Convergence of jump function with respect to the Skorokhod metric
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Consider the space $D := D([0, 1], mathbbN)$ of cadlag functions $f : [0, 1] to mathbbN$ equipped with the Skorokhod $J_1$-metric $d(f,g) := inf_lambda in Lambda max lVert lambda - id rVert, lVert f circ lambda - g rVert $ where $Lambda$ is the set of continuous strictly increasing functions $lambda : [0,1] to [0,1]$ with $lambda(0) = 0$ and $lambda(1) = 1$ and $lVert cdot rVert$ the supremum norm.
Let $x_n, x in D$ with $x_n to x$ in this metric. Denote by $tau_n$ and $tau$ the timepoint of the first jump of $x_n$ and $x$ respectively. Is it true that $tau_n to tau$ and $x_n(tau_n) to x(tau)$?
general-topology skorohod-space
edited Sep 4 at 15:11
saz
73.9k553113
asked Sep 3 at 13:48
yadaddy
1,231816
add a comment |Â
up vote
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Consider the space $D := D([0, 1], mathbbN)$ of cadlag functions $f : [0, 1] to mathbbN$ equipped with the Skorokhod $J_1$-metric $d(f,g) := inf_lambda in Lambda max lVert lambda - id rVert, lVert f circ lambda - g rVert $ where $Lambda$ is the set of continuous strictly increasing functions $lambda : [0,1] to [0,1]$ with $lambda(0) = 0$ and $lambda(1) = 1$ and $lVert cdot rVert$ the supremum norm.
Let $x_n, x in D$ with $x_n to x$ in this metric. Denote by $tau_n$ and $tau$ the timepoint of the first jump of $x_n$ and $x$ respectively. Is it true that $tau_n to tau$ and $x_n(tau_n) to x(tau)$?
general-topology skorohod-space
edited Sep 4 at 15:11
saz
73.9k553113
asked Sep 3 at 13:48
yadaddy
1,231816
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider the space $D := D([0, 1], mathbbN)$ of cadlag functions $f : [0, 1] to mathbbN$ equipped with the Skorokhod $J_1$-metric $d(f,g) := inf_lambda in Lambda max lVert lambda - id rVert, lVert f circ lambda - g rVert $ where $Lambda$ is the set of continuous strictly increasing functions $lambda : [0,1] to [0,1]$ with $lambda(0) = 0$ and $lambda(1) = 1$ and $lVert cdot rVert$ the supremum norm.
Let $x_n, x in D$ with $x_n to x$ in this metric. Denote by $tau_n$ and $tau$ the timepoint of the first jump of $x_n$ and $x$ respectively. Is it true that $tau_n to tau$ and $x_n(tau_n) to x(tau)$?
general-topology skorohod-space
edited Sep 4 at 15:11
saz
73.9k553113
asked Sep 3 at 13:48
yadaddy
1,231816
Consider the space $D := D([0, 1], mathbbN)$ of cadlag functions $f : [0, 1] to mathbbN$ equipped with the Skorokhod $J_1$-metric $d(f,g) := inf_lambda in Lambda max lVert lambda - id rVert, lVert f circ lambda - g rVert $ where $Lambda$ is the set of continuous strictly increasing functions $lambda : [0,1] to [0,1]$ with $lambda(0) = 0$ and $lambda(1) = 1$ and $lVert cdot rVert$ the supremum norm.
Let $x_n, x in D$ with $x_n to x$ in this metric. Denote by $tau_n$ and $tau$ the timepoint of the first jump of $x_n$ and $x$ respectively. Is it true that $tau_n to tau$ and $x_n(tau_n) to x(tau)$?
general-topology skorohod-space
general-topology skorohod-space
edited Sep 4 at 15:11
saz
73.9k553113
asked Sep 3 at 13:48
yadaddy
1,231816
edited Sep 4 at 15:11
saz
73.9k553113
asked Sep 3 at 13:48
yadaddy
1,231816
edited Sep 4 at 15:11
saz
73.9k553113
edited Sep 4 at 15:11
saz
73.9k553113
edited Sep 4 at 15:11
saz
73.9k553113
73.9k553113
asked Sep 3 at 13:48
yadaddy
1,231816
asked Sep 3 at 13:48
yadaddy
1,231816
asked Sep 3 at 13:48
yadaddy
1,231816
1,231816
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Lemma: If $x_n to x$ in $J_1$ topology and $x$ is not constant, then $tau_n to tau$ and $x_n(tau_n) to x(tau)$.
