Mixing
$sum_{n=1}^{infty}p_{n}<infty iff X_{n} to 0$ a.s.
$begingroup$
Let $(p_{n})_{n}subset [0,1]$ and $(X_{n})_{n}$ independent random variables, so that $X_{n}$~ $Ber(p_{n})$
Prove that:
$sum_{n=1}^{infty}p_{n}<infty iff X_{n} to 0$ a.s
Ideas:
"$Rightarrow$"
$sum_{n=1}^{infty}P(X_{n}=1)=sum_{n=1}^{infty}p_{n}<infty$
I would attempt to use the Lemma of Borel-Cantelli, but have no idea how to use the independence aspect.
"$Leftarrow$"
$P(X_{n} to 0)=P(omegain Omega: exists N in mathbb N, X_{n}(omega)=0, forall n geq N)=1$. Let fix this $N$ accordingly, this then means
$sum_{n=1}^{infty}P(X_{n}=1)=sum_{n=1}^{N-1}P(X_{n}=1)+sum_{n=N}^{infty}P(X_{n}=1)=sum_{n=1}^{N-1}P(X_{n}=1)+0=sum_{n=1}^{N-1}p_{n}<infty$
da $p_{n} in [0,1], forall n in mathbb N$
probability-theory random-variables independence borel-cantelli-lemmas
asked Jan 6 at 23:36
SABOYSABOY
614311
$endgroup$
add a comment |
$begingroup$
Let $(p_{n})_{n}subset [0,1]$ and $(X_{n})_{n}$ independent random variables, so that $X_{n}$~ $Ber(p_{n})$
Prove that:
$sum_{n=1}^{infty}p_{n}<infty iff X_{n} to 0$ a.s
Ideas:
"$Rightarrow$"
$sum_{n=1}^{infty}P(X_{n}=1)=sum_{n=1}^{infty}p_{n}<infty$
I would attempt to use the Lemma of Borel-Cantelli, but have no idea how to use the independence aspect.
"$Leftarrow$"
$P(X_{n} to 0)=P(omegain Omega: exists N in mathbb N, X_{n}(omega)=0, forall n geq N)=1$. Let fix this $N$ accordingly, this then means
$sum_{n=1}^{infty}P(X_{n}=1)=sum_{n=1}^{N-1}P(X_{n}=1)+sum_{n=N}^{infty}P(X_{n}=1)=sum_{n=1}^{N-1}P(X_{n}=1)+0=sum_{n=1}^{N-1}p_{n}<infty$
da $p_{n} in [0,1], forall n in mathbb N$
probability-theory random-variables independence borel-cantelli-lemmas
asked Jan 6 at 23:36
SABOYSABOY
614311
$endgroup$
$begingroup$
Can you use the Kolmogorov 0-1 law?
$endgroup$
– ncmathsadist
Jan 6 at 23:38
$begingroup$
Yes, but why would the event $X_{n} to 0$ be asymptotic?
$endgroup$
– SABOY
Jan 6 at 23:42
add a comment |
$begingroup$
Let $(p_{n})_{n}subset [0,1]$ and $(X_{n})_{n}$ independent random variables, so that $X_{n}$~ $Ber(p_{n})$
Prove that:
$sum_{n=1}^{infty}p_{n}<infty iff X_{n} to 0$ a.s
Ideas:
"$Rightarrow$"
$sum_{n=1}^{infty}P(X_{n}=1)=sum_{n=1}^{infty}p_{n}<infty$
I would attempt to use the Lemma of Borel-Cantelli, but have no idea how to use the independence aspect.
