A hitting time bigger than another hitting time
My question is :
Show that if $tau'$ is another hitting time for the filtration $mathcal{F}_n$, then
$$tau := inf{ngeqtau' : X_n in B, mbox{where $B$ is a Borel set}}$$
is also a hitting time.
Does someone know how to prove this ?
Thank you.
measure-theory stopping-times
edited Nov 30 '18 at 13:30
saz
78.5k758123
asked Nov 30 '18 at 2:05
chadichadi
192
add a comment |
My question is :
Show that if $tau'$ is another hitting time for the filtration $mathcal{F}_n$, then
$$tau := inf{ngeqtau' : X_n in B, mbox{where $B$ is a Borel set}}$$
is also a hitting time.
Does someone know how to prove this ?
Thank you.
measure-theory stopping-times
edited Nov 30 '18 at 13:30
saz
78.5k758123
asked Nov 30 '18 at 2:05
chadichadi
192
add a comment |
My question is :
Show that if $tau'$ is another hitting time for the filtration $mathcal{F}_n$, then
$$tau := inf{ngeqtau' : X_n in B, mbox{where $B$ is a Borel set}}$$
is also a hitting time.
Does someone know how to prove this ?
Thank you.
measure-theory stopping-times
edited Nov 30 '18 at 13:30
saz
78.5k758123
asked Nov 30 '18 at 2:05
chadichadi
192
My question is :
Show that if $tau'$ is another hitting time for the filtration $mathcal{F}_n$, then
$$tau := inf{ngeqtau' : X_n in B, mbox{where $B$ is a Borel set}}$$
is also a hitting time.
Does someone know how to prove this ?
Thank you.
measure-theory stopping-times
measure-theory stopping-times
edited Nov 30 '18 at 13:30
saz
78.5k758123
asked Nov 30 '18 at 2:05
chadichadi
192
edited Nov 30 '18 at 13:30
saz
78.5k758123
asked Nov 30 '18 at 2:05
chadichadi
192
edited Nov 30 '18 at 13:30
saz
78.5k758123
edited Nov 30 '18 at 13:30
saz
78.5k758123
edited Nov 30 '18 at 13:30
saz
78.5k758123
78.5k758123
asked Nov 30 '18 at 2:05
chadichadi
192
asked Nov 30 '18 at 2:05
chadichadi
192
asked Nov 30 '18 at 2:05
chadichadi
192
192
add a comment |
add a comment |
1 Answer
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Hint: $${tau leq k} = {tau' leq tau leq k} = bigcup_{j=0}^k ({tau'=j} cap {j leq tau leq k})$$
and so
$${tau leq k} = bigcup_{j=0}^k left( {tau'=j} capbigcup_{i=j}^k {X_i in B} right) in mathcal{F}_k$$
answered Nov 30 '18 at 6:59
sazsaz
78.5k758123
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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Hint: $${tau leq k} = {tau' leq tau leq k} = bigcup_{j=0}^k ({tau'=j} cap {j leq tau leq k})$$
and so
$${tau leq k} = bigcup_{j=0}^k left( {tau'=j} capbigcup_{i=j}^k {X_i in B} right) in mathcal{F}_k$$
answered Nov 30 '18 at 6:59
sazsaz
78.5k758123
add a comment |
Hint: $${tau leq k} = {tau' leq tau leq k} = bigcup_{j=0}^k ({tau'=j} cap {j leq tau leq k})$$
and so
$${tau leq k} = bigcup_{j=0}^k left( {tau'=j} capbigcup_{i=j}^k {X_i in B} right) in mathcal{F}_k$$
answered Nov 30 '18 at 6:59
sazsaz
78.5k758123
add a comment |
Hint: $${tau leq k} = {tau' leq tau leq k} = bigcup_{j=0}^k ({tau'=j} cap {j leq tau leq k})$$
and so
$${tau leq k} = bigcup_{j=0}^k left( {tau'=j} capbigcup_{i=j}^k {X_i in B} right) in mathcal{F}_k$$
answered Nov 30 '18 at 6:59
sazsaz
78.5k758123
Hint: $${tau leq k} = {tau' leq tau leq k} = bigcup_{j=0}^k ({tau'=j} cap {j leq tau leq k})$$
and so
$${tau leq k} = bigcup_{j=0}^k left( {tau'=j} capbigcup_{i=j}^k {X_i in B} right) in mathcal{F}_k$$
answered Nov 30 '18 at 6:59
sazsaz
78.5k758123
edited Nov 30 '18 at 13:30
answered Nov 30 '18 at 6:59
sazsaz
78.5k758123
answered Nov 30 '18 at 6:59
sazsaz
78.5k758123
answered Nov 30 '18 at 6:59
sazsaz
78.5k758123
78.5k758123
add a comment |
add a comment |
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