The distribution of a stopping time
Let $(X_n)_{ngeq0}$ be a sequence of real $i.i.d$ random variables and $tau = inf{ngeq0 : X_nin S}$ with $S in mathcal{B}(mathbb{R}) $
I am trying to find $tau$'s distribution.
Obviously, $tau$ is a stopping time in regards to the natural filtration $sigma(X_0,...,X_n)$ but that's all I could come up with.
Any help would be greatly appreciated.
stochastic-processes martingales stopping-times
edited Dec 28 '18 at 22:38
aghostinthefigures
1,2301216
asked Dec 28 '18 at 19:00
Noah BishopNoah Bishop
494
add a comment |
Let $(X_n)_{ngeq0}$ be a sequence of real $i.i.d$ random variables and $tau = inf{ngeq0 : X_nin S}$ with $S in mathcal{B}(mathbb{R}) $
I am trying to find $tau$'s distribution.
Obviously, $tau$ is a stopping time in regards to the natural filtration $sigma(X_0,...,X_n)$ but that's all I could come up with.
Any help would be greatly appreciated.
stochastic-processes martingales stopping-times
edited Dec 28 '18 at 22:38
aghostinthefigures
1,2301216
asked Dec 28 '18 at 19:00
Noah BishopNoah Bishop
494
2
$tau$ is a geometric random variable, no?
– Mike Earnest
Dec 28 '18 at 19:27
1
so there's no further information on $X_n$ except it being real-valued and i.i.d. ?
– Hayk
Dec 28 '18 at 20:12
add a comment |
Let $(X_n)_{ngeq0}$ be a sequence of real $i.i.d$ random variables and $tau = inf{ngeq0 : X_nin S}$ with $S in mathcal{B}(mathbb{R}) $
I am trying to find $tau$'s distribution.
Obviously, $tau$ is a stopping time in regards to the natural filtration $sigma(X_0,...,X_n)$ but that's all I could come up with.
Any help would be greatly appreciated.
stochastic-processes martingales stopping-times
edited Dec 28 '18 at 22:38
aghostinthefigures
1,2301216
asked Dec 28 '18 at 19:00
Noah BishopNoah Bishop
494
Let $(X_n)_{ngeq0}$ be a sequence of real $i.i.d$ random variables and $tau = inf{ngeq0 : X_nin S}$ with $S in mathcal{B}(mathbb{R}) $
I am trying to find $tau$'s distribution.
Obviously, $tau$ is a stopping time in regards to the natural filtration $sigma(X_0,...,X_n)$ but that's all I could come up with.
Any help would be greatly appreciated.
stochastic-processes martingales stopping-times
stochastic-processes martingales stopping-times
edited Dec 28 '18 at 22:38
aghostinthefigures
1,2301216
asked Dec 28 '18 at 19:00
Noah BishopNoah Bishop
494
edited Dec 28 '18 at 22:38
aghostinthefigures
1,2301216
asked Dec 28 '18 at 19:00
Noah BishopNoah Bishop
494
edited Dec 28 '18 at 22:38
aghostinthefigures
1,2301216
edited Dec 28 '18 at 22:38
aghostinthefigures
1,2301216
edited Dec 28 '18 at 22:38
aghostinthefigures
1,2301216
1,2301216
asked Dec 28 '18 at 19:00
Noah BishopNoah Bishop
494
asked Dec 28 '18 at 19:00
Noah BishopNoah Bishop
494
asked Dec 28 '18 at 19:00
Noah BishopNoah Bishop
494
494
2
$tau$ is a geometric random variable, no?
– Mike Earnest
Dec 28 '18 at 19:27
1
so there's no further information on $X_n$ except it being real-valued and i.i.d. ?
– Hayk
Dec 28 '18 at 20:12
add a comment |
2
$tau$ is a geometric random variable, no?
– Mike Earnest
Dec 28 '18 at 19:27
1
so there's no further information on $X_n$ except it being real-valued and i.i.d. ?
– Hayk
Dec 28 '18 at 20:12
2
2
$tau$ is a geometric random variable, no?
