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Expected time until random walk with positive drift crosses a positive boundary
Clash Royale CLAN TAG#URR8PPP
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Let $x_1,x_2,...$ be i.i.d real-valued random variables with $mathbbE[x_i] > 0$. Let $S_n = sumlimits_i=1^nx_i$ be the partial sums of the r.v.s let and $N= inf n mid S_n > h$ be the first time the sum crosses the positive threshold $h$. My question has two related parts.
I can show that if the $x_i$ has bounded variance then $P(N < infty) = 1$. However, it is intuitive to me that this should be true even if the second moment of the random variable does not exist. So how can I prove that $P(N< infty)$ just assuming $mathbbE[|x_i|] < infty$? (of course still assuming positive drift)
I have seen it used in many papers that $mathbbE[N] < infty$. However, I can find no proof of this in any books that I looked at. I have seen proofs that make the additional assumption of a lower barrier, but they do not apply in my case (technically my lower barrier is $-infty$). How to prove that the expected time is finite?
probability-theory random-walk stopping-times
asked Jul 20 at 19:11
ttb
482211
add a comment |Â
up vote
0
down vote
favorite
Let $x_1,x_2,...$ be i.i.d real-valued random variables with $mathbbE[x_i] > 0$. Let $S_n = sumlimits_i=1^nx_i$ be the partial sums of the r.v.s let and $N= inf n mid S_n > h$ be the first time the sum crosses the positive threshold $h$. My question has two related parts.
I can show that if the $x_i$ has bounded variance then $P(N < infty) = 1$. However, it is intuitive to me that this should be true even if the second moment of the random variable does not exist. So how can I prove that $P(N< infty)$ just assuming $mathbbE[|x_i|] < infty$? (of course still assuming positive drift)
I have seen it used in many papers that $mathbbE[N] < infty$. However, I can find no proof of this in any books that I looked at. I have seen proofs that make the additional assumption of a lower barrier, but they do not apply in my case (technically my lower barrier is $-infty$). How to prove that the expected time is finite?
probability-theory random-walk stopping-times
asked Jul 20 at 19:11
ttb
482211
1
1. Law of large numbers. 2. LLN can be used too (a bit trickier).
– zhoraster
Jul 20 at 19:33
@zhoraster That's interesting, care to expand on how LLN can be used for #2? One is to show $P(S_n > h)$ decays sufficiently quickly, such as $O(1/n^2)$, but I don't know how to go about that.
– ttb
Jul 20 at 19:45
1
Loosely, let $$tau_k = infnge tau_k-1: S_n - S_tau_k-1 > h. $$ Then $tau_k$ is a random walk with positive jumps. If the jumps had infinite expectation, then $tau_n/nto infty$, $nto infty$. But this would contradict LLN for $S_n$ (as they would grow sublinearly).
– zhoraster
Jul 20 at 20:04
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $x_1,x_2,...$ be i.i.d real-valued random variables with $mathbbE[x_i] > 0$. Let $S_n = sumlimits_i=1^nx_i$ be the partial sums of the r.v.s let and $N= inf n mid S_n > h$ be the first time the sum crosses the positive threshold $h$. My question has two related parts.
I can show that if the $x_i$ has bounded variance then $P(N < infty) = 1$. However, it is intuitive to me that this should be true even if the second moment of the random variable does not exist. So how can I prove that $P(N< infty)$ just assuming $mathbbE[|x_i|] < infty$? (of course still assuming positive drift)
I have seen it used in many papers that $mathbbE[N] < infty$. However, I can find no proof of this in any books that I looked at. I have seen proofs that make the additional assumption of a lower barrier, but they do not apply in my case (technically my lower barrier is $-infty$). How to prove that the expected time is finite?
probability-theory random-walk stopping-times
asked Jul 20 at 19:11
ttb
482211
Let $x_1,x_2,...$ be i.i.d real-valued random variables with $mathbbE[x_i] > 0$. Let $S_n = sumlimits_i=1^nx_i$ be the partial sums of the r.v.s let and $N= inf n mid S_n > h$ be the first time the sum crosses the positive threshold $h$. My question has two related parts.
I can show that if the $x_i$ has bounded variance then $P(N < infty) = 1$. However, it is intuitive to me that this should be true even if the second moment of the random variable does not exist. So how can I prove that $P(N< infty)$ just assuming $mathbbE[|x_i|] < infty$? (of course still assuming positive drift)
I have seen it used in many papers that $mathbbE[N] < infty$. However, I can find no proof of this in any books that I looked at. I have seen proofs that make the additional assumption of a lower barrier, but they do not apply in my case (technically my lower barrier is $-infty$). How to prove that the expected time is finite?
probability-theory random-walk stopping-times
asked Jul 20 at 19:11
ttb
482211
asked Jul 20 at 19:11
ttb
482211
asked Jul 20 at 19:11
ttb
482211
asked Jul 20 at 19:11
ttb
482211
482211
1
1. Law of large numbers. 2. LLN can be used too (a bit trickier).
– zhoraster
Jul 20 at 19:33
@zhoraster That's interesting, care to expand on how LLN can be used for #2? One is to show $P(S_n > h)$ decays sufficiently quickly, such as $O(1/n^2)$, but I don't know how to go about that.