"$Leftarrow$"
$P(X_{n} to 0)=P(omegain Omega: exists N in mathbb N, X_{n}(omega)=0, forall n geq N)=1$. Let fix this $N$ accordingly, this then means
$sum_{n=1}^{infty}P(X_{n}=1)=sum_{n=1}^{N-1}P(X_{n}=1)+sum_{n=N}^{infty}P(X_{n}=1)=sum_{n=1}^{N-1}P(X_{n}=1)+0=sum_{n=1}^{N-1}p_{n}<infty$
da $p_{n} in [0,1], forall n in mathbb N$
probability-theory random-variables independence borel-cantelli-lemmas
asked Jan 6 at 23:36
SABOYSABOY
614311
$endgroup$
Let $(p_{n})_{n}subset [0,1]$ and $(X_{n})_{n}$ independent random variables, so that $X_{n}$~ $Ber(p_{n})$
Prove that:
$sum_{n=1}^{infty}p_{n}<infty iff X_{n} to 0$ a.s
Ideas:
"$Rightarrow$"
$sum_{n=1}^{infty}P(X_{n}=1)=sum_{n=1}^{infty}p_{n}<infty$
I would attempt to use the Lemma of Borel-Cantelli, but have no idea how to use the independence aspect.
"$Leftarrow$"
$P(X_{n} to 0)=P(omegain Omega: exists N in mathbb N, X_{n}(omega)=0, forall n geq N)=1$. Let fix this $N$ accordingly, this then means
$sum_{n=1}^{infty}P(X_{n}=1)=sum_{n=1}^{N-1}P(X_{n}=1)+sum_{n=N}^{infty}P(X_{n}=1)=sum_{n=1}^{N-1}P(X_{n}=1)+0=sum_{n=1}^{N-1}p_{n}<infty$
da $p_{n} in [0,1], forall n in mathbb N$
probability-theory random-variables independence borel-cantelli-lemmas
probability-theory random-variables independence borel-cantelli-lemmas
asked Jan 6 at 23:36
SABOYSABOY
614311
asked Jan 6 at 23:36
SABOYSABOY
614311
asked Jan 6 at 23:36
SABOYSABOY
614311
asked Jan 6 at 23:36
SABOYSABOY
614311
asked Jan 6 at 23:36
SABOYSABOY
614311
614311
$begingroup$
Can you use the Kolmogorov 0-1 law?
$endgroup$
– ncmathsadist
Jan 6 at 23:38
$begingroup$
Yes, but why would the event $X_{n} to 0$ be asymptotic?
$endgroup$
– SABOY
Jan 6 at 23:42
add a comment |
$begingroup$
Can you use the Kolmogorov 0-1 law?
$endgroup$
– ncmathsadist
Jan 6 at 23:38
$begingroup$
Yes, but why would the event $X_{n} to 0$ be asymptotic?
$endgroup$
– SABOY
Jan 6 at 23:42
$begingroup$
Can you use the Kolmogorov 0-1 law?
$endgroup$
– ncmathsadist
Jan 6 at 23:38
$begingroup$
Can you use the Kolmogorov 0-1 law?
$endgroup$
– ncmathsadist
Jan 6 at 23:38
$begingroup$
Yes, but why would the event $X_{n} to 0$ be asymptotic?
$endgroup$
– SABOY
Jan 6 at 23:42
$begingroup$
Yes, but why would the event $X_{n} to 0$ be asymptotic?
$endgroup$
– SABOY
Jan 6 at 23:42
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that
$
X_nstackrel{text{a.s}}{to} 0
$
iff $P(|X_n|>varepsilon ; text{i.o})=0$ for every $varepsilon>0$. Since we are dealing with bernoulli random variables we only care about $0<varepsilon<1$. So fix $0<varepsilon<1$.
In this case, $sum_n P(|X_n|>varepsilon)=sum_n p_n$. By the first Borel Cantelli lemma, if $sum_n p_n<infty$, then $X_nto 0$ almost surely. By the second Borel cantelli lemma (the $X_n$ are independent), if $sum_n p_n=infty$, then $P(|X_n|>varepsilon ; text{i.o})=1$, so $X_nnotto0$ almost surely.
answered Jan 6 at 23:51
Foobaz JohnFoobaz John
21.9k41352
$endgroup$
$begingroup$
How is $X_{n} to 0$ the $lim sup $ of $|X_{n}|>epsilon$ ?
