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General random walk expected first exit time via martingale
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$x_t$ is a random walk with nonnegative integer time $t$ on the integer set $mathbf Z$ with transition probability $P(x_t+1mid x_t)=pmathbf 1(x_t+1-x_t=1)+qmathbf 1(x_t+1-x_t=-1)$, $p>0,,q>0$, $p+q=1$ and $pne q$. I know $y_t := big(frac qpbig)^x_t$ is a martingale. Given $x_0=0$, $tau:=mint:x_t=a>0,vee x_t=-b$ where $a,bin mathbf N$.
Can anyone remind me of the martingales that give the expected value and the variance of the first exit time $tau$, perhaps utilizing $y_t$? What is the relationship between the martingales and the probability generating function of the first exit time?
markov-chains martingales random-walk stopping-times
asked Jul 14 at 23:25
Hans
4,12821028
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$x_t$ is a random walk with nonnegative integer time $t$ on the integer set $mathbf Z$ with transition probability $P(x_t+1mid x_t)=pmathbf 1(x_t+1-x_t=1)+qmathbf 1(x_t+1-x_t=-1)$, $p>0,,q>0$, $p+q=1$ and $pne q$. I know $y_t := big(frac qpbig)^x_t$ is a martingale. Given $x_0=0$, $tau:=mint:x_t=a>0,vee x_t=-b$ where $a,bin mathbf N$.
Can anyone remind me of the martingales that give the expected value and the variance of the first exit time $tau$, perhaps utilizing $y_t$? What is the relationship between the martingales and the probability generating function of the first exit time?
markov-chains martingales random-walk stopping-times
asked Jul 14 at 23:25
Hans
4,12821028
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$x_t$ is a random walk with nonnegative integer time $t$ on the integer set $mathbf Z$ with transition probability $P(x_t+1mid x_t)=pmathbf 1(x_t+1-x_t=1)+qmathbf 1(x_t+1-x_t=-1)$, $p>0,,q>0$, $p+q=1$ and $pne q$. I know $y_t := big(frac qpbig)^x_t$ is a martingale. Given $x_0=0$, $tau:=mint:x_t=a>0,vee x_t=-b$ where $a,bin mathbf N$.
Can anyone remind me of the martingales that give the expected value and the variance of the first exit time $tau$, perhaps utilizing $y_t$? What is the relationship between the martingales and the probability generating function of the first exit time?
markov-chains martingales random-walk stopping-times
asked Jul 14 at 23:25
Hans
4,12821028
$x_t$ is a random walk with nonnegative integer time $t$ on the integer set $mathbf Z$ with transition probability $P(x_t+1mid x_t)=pmathbf 1(x_t+1-x_t=1)+qmathbf 1(x_t+1-x_t=-1)$, $p>0,,q>0$, $p+q=1$ and $pne q$. I know $y_t := big(frac qpbig)^x_t$ is a martingale. Given $x_0=0$, $tau:=mint:x_t=a>0,vee x_t=-b$ where $a,bin mathbf N$.
Can anyone remind me of the martingales that give the expected value and the variance of the first exit time $tau$, perhaps utilizing $y_t$? What is the relationship between the martingales and the probability generating function of the first exit time?
markov-chains martingales random-walk stopping-times
asked Jul 14 at 23:25
Hans
4,12821028
edited Jul 15 at 19:04
asked Jul 14 at 23:25
Hans
4,12821028
asked Jul 14 at 23:25
Hans
4,12821028
asked Jul 14 at 23:25
Hans
4,12821028
4,12821028
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add a comment |Â
1 Answer
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It is so very simple. The extra martingale is merely the natural "speed function" $z_t:=x_t-t(p-q)$. It is obvious that $z_t$ is a martingale. Now let $tau:=inft: x_tge a,wedge, x_tle -b $. By Doob's optional stopping theorem, $big(frac pqbig)^x_0=mathbf E[y_tau]=Pbig(frac pqbig)^a+Qbig(frac pqbig)^-b$, where $P=textProb(x_tau=a)$ and $P+Q=1$. We can solve for $P$ and $Q$. Applying again Doob's optional stopping theorem $x_0=mathbf E[z_tau]=aP-bQ-(p-q)mathbf E[tau]$. We can now solve for $mathbf E[tau]$.