– Mike Earnest
Dec 28 '18 at 19:27
$tau$ is a geometric random variable, no?
– Mike Earnest
Dec 28 '18 at 19:27
1
1
so there's no further information on $X_n$ except it being real-valued and i.i.d. ?
– Hayk
Dec 28 '18 at 20:12
so there's no further information on $X_n$ except it being real-valued and i.i.d. ?
– Hayk
Dec 28 '18 at 20:12
add a comment |
1 Answer
1
active
oldest
votes
As Mike pointed out, $tau$ is a geometric random variable. To see this, let $n geq 0$ then
$$P(tau = n) = P(X_0 notin S, X_1 notin S, ldots, X_{n-1} notin S, X_n in S) \ = P(X_0 notin S)P(X_1 notin S) times cdots times P(X_{n-1} notin S) , P(X_n in S) \ = (1-P(X_0 in S))^{n-1} , P(X_0 in S)$$
which proves that $tau$ is a geometric random variable with parameter $p = P(X_0 in S)$.
answered Dec 28 '18 at 20:25
MichhMichh
22316
add a comment |
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1 Answer
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As Mike pointed out, $tau$ is a geometric random variable. To see this, let $n geq 0$ then
$$P(tau = n) = P(X_0 notin S, X_1 notin S, ldots, X_{n-1} notin S, X_n in S) \ = P(X_0 notin S)P(X_1 notin S) times cdots times P(X_{n-1} notin S) , P(X_n in S) \ = (1-P(X_0 in S))^{n-1} , P(X_0 in S)$$
which proves that $tau$ is a geometric random variable with parameter $p = P(X_0 in S)$.
answered Dec 28 '18 at 20:25
MichhMichh
22316
add a comment |
As Mike pointed out, $tau$ is a geometric random variable. To see this, let $n geq 0$ then
$$P(tau = n) = P(X_0 notin S, X_1 notin S, ldots, X_{n-1} notin S, X_n in S) \ = P(X_0 notin S)P(X_1 notin S) times cdots times P(X_{n-1} notin S) , P(X_n in S) \ = (1-P(X_0 in S))^{n-1} , P(X_0 in S)$$
which proves that $tau$ is a geometric random variable with parameter $p = P(X_0 in S)$.
answered Dec 28 '18 at 20:25
MichhMichh
22316
add a comment |
As Mike pointed out, $tau$ is a geometric random variable. To see this, let $n geq 0$ then
$$P(tau = n) = P(X_0 notin S, X_1 notin S, ldots, X_{n-1} notin S, X_n in S) \ = P(X_0 notin S)P(X_1 notin S) times cdots times P(X_{n-1} notin S) , P(X_n in S) \ = (1-P(X_0 in S))^{n-1} , P(X_0 in S)$$
which proves that $tau$ is a geometric random variable with parameter $p = P(X_0 in S)$.
answered Dec 28 '18 at 20:25
MichhMichh
22316
As Mike pointed out, $tau$ is a geometric random variable. To see this, let $n geq 0$ then
$$P(tau = n) = P(X_0 notin S, X_1 notin S, ldots, X_{n-1} notin S, X_n in S) \ = P(X_0 notin S)P(X_1 notin S) times cdots times P(X_{n-1} notin S) , P(X_n in S) \ = (1-P(X_0 in S))^{n-1} , P(X_0 in S)$$
which proves that $tau$ is a geometric random variable with parameter $p = P(X_0 in S)$.
answered Dec 28 '18 at 20:25
MichhMichh
22316
answered Dec 28 '18 at 20:25
MichhMichh
22316
answered Dec 28 '18 at 20:25
MichhMichh
22316
answered Dec 28 '18 at 20:25
MichhMichh
22316
22316
add a comment |
add a comment |
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2
$tau$ is a geometric random variable, no?
– Mike Earnest
Dec 28 '18 at 19:27
1
so there's no further information on $X_n$ except it being real-valued and i.i.d. ?
– Hayk
Dec 28 '18 at 20:12