$endgroup$
– SABOY
Jan 7 at 0:19
$begingroup$
Oh wait $X_{n} to 0$ is the complement of $lim sup |X_{n}| > epsilon$
$endgroup$
– SABOY
Jan 7 at 0:22
add a comment |
$begingroup$
Since $X_n$'s are $0,1$ valued we see that $X_n to 0$ a.s iff $P{X_n=1 , text {i.o.}}=0$. By independence and Borel Cantelli this is so iff $sum P{X_n=1} <infty$ or $sum p_n<infty$
answered Jan 7 at 0:12
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that
$
X_nstackrel{text{a.s}}{to} 0
$
iff $P(|X_n|>varepsilon ; text{i.o})=0$ for every $varepsilon>0$. Since we are dealing with bernoulli random variables we only care about $0<varepsilon<1$. So fix $0<varepsilon<1$.
In this case, $sum_n P(|X_n|>varepsilon)=sum_n p_n$. By the first Borel Cantelli lemma, if $sum_n p_n<infty$, then $X_nto 0$ almost surely. By the second Borel cantelli lemma (the $X_n$ are independent), if $sum_n p_n=infty$, then $P(|X_n|>varepsilon ; text{i.o})=1$, so $X_nnotto0$ almost surely.
answered Jan 6 at 23:51
Foobaz JohnFoobaz John
21.9k41352
$endgroup$
$begingroup$
How is $X_{n} to 0$ the $lim sup $ of $|X_{n}|>epsilon$ ?
$endgroup$
– SABOY
Jan 7 at 0:19
$begingroup$
Oh wait $X_{n} to 0$ is the complement of $lim sup |X_{n}| > epsilon$
$endgroup$
– SABOY
Jan 7 at 0:22
add a comment |
$begingroup$
Note that
$
X_nstackrel{text{a.s}}{to} 0
$
iff $P(|X_n|>varepsilon ; text{i.o})=0$ for every $varepsilon>0$. Since we are dealing with bernoulli random variables we only care about $0<varepsilon<1$. So fix $0<varepsilon<1$.
In this case, $sum_n P(|X_n|>varepsilon)=sum_n p_n$. By the first Borel Cantelli lemma, if $sum_n p_n<infty$, then $X_nto 0$ almost surely. By the second Borel cantelli lemma (the $X_n$ are independent), if $sum_n p_n=infty$, then $P(|X_n|>varepsilon ; text{i.o})=1$, so $X_nnotto0$ almost surely.
answered Jan 6 at 23:51
Foobaz JohnFoobaz John
21.9k41352
$endgroup$
$begingroup$
How is $X_{n} to 0$ the $lim sup $ of $|X_{n}|>epsilon$ ?
$endgroup$
– SABOY
Jan 7 at 0:19
$begingroup$
Oh wait $X_{n} to 0$ is the complement of $lim sup |X_{n}| > epsilon$
$endgroup$
– SABOY
Jan 7 at 0:22
add a comment |
$begingroup$
Note that
$
X_nstackrel{text{a.s}}{to} 0
$
iff $P(|X_n|>varepsilon ; text{i.o})=0$ for every $varepsilon>0$. Since we are dealing with bernoulli random variables we only care about $0<varepsilon<1$. So fix $0<varepsilon<1$.
In this case, $sum_n P(|X_n|>varepsilon)=sum_n p_n$. By the first Borel Cantelli lemma, if $sum_n p_n<infty$, then $X_nto 0$ almost surely. By the second Borel cantelli lemma (the $X_n$ are independent), if $sum_n p_n=infty$, then $P(|X_n|>varepsilon ; text{i.o})=1$, so $X_nnotto0$ almost surely.
answered Jan 6 at 23:51
Foobaz JohnFoobaz John
21.9k41352
$endgroup$
Note that
$
X_nstackrel{text{a.s}}{to} 0
$
iff $P(|X_n|>varepsilon ; text{i.o})=0$ for every $varepsilon>0$. Since we are dealing with bernoulli random variables we only care about $0<varepsilon<1$. So fix $0<varepsilon<1$.
