Mixing
IID random variables $(X_n)$ have $sum e^{X_n} c^n < infty$ a.s.
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I'm working on the following exercise:
Let $X_1, X_2, ldots$ be i.i.d. nonnegative random variables. By virtue of the Borel-Cantelli lemma, show that for every $c in (0,1)$,
$$
sum_{n=1}^infty e^{X_n} c^n begin{cases}
< infty textrm{ a.s.} & textrm{if } mathbb E[X_1] < infty; \
= infty textrm{ a.s.} & textrm{if } mathbb E[X_1] = infty
end{cases}
$$
I'm trying to show $sum_{n=1}^infty mathbb Pleft[sum_{k=1}^n e^{X_k} c^k geq Mright] < infty$ for some large $M > 0$. For then, Borel-Cantelli gives us that $$ mathbb Pleft[limsup left{ sum_{k=1}^n e^{X_k} c^k geq Mright}right] = mathbb Pleft[sum_{k=1}^infty e^{X_k} c^k geq Mright] = 0$$ and we're done. But I don't know how to show $sum_{n=1}^infty mathbb Pleft[sum_{k=1}^n e^{X_k} c^k geq Mright] < infty$. Any suggestions?
probability probability-theory measure-theory convergence borel-cantelli-lemmas
asked 21 hours ago
D Ford
492212
add a comment |
up vote
2
down vote
favorite
I'm working on the following exercise:
Let $X_1, X_2, ldots$ be i.i.d. nonnegative random variables. By virtue of the Borel-Cantelli lemma, show that for every $c in (0,1)$,
$$
sum_{n=1}^infty e^{X_n} c^n begin{cases}
< infty textrm{ a.s.} & textrm{if } mathbb E[X_1] < infty; \
= infty textrm{ a.s.} & textrm{if } mathbb E[X_1] = infty
end{cases}
$$
I'm trying to show $sum_{n=1}^infty mathbb Pleft[sum_{k=1}^n e^{X_k} c^k geq Mright] < infty$ for some large $M > 0$. For then, Borel-Cantelli gives us that $$ mathbb Pleft[limsup left{ sum_{k=1}^n e^{X_k} c^k geq Mright}right] = mathbb Pleft[sum_{k=1}^infty e^{X_k} c^k geq Mright] = 0$$ and we're done. But I don't know how to show $sum_{n=1}^infty mathbb Pleft[sum_{k=1}^n e^{X_k} c^k geq Mright] < infty$. Any suggestions?
probability probability-theory measure-theory convergence borel-cantelli-lemmas
asked 21 hours ago
D Ford
492212
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm working on the following exercise:
Let $X_1, X_2, ldots$ be i.i.d. nonnegative random variables. By virtue of the Borel-Cantelli lemma, show that for every $c in (0,1)$,
$$
sum_{n=1}^infty e^{X_n} c^n begin{cases}
< infty textrm{ a.s.} & textrm{if } mathbb E[X_1] < infty; \
= infty textrm{ a.s.} & textrm{if } mathbb E[X_1] = infty
end{cases}
$$
I'm trying to show $sum_{n=1}^infty mathbb Pleft[sum_{k=1}^n e^{X_k} c^k geq Mright] < infty$ for some large $M > 0$. For then, Borel-Cantelli gives us that $$ mathbb Pleft[limsup left{ sum_{k=1}^n e^{X_k} c^k geq Mright}right] = mathbb Pleft[sum_{k=1}^infty e^{X_k} c^k geq Mright] = 0$$ and we're done. But I don't know how to show $sum_{n=1}^infty mathbb Pleft[sum_{k=1}^n e^{X_k} c^k geq Mright] < infty$. Any suggestions?
