Calculate limit: Make a stopping time $T$ bounded $Tland n$ and take the limit $n to infty$
$begingroup$
Say we have a martingale $X$ and a stopping time $T$.
Instead of directly studying the stopped process $X_T$, many proofs employ a trick,
namely, one considers the bounded stopping time $Tland n$ and instead studies $X_{Tland n}$ and then letting $n to infty$ at some point.
I'm trying to prove that
$$ lim_n X_{T land n} = X_T mathbf{1}_{{T < infty}}.$$
Intuitively it makes sense but I'm stuck trying to prove it.
Edit Update: Like saz noted in the comments below the above limit is not true in general if $P(T = infty) > 0$.
I consider this solved as it cleared up my misunderstanding.
probability-theory stochastic-processes martingales
asked Dec 8 '18 at 15:15
EberhardtEberhardt
103117
$endgroup$
add a comment |
$begingroup$
Say we have a martingale $X$ and a stopping time $T$.
Instead of directly studying the stopped process $X_T$, many proofs employ a trick,
namely, one considers the bounded stopping time $Tland n$ and instead studies $X_{Tland n}$ and then letting $n to infty$ at some point.
I'm trying to prove that
$$ lim_n X_{T land n} = X_T mathbf{1}_{{T < infty}}.$$
Intuitively it makes sense but I'm stuck trying to prove it.
Edit Update: Like saz noted in the comments below the above limit is not true in general if $P(T = infty) > 0$.
I consider this solved as it cleared up my misunderstanding.
probability-theory stochastic-processes martingales
asked Dec 8 '18 at 15:15
EberhardtEberhardt
103117
$endgroup$
2
$begingroup$
If $mathbb{P}(T=infty)>0$ then this does, in general, not hold true. You will need to multiply the left-hand side with $1_{{T<infty}}$ ...
$endgroup$
– saz
Dec 8 '18 at 15:24
$begingroup$
Thank you @saz ! Your comment made me realize I had made an erroneous assumption earlier in a proof I was working on.
$endgroup$
– Eberhardt
Dec 8 '18 at 16:21
1
$begingroup$
So.. did this solve/answer your question? You might want to notice that if $T(omega)<infty$ then $T(omega) wedge n = T(omega)$ for $n geq N(omega)$ sufficiently large, and therefore the convergence is really not difficult to show.
$endgroup$
– saz
Dec 8 '18 at 16:39
$begingroup$
Yes, I consider it solved. As you pointed out it's not true in general. So no wonder I struggled to prove my false claim. If we add the factor $mathbf{1}_{{T < infty}}$ to the left-hand side it becomes much easier.
$endgroup$
– Eberhardt
Dec 8 '18 at 17:02
add a comment |
$begingroup$
Say we have a martingale $X$ and a stopping time $T$.
Instead of directly studying the stopped process $X_T$, many proofs employ a trick,
namely, one considers the bounded stopping time $Tland n$ and instead studies $X_{Tland n}$ and then letting $n to infty$ at some point.
I'm trying to prove that
$$ lim_n X_{T land n} = X_T mathbf{1}_{{T < infty}}.$$
Intuitively it makes sense but I'm stuck trying to prove it.
Edit Update: Like saz noted in the comments below the above limit is not true in general if $P(T = infty) > 0$.
I consider this solved as it cleared up my misunderstanding.
probability-theory stochastic-processes martingales
asked Dec 8 '18 at 15:15
EberhardtEberhardt
103117
$endgroup$
Say we have a martingale $X$ and a stopping time $T$.
Instead of directly studying the stopped process $X_T$, many proofs employ a trick,
namely, one considers the bounded stopping time $Tland n$ and instead studies $X_{Tland n}$ and then letting $n to infty$ at some point.
I'm trying to prove that
$$ lim_n X_{T land n} = X_T mathbf{1}_{{T < infty}}.$$
Intuitively it makes sense but I'm stuck trying to prove it.
Edit Update: Like saz noted in the comments below the above limit is not true in general if $P(T = infty) > 0$.
I consider this solved as it cleared up my misunderstanding.
probability-theory stochastic-processes martingales
probability-theory stochastic-processes martingales
asked Dec 8 '18 at 15:15
EberhardtEberhardt
103117
asked Dec 8 '18 at 15:15
EberhardtEberhardt
103117
edited Dec 8 '18 at 17:02
Eberhardt
asked Dec 8 '18 at 15:15
EberhardtEberhardt
103117
asked Dec 8 '18 at 15:15
EberhardtEberhardt
103117
asked Dec 8 '18 at 15:15
EberhardtEberhardt
103117
103117
2
$begingroup$
If $mathbb{P}(T=infty)>0$ then this does, in general, not hold true. You will need to multiply the left-hand side with $1_{{T<infty}}$ ...
$endgroup$
– saz
Dec 8 '18 at 15:24
$begingroup$
Thank you @saz ! Your comment made me realize I had made an erroneous assumption earlier in a proof I was working on.
