Second lemma of Borel-Cantelli: Normal distribution
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In my textbook is an example of an application of Borel-Cantelli's lemma which I don't understand.
Let $X_n$, $ngeq 1$, be a sequence of independent $N(0, sigma^2)$- distributed random variables, with $sigma > 0$. From the second lemma of Borel-Cantelli it follows that: P- almost surely (P-a.s.) $limsup_n X_n = infty$.
For sake of completeness: $N(0, sigma^2$) denotes the normal distribution with mean $=0$ and variance $sigma$. Our 2nd lemma of Borel Cantelli says:
Let $A_n$, $n geq 1$, be a sequence of independent events on a probability space. Then: $$sum_n P(A_n) = infty Rightarrow Pleft[limsup A_nright]=1.$$
What I don't see is why the normal distribution of $X_n$ implies that $sum_nP({X_n leq x}) = infty$ (or does it not?) and how from this we then can apply Borel Cantelli's lemma.
probability-theory random-variables limsup-and-liminf borel-cantelli-lemmas
edited Jan 22 at 9:44
Davide Giraudo
127k16151264
asked Sep 30 '17 at 10:45
QuasarQuasar
1278
$endgroup$
add a comment |
$begingroup$
In my textbook is an example of an application of Borel-Cantelli's lemma which I don't understand.
Let $X_n$, $ngeq 1$, be a sequence of independent $N(0, sigma^2)$- distributed random variables, with $sigma > 0$. From the second lemma of Borel-Cantelli it follows that: P- almost surely (P-a.s.) $limsup_n X_n = infty$.
For sake of completeness: $N(0, sigma^2$) denotes the normal distribution with mean $=0$ and variance $sigma$. Our 2nd lemma of Borel Cantelli says:
Let $A_n$, $n geq 1$, be a sequence of independent events on a probability space. Then: $$sum_n P(A_n) = infty Rightarrow Pleft[limsup A_nright]=1.$$
What I don't see is why the normal distribution of $X_n$ implies that $sum_nP({X_n leq x}) = infty$ (or does it not?) and how from this we then can apply Borel Cantelli's lemma.
probability-theory random-variables limsup-and-liminf borel-cantelli-lemmas
edited Jan 22 at 9:44
Davide Giraudo
127k16151264
asked Sep 30 '17 at 10:45
QuasarQuasar
1278
$endgroup$
1
$begingroup$
Note that $P(X_n)$ doesn't make any sense. $X_n$ is a random variable; $P(A)$ is only defined for events $A$.
$endgroup$
– saz
Sep 30 '17 at 11:00
$begingroup$
@saz You are right. I have corrected it.
$endgroup$
– Quasar
Oct 1 '17 at 8:33
$begingroup$
You are interested in $X_n$ having large values, right? So it would make more sense to study $sum_n P({X_n color{red}{geq} x})$ ... Use that the random variables are identically distributed!
$endgroup$
– saz
Oct 1 '17 at 8:43
add a comment |
$begingroup$
In my textbook is an example of an application of Borel-Cantelli's lemma which I don't understand.
Let $X_n$, $ngeq 1$, be a sequence of independent $N(0, sigma^2)$- distributed random variables, with $sigma > 0$. From the second lemma of Borel-Cantelli it follows that: P- almost surely (P-a.s.) $limsup_n X_n = infty$.
For sake of completeness: $N(0, sigma^2$) denotes the normal distribution with mean $=0$ and variance $sigma$. Our 2nd lemma of Borel Cantelli says:
Let $A_n$, $n geq 1$, be a sequence of independent events on a probability space. Then: $$sum_n P(A_n) = infty Rightarrow Pleft[limsup A_nright]=1.$$
What I don't see is why the normal distribution of $X_n$ implies that $sum_nP({X_n leq x}) = infty$ (or does it not?) and how from this we then can apply Borel Cantelli's lemma.
probability-theory random-variables limsup-and-liminf borel-cantelli-lemmas
edited Jan 22 at 9:44
Davide Giraudo
127k16151264
asked Sep 30 '17 at 10:45
QuasarQuasar
1278
$endgroup$
In my textbook is an example of an application of Borel-Cantelli's lemma which I don't understand.