probability probability-theory measure-theory convergence borel-cantelli-lemmas
asked 21 hours ago
D Ford
492212
I'm working on the following exercise:
Let $X_1, X_2, ldots$ be i.i.d. nonnegative random variables. By virtue of the Borel-Cantelli lemma, show that for every $c in (0,1)$,
$$
sum_{n=1}^infty e^{X_n} c^n begin{cases}
< infty textrm{ a.s.} & textrm{if } mathbb E[X_1] < infty; \
= infty textrm{ a.s.} & textrm{if } mathbb E[X_1] = infty
end{cases}
$$
I'm trying to show $sum_{n=1}^infty mathbb Pleft[sum_{k=1}^n e^{X_k} c^k geq Mright] < infty$ for some large $M > 0$. For then, Borel-Cantelli gives us that $$ mathbb Pleft[limsup left{ sum_{k=1}^n e^{X_k} c^k geq Mright}right] = mathbb Pleft[sum_{k=1}^infty e^{X_k} c^k geq Mright] = 0$$ and we're done. But I don't know how to show $sum_{n=1}^infty mathbb Pleft[sum_{k=1}^n e^{X_k} c^k geq Mright] < infty$. Any suggestions?
probability probability-theory measure-theory convergence borel-cantelli-lemmas
probability probability-theory measure-theory convergence borel-cantelli-lemmas
asked 21 hours ago
D Ford
492212
asked 21 hours ago
D Ford
492212
asked 21 hours ago
D Ford
492212
asked 21 hours ago
D Ford
492212
asked 21 hours ago
D Ford
492212
492212
add a comment |
add a comment |
1 Answer
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0
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If $mathbb{E}(X_1)< infty$ then it follows from the strong law of large numbers that $S_n := sum_{j=1}^n X_j$ satisfies
$$lim_{n to infty} frac{S_n}{n} = mathbb{E}(X_1) quad text{a.s.};$$
hence
$$lim_{n to infty} frac{X_n}{n} = lim_{n to infty} left( frac{S_n}{n} - frac{S_{n-1}}{n} right)=0 quad text{a.s.}$$
Consequently, there exists for almost all $omega in Omega$ some $N in mathbb{N}$ such that $$left| frac{X_n(omega)}{n} right| leq -log(sqrt{c}) quad text{for all $n geq N$}$$ and so $$sum_{n geq N} e^{X_n(omega)} c^n leq sum_{n geq N} sqrt{c}^n < infty.$$
If $mathbb{E}(X_1)=infty$ then
$$sum_{n geq 1} mathbb{P}(X_n geq n)=sum_{n geq 1} mathbb{P}(X_1 geq n) =infty,$$
and therefore it follows from the Borel Cantelli lemma that $$mathbb{P}(X_n geq n , , text{infinitely often})=1,$$ i.e. $$e^{X_n} geq e^n quad text{for infinitely many $n$ with probability 1.}$$ This implies $sum_{n geq 1} e^{X_n} c^n = infty$ almost surely for $c:= 1/e$.
answered 21 hours ago
saz
76.3k755117
1
For the fact that $mathbb{E}[X_1] < infty$ implies $X_n/n to 0$ a.s., it may as well be proved by Borel-Cantelli: $$forall epsilon > 0 : quad sum_{n=1}^{infty} mathbb{P}[X_n > epsilon n] leq frac{1}{epsilon}mathbb{E}[X_1] < infty. $$ I am very certain that you already know this, but just wanted to mention in case OP needs an alternative approach.
– Sangchul Lee
18 hours ago
@SangchulLee Yes of course; thanks for your comment.
– saz
18 hours ago
2
Sorry but $0<c<1$ hence $|log(c)|=-log c$ and $$left| frac{X_n(omega)}{n} right| leq |log(c)|$$ implies $$e^{X_n(omega)} c^n leq e^{-nlog c}c^n=1$$ and not what you write. I am afraid the intermediate step you need is rather something like $$frac{X_n(omega)}{n}le-frac12log c$$
– Did
18 hours ago
@Did Ah right, I messed that up; thanks a lot for your comment.