– ttb
Jul 20 at 19:45
1
Loosely, let $$tau_k = infnge tau_k-1: S_n - S_tau_k-1 > h. $$ Then $tau_k$ is a random walk with positive jumps. If the jumps had infinite expectation, then $tau_n/nto infty$, $nto infty$. But this would contradict LLN for $S_n$ (as they would grow sublinearly).
– zhoraster
Jul 20 at 20:04
add a comment |Â
1
1. Law of large numbers. 2. LLN can be used too (a bit trickier).
– zhoraster
Jul 20 at 19:33
@zhoraster That's interesting, care to expand on how LLN can be used for #2? One is to show $P(S_n > h)$ decays sufficiently quickly, such as $O(1/n^2)$, but I don't know how to go about that.
– ttb
Jul 20 at 19:45
1
Loosely, let $$tau_k = infnge tau_k-1: S_n - S_tau_k-1 > h. $$ Then $tau_k$ is a random walk with positive jumps. If the jumps had infinite expectation, then $tau_n/nto infty$, $nto infty$. But this would contradict LLN for $S_n$ (as they would grow sublinearly).
– zhoraster
Jul 20 at 20:04
1
1
1. Law of large numbers. 2. LLN can be used too (a bit trickier).
– zhoraster
Jul 20 at 19:33
1. Law of large numbers. 2. LLN can be used too (a bit trickier).
– zhoraster
Jul 20 at 19:33
@zhoraster That's interesting, care to expand on how LLN can be used for #2? One is to show $P(S_n > h)$ decays sufficiently quickly, such as $O(1/n^2)$, but I don't know how to go about that.
– ttb
Jul 20 at 19:45
@zhoraster That's interesting, care to expand on how LLN can be used for #2? One is to show $P(S_n > h)$ decays sufficiently quickly, such as $O(1/n^2)$, but I don't know how to go about that.
– ttb
Jul 20 at 19:45
1
1
Loosely, let $$tau_k = infnge tau_k-1: S_n - S_tau_k-1 > h. $$ Then $tau_k$ is a random walk with positive jumps. If the jumps had infinite expectation, then $tau_n/nto infty$, $nto infty$. But this would contradict LLN for $S_n$ (as they would grow sublinearly).
– zhoraster
Jul 20 at 20:04
Loosely, let $$tau_k = infnge tau_k-1: S_n - S_tau_k-1 > h. $$ Then $tau_k$ is a random walk with positive jumps. If the jumps had infinite expectation, then $tau_n/nto infty$, $nto infty$. But this would contradict LLN for $S_n$ (as they would grow sublinearly).
– zhoraster
Jul 20 at 20:04
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
By the strong law of large numbers $S_nto infty$ a.s. Thus, $mathsfP(N<infty)=1$ (for $h>0$). As for the expectation, for any $rge 1$, $mathsfEN^r<infty$ iff $mathsfE|x_1^-|^r<infty$ (see this paper for details).
answered Jul 20 at 19:37
d.k.o.
7,709526
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
By the strong law of large numbers $S_nto infty$ a.s. Thus, $mathsfP(N<infty)=1$ (for $h>0$). As for the expectation, for any $rge 1$, $mathsfEN^r<infty$ iff $mathsfE|x_1^-|^r<infty$ (see this paper for details).
answered Jul 20 at 19:37
d.k.o.
7,709526
add a comment |Â
up vote
1
down vote
accepted
By the strong law of large numbers $S_nto infty$ a.s. Thus, $mathsfP(N<infty)=1$ (for $h>0$). As for the expectation, for any $rge 1$, $mathsfEN^r<infty$ iff $mathsfE|x_1^-|^r<infty$ (see this paper for details).
answered Jul 20 at 19:37
d.k.o.
7,709526
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
By the strong law of large numbers $S_nto infty$ a.s. Thus, $mathsfP(N<infty)=1$ (for $h>0$). As for the expectation, for any $rge 1$, $mathsfEN^r<infty$ iff $mathsfE|x_1^-|^r<infty$ (see this paper for details).
answered Jul 20 at 19:37
d.k.o.
7,709526
By the strong law of large numbers $S_nto infty$ a.s. Thus, $mathsfP(N<infty)=1$ (for $h>0$). As for the expectation, for any $rge 1$, $mathsfEN^r<infty$ iff $mathsfE|x_1^-|^r<infty$ (see this paper for details).
answered Jul 20 at 19:37
d.k.o.
7,709526
answered Jul 20 at 19:37
d.k.o.
7,709526
answered Jul 20 at 19:37
d.k.o.
7,709526
answered Jul 20 at 19:37
d.k.o.
7,709526
7,709526
add a comment |Â
add a comment |Â
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1
1. Law of large numbers. 2. LLN can be used too (a bit trickier).
– zhoraster
Jul 20 at 19:33
@zhoraster That's interesting, care to expand on how LLN can be used for #2? One is to show $P(S_n > h)$ decays sufficiently quickly, such as $O(1/n^2)$, but I don't know how to go about that.
– ttb
Jul 20 at 19:45
1
Loosely, let $$tau_k = infnge tau_k-1: S_n - S_tau_k-1 > h. $$ Then $tau_k$ is a random walk with positive jumps. If the jumps had infinite expectation, then $tau_n/nto infty$, $nto infty$. But this would contradict LLN for $S_n$ (as they would grow sublinearly).
– zhoraster
Jul 20 at 20:04