequence are independent of finite number of termsSequence of Independent random variables$liminf$ and $limsup$ of events and random variablesIs it correct to say that ($colorred( limsup |W_k|/kcolorred) le 1) supseteq limsup colorred(|W_k|/k le 1colorred)$?Further clarification on the connection between limsup and liminf for sequences of sets and real numbers$limsup$ sequence independent $mathcalN(0,sigma^2)$soft question about a sequence of random variables that a.s. convergesInterchangeability of $limsup$
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limsup of a sequence of random variables (definition)
convergence of sequence of random variables and cauchy sequencesConvergence of discrete random variables, show $fracS_nsqrtnto0$ a.s.Prove that limsup and liminf of an independent sequence are independent of finite number of termsSequence of Independent random variables$liminf$ and $limsup$ of events and random variablesIs it correct to say that ($colorred( limsup |W_k|/kcolorred) le 1) supseteq limsup colorred(|W_k|/k le 1colorred)$?Further clarification on the connection between limsup and liminf for sequences of sets and real numbers$limsup$ sequence independent $mathcalN(0,sigma^2)$soft question about a sequence of random variables that a.s. convergesInterchangeability of $limsup$
$begingroup$
Let $X_n$ be a sequence of random variables.
First, $limsup X_n=inf_nsup_mge nX_m$. So,
$$limsup X_nle c=bigcup_nbigcap_mge nX_mle c$$
Is it correct to say that if $PX_nle c text i.o.equiv PlimsupX_nle c=1$ then $limsup X_nle c text a.s.$?
Edit 1:
By the same way
$$liminf X_nge c=bigcup_nbigcap_mge nX_mge c$$
So, $PX_nge c text i.o.=1 Rightarrow liminf X_nge c text a.s.$
Edit 2:
Following @Did's comments, the answer follows from
$$[X_ngeqslant c texti.o.]subseteq[limsup X_ngeqslant c] text and [X_nleqslant c texti.o.]subseteq[liminf X_nleqslant c]$$
probability-theory random-variables limsup-and-liminf
edited Nov 17 '15 at 16:06
BCLC
1
asked Oct 4 '14 at 8:22
d.k.o.d.k.o.
10.4k630
$endgroup$
|
show 3 more comments
$begingroup$
Let $X_n$ be a sequence of random variables.
First, $limsup X_n=inf_nsup_mge nX_m$. So,
$$limsup X_nle c=bigcup_nbigcap_mge nX_mle c$$
Is it correct to say that if $PX_nle c text i.o.equiv PlimsupX_nle c=1$ then $limsup X_nle c text a.s.$?
Edit 1:
By the same way
$$liminf X_nge c=bigcup_nbigcap_mge nX_mge c$$
So, $PX_nge c text i.o.=1 Rightarrow liminf X_nge c text a.s.$
Edit 2:
Following @Did's comments, the answer follows from
$$[X_ngeqslant c texti.o.]subseteq[limsup X_ngeqslant c] text and [X_nleqslant c texti.o.]subseteq[liminf X_nleqslant c]$$
probability-theory random-variables limsup-and-liminf
edited Nov 17 '15 at 16:06
BCLC
1
asked Oct 4 '14 at 8:22
d.k.o.d.k.o.
10.4k630
$endgroup$
$begingroup$
your question needs repairs.
$endgroup$
– drhab
Oct 4 '14 at 9:40
$begingroup$
@drhab pls explain...
$endgroup$
– d.k.o.
Oct 4 '14 at 9:55
$begingroup$
$Pleft X_ngeq c: i.oright Rightarrow$... How can a value on its own imply something?
$endgroup$
– drhab
Oct 4 '14 at 9:59
2
$begingroup$
The third line is wrong. In general, $$limsup X_nle cnebigcup_nbigcap_mge nX_mle c=liminfX_nle c.$$ (I corrected the typo $X_n$ for $X_m$.) It is not even always true that $$limsup X_nle c=limsupX_nle c.$$ For example, if $X_n(omega)=c+1/n$ for every $n$, then $limsup X_n(omega)=lim X_n(omega)=c$ hence $omega$ is in the event $limsup X_nle c$ but $X_n(omega)gt c$ for every $n$ hence $omega$ is neither in the event $limsupX_nle c$ nor in the event $liminfX_nle c$.