– saz
9 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If $mathbb{E}(X_1)< infty$ then it follows from the strong law of large numbers that $S_n := sum_{j=1}^n X_j$ satisfies
$$lim_{n to infty} frac{S_n}{n} = mathbb{E}(X_1) quad text{a.s.};$$
hence
$$lim_{n to infty} frac{X_n}{n} = lim_{n to infty} left( frac{S_n}{n} - frac{S_{n-1}}{n} right)=0 quad text{a.s.}$$
Consequently, there exists for almost all $omega in Omega$ some $N in mathbb{N}$ such that $$left| frac{X_n(omega)}{n} right| leq -log(sqrt{c}) quad text{for all $n geq N$}$$ and so $$sum_{n geq N} e^{X_n(omega)} c^n leq sum_{n geq N} sqrt{c}^n < infty.$$
If $mathbb{E}(X_1)=infty$ then
$$sum_{n geq 1} mathbb{P}(X_n geq n)=sum_{n geq 1} mathbb{P}(X_1 geq n) =infty,$$
and therefore it follows from the Borel Cantelli lemma that $$mathbb{P}(X_n geq n , , text{infinitely often})=1,$$ i.e. $$e^{X_n} geq e^n quad text{for infinitely many $n$ with probability 1.}$$ This implies $sum_{n geq 1} e^{X_n} c^n = infty$ almost surely for $c:= 1/e$.
answered 21 hours ago
saz
76.3k755117
1
For the fact that $mathbb{E}[X_1] < infty$ implies $X_n/n to 0$ a.s., it may as well be proved by Borel-Cantelli: $$forall epsilon > 0 : quad sum_{n=1}^{infty} mathbb{P}[X_n > epsilon n] leq frac{1}{epsilon}mathbb{E}[X_1] < infty. $$ I am very certain that you already know this, but just wanted to mention in case OP needs an alternative approach.
– Sangchul Lee
18 hours ago
@SangchulLee Yes of course; thanks for your comment.
– saz
18 hours ago
2
Sorry but $0<c<1$ hence $|log(c)|=-log c$ and $$left| frac{X_n(omega)}{n} right| leq |log(c)|$$ implies $$e^{X_n(omega)} c^n leq e^{-nlog c}c^n=1$$ and not what you write. I am afraid the intermediate step you need is rather something like $$frac{X_n(omega)}{n}le-frac12log c$$
– Did
18 hours ago
@Did Ah right, I messed that up; thanks a lot for your comment.
– saz
9 hours ago
add a comment |
up vote
0
down vote
If $mathbb{E}(X_1)< infty$ then it follows from the strong law of large numbers that $S_n := sum_{j=1}^n X_j$ satisfies
$$lim_{n to infty} frac{S_n}{n} = mathbb{E}(X_1) quad text{a.s.};$$
hence
$$lim_{n to infty} frac{X_n}{n} = lim_{n to infty} left( frac{S_n}{n} - frac{S_{n-1}}{n} right)=0 quad text{a.s.}$$
Consequently, there exists for almost all $omega in Omega$ some $N in mathbb{N}$ such that $$left| frac{X_n(omega)}{n} right| leq -log(sqrt{c}) quad text{for all $n geq N$}$$ and so $$sum_{n geq N} e^{X_n(omega)} c^n leq sum_{n geq N} sqrt{c}^n < infty.$$
If $mathbb{E}(X_1)=infty$ then
$$sum_{n geq 1} mathbb{P}(X_n geq n)=sum_{n geq 1} mathbb{P}(X_1 geq n) =infty,$$
and therefore it follows from the Borel Cantelli lemma that $$mathbb{P}(X_n geq n , , text{infinitely often})=1,$$ i.e. $$e^{X_n} geq e^n quad text{for infinitely many $n$ with probability 1.}$$ This implies $sum_{n geq 1} e^{X_n} c^n = infty$ almost surely for $c:= 1/e$.
answered 21 hours ago
saz
76.3k755117
1
For the fact that $mathbb{E}[X_1] < infty$ implies $X_n/n to 0$ a.s., it may as well be proved by Borel-Cantelli: $$forall epsilon > 0 : quad sum_{n=1}^{infty} mathbb{P}[X_n > epsilon n] leq frac{1}{epsilon}mathbb{E}[X_1] < infty. $$ I am very certain that you already know this, but just wanted to mention in case OP needs an alternative approach.