In this case, $sum_n P(|X_n|>varepsilon)=sum_n p_n$. By the first Borel Cantelli lemma, if $sum_n p_n<infty$, then $X_nto 0$ almost surely. By the second Borel cantelli lemma (the $X_n$ are independent), if $sum_n p_n=infty$, then $P(|X_n|>varepsilon ; text{i.o})=1$, so $X_nnotto0$ almost surely.
answered Jan 6 at 23:51
Foobaz JohnFoobaz John
21.9k41352
edited Jan 6 at 23:57
answered Jan 6 at 23:51
Foobaz JohnFoobaz John
21.9k41352
answered Jan 6 at 23:51
Foobaz JohnFoobaz John
21.9k41352
answered Jan 6 at 23:51
Foobaz JohnFoobaz John
21.9k41352
21.9k41352
$begingroup$
How is $X_{n} to 0$ the $lim sup $ of $|X_{n}|>epsilon$ ?
$endgroup$
– SABOY
Jan 7 at 0:19
$begingroup$
Oh wait $X_{n} to 0$ is the complement of $lim sup |X_{n}| > epsilon$
$endgroup$
– SABOY
Jan 7 at 0:22
add a comment |
$begingroup$
How is $X_{n} to 0$ the $lim sup $ of $|X_{n}|>epsilon$ ?
$endgroup$
– SABOY
Jan 7 at 0:19
$begingroup$
Oh wait $X_{n} to 0$ is the complement of $lim sup |X_{n}| > epsilon$
$endgroup$
– SABOY
Jan 7 at 0:22
$begingroup$
How is $X_{n} to 0$ the $lim sup $ of $|X_{n}|>epsilon$ ?
$endgroup$
– SABOY
Jan 7 at 0:19
$begingroup$
How is $X_{n} to 0$ the $lim sup $ of $|X_{n}|>epsilon$ ?
$endgroup$
– SABOY
Jan 7 at 0:19
$begingroup$
Oh wait $X_{n} to 0$ is the complement of $lim sup |X_{n}| > epsilon$
$endgroup$
– SABOY
Jan 7 at 0:22
$begingroup$
Oh wait $X_{n} to 0$ is the complement of $lim sup |X_{n}| > epsilon$
$endgroup$
– SABOY
Jan 7 at 0:22
add a comment |
$begingroup$
Since $X_n$'s are $0,1$ valued we see that $X_n to 0$ a.s iff $P{X_n=1 , text {i.o.}}=0$. By independence and Borel Cantelli this is so iff $sum P{X_n=1} <infty$ or $sum p_n<infty$
answered Jan 7 at 0:12
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
add a comment |
$begingroup$
Since $X_n$'s are $0,1$ valued we see that $X_n to 0$ a.s iff $P{X_n=1 , text {i.o.}}=0$. By independence and Borel Cantelli this is so iff $sum P{X_n=1} <infty$ or $sum p_n<infty$
answered Jan 7 at 0:12
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
add a comment |
$begingroup$
Since $X_n$'s are $0,1$ valued we see that $X_n to 0$ a.s iff $P{X_n=1 , text {i.o.}}=0$. By independence and Borel Cantelli this is so iff $sum P{X_n=1} <infty$ or $sum p_n<infty$
answered Jan 7 at 0:12
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
Since $X_n$'s are $0,1$ valued we see that $X_n to 0$ a.s iff $P{X_n=1 , text {i.o.}}=0$. By independence and Borel Cantelli this is so iff $sum P{X_n=1} <infty$ or $sum p_n<infty$
answered Jan 7 at 0:12
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Jan 7 at 0:12
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Jan 7 at 0:12
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Jan 7 at 0:12
Kavi Rama MurthyKavi Rama Murthy
56k42158
56k42158
add a comment |
add a comment |
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$begingroup$
Can you use the Kolmogorov 0-1 law?
$endgroup$
– ncmathsadist
Jan 6 at 23:38
$begingroup$
Yes, but why would the event $X_{n} to 0$ be asymptotic?
$endgroup$
– SABOY
Jan 6 at 23:42