answered Jul 20 at 5:20
Hans
4,12821028
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
It is so very simple. The extra martingale is merely the natural "speed function" $z_t:=x_t-t(p-q)$. It is obvious that $z_t$ is a martingale. Now let $tau:=inft: x_tge a,wedge, x_tle -b $. By Doob's optional stopping theorem, $big(frac pqbig)^x_0=mathbf E[y_tau]=Pbig(frac pqbig)^a+Qbig(frac pqbig)^-b$, where $P=textProb(x_tau=a)$ and $P+Q=1$. We can solve for $P$ and $Q$. Applying again Doob's optional stopping theorem $x_0=mathbf E[z_tau]=aP-bQ-(p-q)mathbf E[tau]$. We can now solve for $mathbf E[tau]$.
answered Jul 20 at 5:20
Hans
4,12821028
add a comment |Â
up vote
0
down vote
It is so very simple. The extra martingale is merely the natural "speed function" $z_t:=x_t-t(p-q)$. It is obvious that $z_t$ is a martingale. Now let $tau:=inft: x_tge a,wedge, x_tle -b $. By Doob's optional stopping theorem, $big(frac pqbig)^x_0=mathbf E[y_tau]=Pbig(frac pqbig)^a+Qbig(frac pqbig)^-b$, where $P=textProb(x_tau=a)$ and $P+Q=1$. We can solve for $P$ and $Q$. Applying again Doob's optional stopping theorem $x_0=mathbf E[z_tau]=aP-bQ-(p-q)mathbf E[tau]$. We can now solve for $mathbf E[tau]$.
answered Jul 20 at 5:20
Hans
4,12821028
add a comment |Â
up vote
0
down vote
up vote
0
down vote
It is so very simple. The extra martingale is merely the natural "speed function" $z_t:=x_t-t(p-q)$. It is obvious that $z_t$ is a martingale. Now let $tau:=inft: x_tge a,wedge, x_tle -b $. By Doob's optional stopping theorem, $big(frac pqbig)^x_0=mathbf E[y_tau]=Pbig(frac pqbig)^a+Qbig(frac pqbig)^-b$, where $P=textProb(x_tau=a)$ and $P+Q=1$. We can solve for $P$ and $Q$. Applying again Doob's optional stopping theorem $x_0=mathbf E[z_tau]=aP-bQ-(p-q)mathbf E[tau]$. We can now solve for $mathbf E[tau]$.
answered Jul 20 at 5:20
Hans
4,12821028
It is so very simple. The extra martingale is merely the natural "speed function" $z_t:=x_t-t(p-q)$. It is obvious that $z_t$ is a martingale. Now let $tau:=inft: x_tge a,wedge, x_tle -b $. By Doob's optional stopping theorem, $big(frac pqbig)^x_0=mathbf E[y_tau]=Pbig(frac pqbig)^a+Qbig(frac pqbig)^-b$, where $P=textProb(x_tau=a)$ and $P+Q=1$. We can solve for $P$ and $Q$. Applying again Doob's optional stopping theorem $x_0=mathbf E[z_tau]=aP-bQ-(p-q)mathbf E[tau]$. We can now solve for $mathbf E[tau]$.
answered Jul 20 at 5:20
Hans
4,12821028
answered Jul 20 at 5:20
Hans
4,12821028
answered Jul 20 at 5:20
Hans
4,12821028
answered Jul 20 at 5:20
Hans
4,12821028
4,12821028
add a comment |Â
add a comment |Â
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