$endgroup$
– Did
Oct 4 '14 at 10:36
2
$begingroup$
$[X_ngeqslant c texti.o.]subseteq[limsup X_ngeqslant c]$ and $[X_nleqslant c texti.o.]subseteq[liminf X_nleqslant c]$ $.
$endgroup$
– Did
Oct 4 '14 at 20:06
|
show 3 more comments
$begingroup$
Let $X_n$ be a sequence of random variables.
First, $limsup X_n=inf_nsup_mge nX_m$. So,
$$limsup X_nle c=bigcup_nbigcap_mge nX_mle c$$
Is it correct to say that if $PX_nle c text i.o.equiv PlimsupX_nle c=1$ then $limsup X_nle c text a.s.$?
Edit 1:
By the same way
$$liminf X_nge c=bigcup_nbigcap_mge nX_mge c$$
So, $PX_nge c text i.o.=1 Rightarrow liminf X_nge c text a.s.$
Edit 2:
Following @Did's comments, the answer follows from
$$[X_ngeqslant c texti.o.]subseteq[limsup X_ngeqslant c] text and [X_nleqslant c texti.o.]subseteq[liminf X_nleqslant c]$$
probability-theory random-variables limsup-and-liminf
edited Nov 17 '15 at 16:06
BCLC
1
asked Oct 4 '14 at 8:22
d.k.o.d.k.o.
10.4k630
$endgroup$
Let $X_n$ be a sequence of random variables.
First, $limsup X_n=inf_nsup_mge nX_m$. So,
$$limsup X_nle c=bigcup_nbigcap_mge nX_mle c$$
Is it correct to say that if $PX_nle c text i.o.equiv PlimsupX_nle c=1$ then $limsup X_nle c text a.s.$?
Edit 1:
By the same way
$$liminf X_nge c=bigcup_nbigcap_mge nX_mge c$$
So, $PX_nge c text i.o.=1 Rightarrow liminf X_nge c text a.s.$
Edit 2:
Following @Did's comments, the answer follows from
$$[X_ngeqslant c texti.o.]subseteq[limsup X_ngeqslant c] text and [X_nleqslant c texti.o.]subseteq[liminf X_nleqslant c]$$
probability-theory random-variables limsup-and-liminf
probability-theory random-variables limsup-and-liminf
edited Nov 17 '15 at 16:06
BCLC
1
asked Oct 4 '14 at 8:22
d.k.o.d.k.o.
10.4k630
edited Nov 17 '15 at 16:06
BCLC
1
asked Oct 4 '14 at 8:22
d.k.o.d.k.o.
10.4k630
edited Nov 17 '15 at 16:06
BCLC
1
edited Nov 17 '15 at 16:06
BCLC
1
edited Nov 17 '15 at 16:06
BCLC
1
1
asked Oct 4 '14 at 8:22
d.k.o.d.k.o.
10.4k630
asked Oct 4 '14 at 8:22
d.k.o.d.k.o.
10.4k630
asked Oct 4 '14 at 8:22
d.k.o.d.k.o.
10.4k630
10.4k630
$begingroup$
your question needs repairs.
$endgroup$
– drhab
Oct 4 '14 at 9:40
$begingroup$
@drhab pls explain...
$endgroup$
– d.k.o.
Oct 4 '14 at 9:55
$begingroup$
$Pleft X_ngeq c: i.oright Rightarrow$... How can a value on its own imply something?
$endgroup$
– drhab
Oct 4 '14 at 9:59
2
$begingroup$
The third line is wrong. In general, $$limsup X_nle cnebigcup_nbigcap_mge nX_mle c=liminfX_nle c.$$ (I corrected the typo $X_n$ for $X_m$.) It is not even always true that $$limsup X_nle c=limsupX_nle c.$$ For example, if $X_n(omega)=c+1/n$ for every $n$, then $limsup X_n(omega)=lim X_n(omega)=c$ hence $omega$ is in the event $limsup X_nle c$ but $X_n(omega)gt c$ for every $n$ hence $omega$ is neither in the event $limsupX_nle c$ nor in the event $liminfX_nle c$.