$endgroup$
– Eberhardt
Dec 8 '18 at 16:21
1
$begingroup$
So.. did this solve/answer your question? You might want to notice that if $T(omega)<infty$ then $T(omega) wedge n = T(omega)$ for $n geq N(omega)$ sufficiently large, and therefore the convergence is really not difficult to show.
$endgroup$
– saz
Dec 8 '18 at 16:39
$begingroup$
Yes, I consider it solved. As you pointed out it's not true in general. So no wonder I struggled to prove my false claim. If we add the factor $mathbf{1}_{{T < infty}}$ to the left-hand side it becomes much easier.
$endgroup$
– Eberhardt
Dec 8 '18 at 17:02
add a comment |
2
$begingroup$
If $mathbb{P}(T=infty)>0$ then this does, in general, not hold true. You will need to multiply the left-hand side with $1_{{T<infty}}$ ...
$endgroup$
– saz
Dec 8 '18 at 15:24
$begingroup$
Thank you @saz ! Your comment made me realize I had made an erroneous assumption earlier in a proof I was working on.
$endgroup$
– Eberhardt
Dec 8 '18 at 16:21
1
$begingroup$
So.. did this solve/answer your question? You might want to notice that if $T(omega)<infty$ then $T(omega) wedge n = T(omega)$ for $n geq N(omega)$ sufficiently large, and therefore the convergence is really not difficult to show.
$endgroup$
– saz
Dec 8 '18 at 16:39
$begingroup$
Yes, I consider it solved. As you pointed out it's not true in general. So no wonder I struggled to prove my false claim. If we add the factor $mathbf{1}_{{T < infty}}$ to the left-hand side it becomes much easier.
$endgroup$
– Eberhardt
Dec 8 '18 at 17:02
2
2
$begingroup$
If $mathbb{P}(T=infty)>0$ then this does, in general, not hold true. You will need to multiply the left-hand side with $1_{{T<infty}}$ ...
$endgroup$
– saz
Dec 8 '18 at 15:24
$begingroup$
If $mathbb{P}(T=infty)>0$ then this does, in general, not hold true. You will need to multiply the left-hand side with $1_{{T<infty}}$ ...
$endgroup$
– saz
Dec 8 '18 at 15:24
$begingroup$
Thank you @saz ! Your comment made me realize I had made an erroneous assumption earlier in a proof I was working on.
$endgroup$
– Eberhardt
Dec 8 '18 at 16:21
$begingroup$
Thank you @saz ! Your comment made me realize I had made an erroneous assumption earlier in a proof I was working on.
$endgroup$
– Eberhardt
Dec 8 '18 at 16:21
1
1
$begingroup$
So.. did this solve/answer your question? You might want to notice that if $T(omega)<infty$ then $T(omega) wedge n = T(omega)$ for $n geq N(omega)$ sufficiently large, and therefore the convergence is really not difficult to show.
$endgroup$
– saz
Dec 8 '18 at 16:39
$begingroup$
So.. did this solve/answer your question? You might want to notice that if $T(omega)<infty$ then $T(omega) wedge n = T(omega)$ for $n geq N(omega)$ sufficiently large, and therefore the convergence is really not difficult to show.
$endgroup$
– saz
Dec 8 '18 at 16:39
$begingroup$
Yes, I consider it solved. As you pointed out it's not true in general. So no wonder I struggled to prove my false claim. If we add the factor $mathbf{1}_{{T < infty}}$ to the left-hand side it becomes much easier.
$endgroup$
– Eberhardt
Dec 8 '18 at 17:02
$begingroup$
Yes, I consider it solved. As you pointed out it's not true in general. So no wonder I struggled to prove my false claim. If we add the factor $mathbf{1}_{{T < infty}}$ to the left-hand side it becomes much easier.
$endgroup$
– Eberhardt
Dec 8 '18 at 17:02
add a comment |
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$begingroup$
If $mathbb{P}(T=infty)>0$ then this does, in general, not hold true. You will need to multiply the left-hand side with $1_{{T<infty}}$ ...
$endgroup$
– saz
Dec 8 '18 at 15:24
$begingroup$
Thank you @saz ! Your comment made me realize I had made an erroneous assumption earlier in a proof I was working on.
$endgroup$
– Eberhardt
Dec 8 '18 at 16:21
1
$begingroup$
So.. did this solve/answer your question? You might want to notice that if $T(omega)<infty$ then $T(omega) wedge n = T(omega)$ for $n geq N(omega)$ sufficiently large, and therefore the convergence is really not difficult to show.
$endgroup$
– saz
Dec 8 '18 at 16:39
$begingroup$
Yes, I consider it solved. As you pointed out it's not true in general. So no wonder I struggled to prove my false claim. If we add the factor $mathbf{1}_{{T < infty}}$ to the left-hand side it becomes much easier.
$endgroup$
– Eberhardt
Dec 8 '18 at 17:02