– Sangchul Lee
18 hours ago
@SangchulLee Yes of course; thanks for your comment.
– saz
18 hours ago
2
Sorry but $0<c<1$ hence $|log(c)|=-log c$ and $$left| frac{X_n(omega)}{n} right| leq |log(c)|$$ implies $$e^{X_n(omega)} c^n leq e^{-nlog c}c^n=1$$ and not what you write. I am afraid the intermediate step you need is rather something like $$frac{X_n(omega)}{n}le-frac12log c$$
– Did
18 hours ago
@Did Ah right, I messed that up; thanks a lot for your comment.
– saz
9 hours ago
add a comment |
up vote
0
down vote
up vote
0
down vote
If $mathbb{E}(X_1)< infty$ then it follows from the strong law of large numbers that $S_n := sum_{j=1}^n X_j$ satisfies
$$lim_{n to infty} frac{S_n}{n} = mathbb{E}(X_1) quad text{a.s.};$$
hence
$$lim_{n to infty} frac{X_n}{n} = lim_{n to infty} left( frac{S_n}{n} - frac{S_{n-1}}{n} right)=0 quad text{a.s.}$$
Consequently, there exists for almost all $omega in Omega$ some $N in mathbb{N}$ such that $$left| frac{X_n(omega)}{n} right| leq -log(sqrt{c}) quad text{for all $n geq N$}$$ and so $$sum_{n geq N} e^{X_n(omega)} c^n leq sum_{n geq N} sqrt{c}^n < infty.$$
If $mathbb{E}(X_1)=infty$ then
$$sum_{n geq 1} mathbb{P}(X_n geq n)=sum_{n geq 1} mathbb{P}(X_1 geq n) =infty,$$
and therefore it follows from the Borel Cantelli lemma that $$mathbb{P}(X_n geq n , , text{infinitely often})=1,$$ i.e. $$e^{X_n} geq e^n quad text{for infinitely many $n$ with probability 1.}$$ This implies $sum_{n geq 1} e^{X_n} c^n = infty$ almost surely for $c:= 1/e$.
answered 21 hours ago
saz
76.3k755117
If $mathbb{E}(X_1)< infty$ then it follows from the strong law of large numbers that $S_n := sum_{j=1}^n X_j$ satisfies
$$lim_{n to infty} frac{S_n}{n} = mathbb{E}(X_1) quad text{a.s.};$$
hence
$$lim_{n to infty} frac{X_n}{n} = lim_{n to infty} left( frac{S_n}{n} - frac{S_{n-1}}{n} right)=0 quad text{a.s.}$$
Consequently, there exists for almost all $omega in Omega$ some $N in mathbb{N}$ such that $$left| frac{X_n(omega)}{n} right| leq -log(sqrt{c}) quad text{for all $n geq N$}$$ and so $$sum_{n geq N} e^{X_n(omega)} c^n leq sum_{n geq N} sqrt{c}^n < infty.$$
If $mathbb{E}(X_1)=infty$ then
$$sum_{n geq 1} mathbb{P}(X_n geq n)=sum_{n geq 1} mathbb{P}(X_1 geq n) =infty,$$
and therefore it follows from the Borel Cantelli lemma that $$mathbb{P}(X_n geq n , , text{infinitely often})=1,$$ i.e. $$e^{X_n} geq e^n quad text{for infinitely many $n$ with probability 1.}$$ This implies $sum_{n geq 1} e^{X_n} c^n = infty$ almost surely for $c:= 1/e$.
answered 21 hours ago
saz
76.3k755117
edited 9 hours ago
answered 21 hours ago
saz
76.3k755117
answered 21 hours ago
saz
76.3k755117
answered 21 hours ago
saz
76.3k755117
76.3k755117
1
For the fact that $mathbb{E}[X_1] < infty$ implies $X_n/n to 0$ a.s., it may as well be proved by Borel-Cantelli: $$forall epsilon > 0 : quad sum_{n=1}^{infty} mathbb{P}[X_n > epsilon n] leq frac{1}{epsilon}mathbb{E}[X_1] < infty. $$ I am very certain that you already know this, but just wanted to mention in case OP needs an alternative approach.