$endgroup$
– Did
Oct 4 '14 at 10:36
2
$begingroup$
$[X_ngeqslant c texti.o.]subseteq[limsup X_ngeqslant c]$ and $[X_nleqslant c texti.o.]subseteq[liminf X_nleqslant c]$ $.
$endgroup$
– Did
Oct 4 '14 at 20:06
|
show 3 more comments
$begingroup$
your question needs repairs.
$endgroup$
– drhab
Oct 4 '14 at 9:40
$begingroup$
@drhab pls explain...
$endgroup$
– d.k.o.
Oct 4 '14 at 9:55
$begingroup$
$Pleft X_ngeq c: i.oright Rightarrow$... How can a value on its own imply something?
$endgroup$
– drhab
Oct 4 '14 at 9:59
2
$begingroup$
The third line is wrong. In general, $$limsup X_nle cnebigcup_nbigcap_mge nX_mle c=liminfX_nle c.$$ (I corrected the typo $X_n$ for $X_m$.) It is not even always true that $$limsup X_nle c=limsupX_nle c.$$ For example, if $X_n(omega)=c+1/n$ for every $n$, then $limsup X_n(omega)=lim X_n(omega)=c$ hence $omega$ is in the event $limsup X_nle c$ but $X_n(omega)gt c$ for every $n$ hence $omega$ is neither in the event $limsupX_nle c$ nor in the event $liminfX_nle c$.
$endgroup$
– Did
Oct 4 '14 at 10:36
2
$begingroup$
$[X_ngeqslant c texti.o.]subseteq[limsup X_ngeqslant c]$ and $[X_nleqslant c texti.o.]subseteq[liminf X_nleqslant c]$ $.
$endgroup$
– Did
Oct 4 '14 at 20:06
$begingroup$
your question needs repairs.
$endgroup$
– drhab
Oct 4 '14 at 9:40
$begingroup$
your question needs repairs.
$endgroup$
– drhab
Oct 4 '14 at 9:40
$begingroup$
@drhab pls explain...
$endgroup$
– d.k.o.
Oct 4 '14 at 9:55
$begingroup$
@drhab pls explain...
$endgroup$
– d.k.o.
Oct 4 '14 at 9:55
$begingroup$
$Pleft X_ngeq c: i.oright Rightarrow$... How can a value on its own imply something?
$endgroup$
– drhab
Oct 4 '14 at 9:59
$begingroup$
$Pleft X_ngeq c: i.oright Rightarrow$... How can a value on its own imply something?
$endgroup$
– drhab
Oct 4 '14 at 9:59
2
2
$begingroup$
The third line is wrong. In general, $$limsup X_nle cnebigcup_nbigcap_mge nX_mle c=liminfX_nle c.$$ (I corrected the typo $X_n$ for $X_m$.) It is not even always true that $$limsup X_nle c=limsupX_nle c.$$ For example, if $X_n(omega)=c+1/n$ for every $n$, then $limsup X_n(omega)=lim X_n(omega)=c$ hence $omega$ is in the event $limsup X_nle c$ but $X_n(omega)gt c$ for every $n$ hence $omega$ is neither in the event $limsupX_nle c$ nor in the event $liminfX_nle c$.
$endgroup$
– Did
Oct 4 '14 at 10:36
$begingroup$
The third line is wrong. In general, $$limsup X_nle cnebigcup_nbigcap_mge nX_mle c=liminfX_nle c.$$ (I corrected the typo $X_n$ for $X_m$.) It is not even always true that $$limsup X_nle c=limsupX_nle c.$$ For example, if $X_n(omega)=c+1/n$ for every $n$, then $limsup X_n(omega)=lim X_n(omega)=c$ hence $omega$ is in the event $limsup X_nle c$ but $X_n(omega)gt c$ for every $n$ hence $omega$ is neither in the event $limsupX_nle c$ nor in the event $liminfX_nle c$.