Let $X_n$, $ngeq 1$, be a sequence of independent $N(0, sigma^2)$- distributed random variables, with $sigma > 0$. From the second lemma of Borel-Cantelli it follows that: P- almost surely (P-a.s.) $limsup_n X_n = infty$.
For sake of completeness: $N(0, sigma^2$) denotes the normal distribution with mean $=0$ and variance $sigma$. Our 2nd lemma of Borel Cantelli says:
Let $A_n$, $n geq 1$, be a sequence of independent events on a probability space. Then: $$sum_n P(A_n) = infty Rightarrow Pleft[limsup A_nright]=1.$$
What I don't see is why the normal distribution of $X_n$ implies that $sum_nP({X_n leq x}) = infty$ (or does it not?) and how from this we then can apply Borel Cantelli's lemma.
probability-theory random-variables limsup-and-liminf borel-cantelli-lemmas
probability-theory random-variables limsup-and-liminf borel-cantelli-lemmas
edited Jan 22 at 9:44
Davide Giraudo
127k16151264
asked Sep 30 '17 at 10:45
QuasarQuasar
1278
edited Jan 22 at 9:44
Davide Giraudo
127k16151264
asked Sep 30 '17 at 10:45
QuasarQuasar
1278
edited Jan 22 at 9:44
Davide Giraudo
127k16151264
edited Jan 22 at 9:44
Davide Giraudo
127k16151264
edited Jan 22 at 9:44
Davide Giraudo
127k16151264
127k16151264
asked Sep 30 '17 at 10:45
QuasarQuasar
1278
asked Sep 30 '17 at 10:45
QuasarQuasar
1278
asked Sep 30 '17 at 10:45
QuasarQuasar
1278
1278
1
$begingroup$
Note that $P(X_n)$ doesn't make any sense. $X_n$ is a random variable; $P(A)$ is only defined for events $A$.
$endgroup$
– saz
Sep 30 '17 at 11:00
$begingroup$
@saz You are right. I have corrected it.
$endgroup$
– Quasar
Oct 1 '17 at 8:33
$begingroup$
You are interested in $X_n$ having large values, right? So it would make more sense to study $sum_n P({X_n color{red}{geq} x})$ ... Use that the random variables are identically distributed!
$endgroup$
– saz
Oct 1 '17 at 8:43
add a comment |
1
$begingroup$
Note that $P(X_n)$ doesn't make any sense. $X_n$ is a random variable; $P(A)$ is only defined for events $A$.
$endgroup$
– saz
Sep 30 '17 at 11:00
$begingroup$
@saz You are right. I have corrected it.
$endgroup$
– Quasar
Oct 1 '17 at 8:33
$begingroup$
You are interested in $X_n$ having large values, right? So it would make more sense to study $sum_n P({X_n color{red}{geq} x})$ ... Use that the random variables are identically distributed!
$endgroup$
– saz
Oct 1 '17 at 8:43
1
1
$begingroup$
Note that $P(X_n)$ doesn't make any sense. $X_n$ is a random variable; $P(A)$ is only defined for events $A$.
$endgroup$
– saz
Sep 30 '17 at 11:00
$begingroup$
Note that $P(X_n)$ doesn't make any sense. $X_n$ is a random variable; $P(A)$ is only defined for events $A$.
$endgroup$
– saz
Sep 30 '17 at 11:00
$begingroup$
@saz You are right. I have corrected it.
$endgroup$
– Quasar
Oct 1 '17 at 8:33
$begingroup$
@saz You are right. I have corrected it.
$endgroup$
– Quasar
Oct 1 '17 at 8:33
$begingroup$
You are interested in $X_n$ having large values, right? So it would make more sense to study $sum_n P({X_n color{red}{geq} x})$ ... Use that the random variables are identically distributed!