– Sangchul Lee
18 hours ago
@SangchulLee Yes of course; thanks for your comment.
– saz
18 hours ago
2
Sorry but $0<c<1$ hence $|log(c)|=-log c$ and $$left| frac{X_n(omega)}{n} right| leq |log(c)|$$ implies $$e^{X_n(omega)} c^n leq e^{-nlog c}c^n=1$$ and not what you write. I am afraid the intermediate step you need is rather something like $$frac{X_n(omega)}{n}le-frac12log c$$
– Did
18 hours ago
@Did Ah right, I messed that up; thanks a lot for your comment.
– saz
9 hours ago
add a comment |
1
For the fact that $mathbb{E}[X_1] < infty$ implies $X_n/n to 0$ a.s., it may as well be proved by Borel-Cantelli: $$forall epsilon > 0 : quad sum_{n=1}^{infty} mathbb{P}[X_n > epsilon n] leq frac{1}{epsilon}mathbb{E}[X_1] < infty. $$ I am very certain that you already know this, but just wanted to mention in case OP needs an alternative approach.
– Sangchul Lee
18 hours ago
@SangchulLee Yes of course; thanks for your comment.
– saz
18 hours ago
2
Sorry but $0<c<1$ hence $|log(c)|=-log c$ and $$left| frac{X_n(omega)}{n} right| leq |log(c)|$$ implies $$e^{X_n(omega)} c^n leq e^{-nlog c}c^n=1$$ and not what you write. I am afraid the intermediate step you need is rather something like $$frac{X_n(omega)}{n}le-frac12log c$$
– Did
18 hours ago
@Did Ah right, I messed that up; thanks a lot for your comment.
– saz
9 hours ago
1
1
For the fact that $mathbb{E}[X_1] < infty$ implies $X_n/n to 0$ a.s., it may as well be proved by Borel-Cantelli: $$forall epsilon > 0 : quad sum_{n=1}^{infty} mathbb{P}[X_n > epsilon n] leq frac{1}{epsilon}mathbb{E}[X_1] < infty. $$ I am very certain that you already know this, but just wanted to mention in case OP needs an alternative approach.
– Sangchul Lee
18 hours ago
For the fact that $mathbb{E}[X_1] < infty$ implies $X_n/n to 0$ a.s., it may as well be proved by Borel-Cantelli: $$forall epsilon > 0 : quad sum_{n=1}^{infty} mathbb{P}[X_n > epsilon n] leq frac{1}{epsilon}mathbb{E}[X_1] < infty. $$ I am very certain that you already know this, but just wanted to mention in case OP needs an alternative approach.
– Sangchul Lee
18 hours ago
@SangchulLee Yes of course; thanks for your comment.
– saz
18 hours ago
@SangchulLee Yes of course; thanks for your comment.
– saz
18 hours ago
2
2
Sorry but $0<c<1$ hence $|log(c)|=-log c$ and $$left| frac{X_n(omega)}{n} right| leq |log(c)|$$ implies $$e^{X_n(omega)} c^n leq e^{-nlog c}c^n=1$$ and not what you write. I am afraid the intermediate step you need is rather something like $$frac{X_n(omega)}{n}le-frac12log c$$
– Did
18 hours ago
Sorry but $0<c<1$ hence $|log(c)|=-log c$ and $$left| frac{X_n(omega)}{n} right| leq |log(c)|$$ implies $$e^{X_n(omega)} c^n leq e^{-nlog c}c^n=1$$ and not what you write. I am afraid the intermediate step you need is rather something like $$frac{X_n(omega)}{n}le-frac12log c$$
– Did
18 hours ago
@Did Ah right, I messed that up; thanks a lot for your comment.
– saz
9 hours ago
@Did Ah right, I messed that up; thanks a lot for your comment.
– saz
9 hours ago
add a comment |
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