$endgroup$
– Did
Oct 4 '14 at 10:36
2
2
$begingroup$
$[X_ngeqslant c texti.o.]subseteq[limsup X_ngeqslant c]$ and $[X_nleqslant c texti.o.]subseteq[liminf X_nleqslant c]$ $.
$endgroup$
– Did
Oct 4 '14 at 20:06
$begingroup$
$[X_ngeqslant c texti.o.]subseteq[limsup X_ngeqslant c]$ and $[X_nleqslant c texti.o.]subseteq[liminf X_nleqslant c]$ $.
$endgroup$
– Did
Oct 4 '14 at 20:06
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
No.
Let $X_n=left(-1right)^n$ (constant random variables) so that: $$Pleft X_nleq0text i.o.right =1=Pleft X_ngeq0text i.o.right$$
However $limsup X_n=1$ and $liminf X_n=-1$ so that: $$Pleft limsup X_nleq0right =0=Pleft liminf X_ngeq0right$$
addendum:
Let $left(a_nright)$ be some sequence in $mathbbR$.
Then:
$$limsup a_n>cimplies a_n>ctext i.o.implieslimsup a_ngeq c$$
answered Oct 4 '14 at 10:12
drhabdrhab
103k545136
$endgroup$
$begingroup$
I think you made a mistake in the addendum. We have $1/n >0$ infinitely often, yet $limsup 1/n = 0$.
$endgroup$
– rabota
Mar 16 at 21:09
$begingroup$
@rabota You are correct. Thank you for attending me. I repaired.
$endgroup$
– drhab
Mar 16 at 22:05
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
No.
Let $X_n=left(-1right)^n$ (constant random variables) so that: $$Pleft X_nleq0text i.o.right =1=Pleft X_ngeq0text i.o.right$$
However $limsup X_n=1$ and $liminf X_n=-1$ so that: $$Pleft limsup X_nleq0right =0=Pleft liminf X_ngeq0right$$
addendum:
Let $left(a_nright)$ be some sequence in $mathbbR$.
Then:
$$limsup a_n>cimplies a_n>ctext i.o.implieslimsup a_ngeq c$$
answered Oct 4 '14 at 10:12
drhabdrhab
103k545136
$endgroup$
$begingroup$
I think you made a mistake in the addendum. We have $1/n >0$ infinitely often, yet $limsup 1/n = 0$.
$endgroup$
– rabota
Mar 16 at 21:09
$begingroup$
@rabota You are correct. Thank you for attending me. I repaired.
$endgroup$
– drhab
Mar 16 at 22:05
add a comment |
$begingroup$
No.
Let $X_n=left(-1right)^n$ (constant random variables) so that: $$Pleft X_nleq0text i.o.right =1=Pleft X_ngeq0text i.o.right$$
However $limsup X_n=1$ and $liminf X_n=-1$ so that: $$Pleft limsup X_nleq0right =0=Pleft liminf X_ngeq0right$$
addendum:
Let $left(a_nright)$ be some sequence in $mathbbR$.
Then:
$$limsup a_n>cimplies a_n>ctext i.o.implieslimsup a_ngeq c$$
answered Oct 4 '14 at 10:12
drhabdrhab
103k545136
$endgroup$
$begingroup$
I think you made a mistake in the addendum. We have $1/n >0$ infinitely often, yet $limsup 1/n = 0$.
$endgroup$
– rabota
Mar 16 at 21:09
$begingroup$
@rabota You are correct. Thank you for attending me. I repaired.
$endgroup$
– drhab
Mar 16 at 22:05
add a comment |
$begingroup$
No.
Let $X_n=left(-1right)^n$ (constant random variables) so that: $$Pleft X_nleq0text i.o.right =1=Pleft X_ngeq0text i.o.right$$
However $limsup X_n=1$ and $liminf X_n=-1$ so that: $$Pleft limsup X_nleq0right =0=Pleft liminf X_ngeq0right$$
addendum:
Let $left(a_nright)$ be some sequence in $mathbbR$.