$endgroup$
– saz
Oct 1 '17 at 8:43
$begingroup$
You are interested in $X_n$ having large values, right? So it would make more sense to study $sum_n P({X_n color{red}{geq} x})$ ... Use that the random variables are identically distributed!
$endgroup$
– saz
Oct 1 '17 at 8:43
add a comment |
1 Answer
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oldest
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Let us fix a positive integer $N$ and define the (independent) events $A_n:=left{X_ngeqslant Nright}$. Since the random variable $X_n$ has the same distribution as $X_1$, $mathbb Pleft(A_nright)=mathbb Pleft(A_1right)gt 0$, which implies that the series $sum_{ngeqslant 1}mathbb Pleft(A_nright)$ is divergent. By the second Borel-Cantelli lemma, $mathbb Pleft(limsup_{nto +infty}A_nright)=1$, which means that there exists a set $Omega_N$ of probability one such that for all $omegainOmega_N$, the set ${ninmathbb N, X_n(omega)geqslant N}$ is finite. This implies that
$$
forall omegainOmega_N, limsup_{nto +infty}X_n(omega)geqslant N.
$$
Let $Omega':=bigcap_{Ngeqslant 1}Omega_N$. Then $Omega'$ has probability one and $limsup_{nto +infty}X_n(omega)=+infty$ for all $omegainOmega'$.
Remark that we do not really need the $X_n$ to have a normal distribution: it suffices that the $X_n$ have the same distribution and that $mathbb Pleft(left{X_1geqslant Nright}right)gt 0$ for all $N$.
answered Jan 21 at 23:07
Davide GiraudoDavide Giraudo
127k16151264
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add a comment |
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$begingroup$
Let us fix a positive integer $N$ and define the (independent) events $A_n:=left{X_ngeqslant Nright}$. Since the random variable $X_n$ has the same distribution as $X_1$, $mathbb Pleft(A_nright)=mathbb Pleft(A_1right)gt 0$, which implies that the series $sum_{ngeqslant 1}mathbb Pleft(A_nright)$ is divergent. By the second Borel-Cantelli lemma, $mathbb Pleft(limsup_{nto +infty}A_nright)=1$, which means that there exists a set $Omega_N$ of probability one such that for all $omegainOmega_N$, the set ${ninmathbb N, X_n(omega)geqslant N}$ is finite. This implies that
$$
forall omegainOmega_N, limsup_{nto +infty}X_n(omega)geqslant N.
$$
Let $Omega':=bigcap_{Ngeqslant 1}Omega_N$. Then $Omega'$ has probability one and $limsup_{nto +infty}X_n(omega)=+infty$ for all $omegainOmega'$.
Remark that we do not really need the $X_n$ to have a normal distribution: it suffices that the $X_n$ have the same distribution and that $mathbb Pleft(left{X_1geqslant Nright}right)gt 0$ for all $N$.
answered Jan 21 at 23:07
Davide GiraudoDavide Giraudo
127k16151264
$endgroup$
add a comment |
$begingroup$
Let us fix a positive integer $N$ and define the (independent) events $A_n:=left{X_ngeqslant Nright}$. Since the random variable $X_n$ has the same distribution as $X_1$, $mathbb Pleft(A_nright)=mathbb Pleft(A_1right)gt 0$, which implies that the series $sum_{ngeqslant 1}mathbb Pleft(A_nright)$ is divergent. By the second Borel-Cantelli lemma, $mathbb Pleft(limsup_{nto +infty}A_nright)=1$, which means that there exists a set $Omega_N$ of probability one such that for all $omegainOmega_N$, the set ${ninmathbb N, X_n(omega)geqslant N}$ is finite. This implies that
$$
forall omegainOmega_N, limsup_{nto +infty}X_n(omega)geqslant N.
$$
Let $Omega':=bigcap_{Ngeqslant 1}Omega_N$. Then $Omega'$ has probability one and $limsup_{nto +infty}X_n(omega)=+infty$ for all $omegainOmega'$.