Then:
$$limsup a_n>cimplies a_n>ctext i.o.implieslimsup a_ngeq c$$
answered Oct 4 '14 at 10:12
drhabdrhab
103k545136
$endgroup$
No.
Let $X_n=left(-1right)^n$ (constant random variables) so that: $$Pleft X_nleq0text i.o.right =1=Pleft X_ngeq0text i.o.right$$
However $limsup X_n=1$ and $liminf X_n=-1$ so that: $$Pleft limsup X_nleq0right =0=Pleft liminf X_ngeq0right$$
addendum:
Let $left(a_nright)$ be some sequence in $mathbbR$.
Then:
$$limsup a_n>cimplies a_n>ctext i.o.implieslimsup a_ngeq c$$
answered Oct 4 '14 at 10:12
drhabdrhab
103k545136
edited Mar 16 at 22:04
answered Oct 4 '14 at 10:12
drhabdrhab
103k545136
answered Oct 4 '14 at 10:12
drhabdrhab
103k545136
answered Oct 4 '14 at 10:12
drhabdrhab
103k545136
103k545136
$begingroup$
I think you made a mistake in the addendum. We have $1/n >0$ infinitely often, yet $limsup 1/n = 0$.
$endgroup$
– rabota
Mar 16 at 21:09
$begingroup$
@rabota You are correct. Thank you for attending me. I repaired.
$endgroup$
– drhab
Mar 16 at 22:05
add a comment |
$begingroup$
I think you made a mistake in the addendum. We have $1/n >0$ infinitely often, yet $limsup 1/n = 0$.
$endgroup$
– rabota
Mar 16 at 21:09
$begingroup$
@rabota You are correct. Thank you for attending me. I repaired.
$endgroup$
– drhab
Mar 16 at 22:05
$begingroup$
I think you made a mistake in the addendum. We have $1/n >0$ infinitely often, yet $limsup 1/n = 0$.
$endgroup$
– rabota
Mar 16 at 21:09
$begingroup$
I think you made a mistake in the addendum. We have $1/n >0$ infinitely often, yet $limsup 1/n = 0$.
$endgroup$
– rabota
Mar 16 at 21:09
$begingroup$
@rabota You are correct. Thank you for attending me. I repaired.
$endgroup$
– drhab
Mar 16 at 22:05
$begingroup$
@rabota You are correct. Thank you for attending me. I repaired.
$endgroup$
– drhab
Mar 16 at 22:05
add a comment |
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$begingroup$
your question needs repairs.
$endgroup$
– drhab
Oct 4 '14 at 9:40
$begingroup$
@drhab pls explain...
$endgroup$
– d.k.o.
Oct 4 '14 at 9:55
$begingroup$
$Pleft X_ngeq c: i.oright Rightarrow$... How can a value on its own imply something?
$endgroup$
– drhab
Oct 4 '14 at 9:59
2
$begingroup$
The third line is wrong. In general, $$limsup X_nle cnebigcup_nbigcap_mge nX_mle c=liminfX_nle c.$$ (I corrected the typo $X_n$ for $X_m$.) It is not even always true that $$limsup X_nle c=limsupX_nle c.$$ For example, if $X_n(omega)=c+1/n$ for every $n$, then $limsup X_n(omega)=lim X_n(omega)=c$ hence $omega$ is in the event $limsup X_nle c$ but $X_n(omega)gt c$ for every $n$ hence $omega$ is neither in the event $limsupX_nle c$ nor in the event $liminfX_nle c$.
$endgroup$
– Did
Oct 4 '14 at 10:36
2
$begingroup$
$[X_ngeqslant c texti.o.]subseteq[limsup X_ngeqslant c]$ and $[X_nleqslant c texti.o.]subseteq[liminf X_nleqslant c]$ $.
$endgroup$
– Did
Oct 4 '14 at 20:06