Remark that we do not really need the $X_n$ to have a normal distribution: it suffices that the $X_n$ have the same distribution and that $mathbb Pleft(left{X_1geqslant Nright}right)gt 0$ for all $N$.
answered Jan 21 at 23:07
Davide GiraudoDavide Giraudo
127k16151264
$endgroup$
add a comment |
$begingroup$
Let us fix a positive integer $N$ and define the (independent) events $A_n:=left{X_ngeqslant Nright}$. Since the random variable $X_n$ has the same distribution as $X_1$, $mathbb Pleft(A_nright)=mathbb Pleft(A_1right)gt 0$, which implies that the series $sum_{ngeqslant 1}mathbb Pleft(A_nright)$ is divergent. By the second Borel-Cantelli lemma, $mathbb Pleft(limsup_{nto +infty}A_nright)=1$, which means that there exists a set $Omega_N$ of probability one such that for all $omegainOmega_N$, the set ${ninmathbb N, X_n(omega)geqslant N}$ is finite. This implies that
$$
forall omegainOmega_N, limsup_{nto +infty}X_n(omega)geqslant N.
$$
Let $Omega':=bigcap_{Ngeqslant 1}Omega_N$. Then $Omega'$ has probability one and $limsup_{nto +infty}X_n(omega)=+infty$ for all $omegainOmega'$.
Remark that we do not really need the $X_n$ to have a normal distribution: it suffices that the $X_n$ have the same distribution and that $mathbb Pleft(left{X_1geqslant Nright}right)gt 0$ for all $N$.
answered Jan 21 at 23:07
Davide GiraudoDavide Giraudo
127k16151264
$endgroup$
Let us fix a positive integer $N$ and define the (independent) events $A_n:=left{X_ngeqslant Nright}$. Since the random variable $X_n$ has the same distribution as $X_1$, $mathbb Pleft(A_nright)=mathbb Pleft(A_1right)gt 0$, which implies that the series $sum_{ngeqslant 1}mathbb Pleft(A_nright)$ is divergent. By the second Borel-Cantelli lemma, $mathbb Pleft(limsup_{nto +infty}A_nright)=1$, which means that there exists a set $Omega_N$ of probability one such that for all $omegainOmega_N$, the set ${ninmathbb N, X_n(omega)geqslant N}$ is finite. This implies that
$$
forall omegainOmega_N, limsup_{nto +infty}X_n(omega)geqslant N.
$$
Let $Omega':=bigcap_{Ngeqslant 1}Omega_N$. Then $Omega'$ has probability one and $limsup_{nto +infty}X_n(omega)=+infty$ for all $omegainOmega'$.
Remark that we do not really need the $X_n$ to have a normal distribution: it suffices that the $X_n$ have the same distribution and that $mathbb Pleft(left{X_1geqslant Nright}right)gt 0$ for all $N$.
answered Jan 21 at 23:07
Davide GiraudoDavide Giraudo
127k16151264
edited Jan 22 at 9:46
answered Jan 21 at 23:07
Davide GiraudoDavide Giraudo
127k16151264
answered Jan 21 at 23:07
Davide GiraudoDavide Giraudo
127k16151264
answered Jan 21 at 23:07
Davide GiraudoDavide Giraudo
127k16151264
127k16151264
add a comment |
add a comment |
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$begingroup$
Note that $P(X_n)$ doesn't make any sense. $X_n$ is a random variable; $P(A)$ is only defined for events $A$.
$endgroup$
– saz
Sep 30 '17 at 11:00
$begingroup$
@saz You are right. I have corrected it.
$endgroup$
– Quasar
Oct 1 '17 at 8:33
$begingroup$
You are interested in $X_n$ having large values, right? So it would make more sense to study $sum_n P({X_n color{red}{geq} x})$ ... Use that the random variables are identically distributed!
$endgroup$
– saz
Oct 1 '17 at 8:43