Proof: Since $x_n to x$ in $J_1$ we can find a sequence $(lambda_n)_n in mathbbN$ of continuous strictly increasing functions $lambda_n: [0,1] to [0,1]$ with $lambda_n(0)=0$, $lambda_n(1)=1$ such that $$|lambda_n-textid|_infty to 0 qquad |x_n circ lambda_n-x|_infty to 0.$$ In particular, there exists $N in mathbbN$ such that $$|x_n circ lambda_n-x|_infty < 1 quad textfor all $n geq N$,$$ and since the functions are taking values in $mathbbN$, this already implies $$ forall n geq N , forall t in [0,1]: quad x_n(lambda_n(t))= x(t). tag1$$ As $x$ is right-continuous, non-constant and takes values in $mathbbN$, we know that the first jump time $tau$ of $x$ satisfies $$tau = inft in (0,1]: x(0) neq x(t) in (0,1].$$ It follows from $(1)$ that $$tau_n = lambda_n(tau)$$ and $$x_n(tau_n) = x_n(lambda_n(tau)) = x(tau)$$ for all $n geq N$. In particular, $x_n(lambda_n(tau)) to x(tau)$ and $$tau_n =lambda_n(tau) xrightarrown to infty tau.$$
answered Sep 4 at 6:08
saz
73.9k553113
Please observe that the range space of the cadlag functions is discrete and countable.
– yadaddy
Sep 4 at 6:11
@yadaddy Sorry, I missed that; see my edited answer.
– saz
Sep 4 at 15:10
Thank you. I see that $x_n(lambda_n(tau)) = x(tau)$ for all $n geq N$, but I don't see why $tau_n = lambda_n(tau)$. The time shifts $lambda_n$ that are given to us by $J_1$-convergence need not be those that map the jump time point $tau$ of $x$ to the jump timepoints $tau_n$ of $x_n$.
– yadaddy
Sep 5 at 7:41
1
@yadaddy For $s in [0,tau)$, we have $x(s)=x(0)$, and it follows from $(1)$ that $x_n(s)=x(0)$ for any $s in [0,lambda_n(tau))$. At $s=lambda_n(tau)$, it follows again from (1) that $x_n$ jumps and the new position is exactly $x(tau)$. If this makes things easier for you, you may note that $lambda_n$ is strictly increasing (hence invertible), and then $(1)$ becomes $x_n(s) = x(lambda_n^-1(s))$. By looking at the "trajectory" of $x$ you can recover all the information about $x_n$.
– saz
Sep 5 at 7:52
Yes, that helped.
– yadaddy
Sep 5 at 8:13
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Lemma: If $x_n to x$ in $J_1$ topology and $x$ is not constant, then $tau_n to tau$ and $x_n(tau_n) to x(tau)$.
Proof: Since $x_n to x$ in $J_1$ we can find a sequence $(lambda_n)_n in mathbbN$ of continuous strictly increasing functions $lambda_n: [0,1] to [0,1]$ with $lambda_n(0)=0$, $lambda_n(1)=1$ such that $$|lambda_n-textid|_infty to 0 qquad |x_n circ lambda_n-x|_infty to 0.$$ In particular, there exists $N in mathbbN$ such that $$|x_n circ lambda_n-x|_infty < 1 quad textfor all $n geq N$,$$ and since the functions are taking values in $mathbbN$, this already implies $$ forall n geq N , forall t in [0,1]: quad x_n(lambda_n(t))= x(t). tag1$$ As $x$ is right-continuous, non-constant and takes values in $mathbbN$, we know that the first jump time $tau$ of $x$ satisfies $$tau = inft in (0,1]: x(0) neq x(t) in (0,1].$$ It follows from $(1)$ that $$tau_n = lambda_n(tau)$$ and $$x_n(tau_n) = x_n(lambda_n(tau)) = x(tau)$$ for all $n geq N$. In particular, $x_n(lambda_n(tau)) to x(tau)$ and $$tau_n =lambda_n(tau) xrightarrown to infty tau.$$
answered Sep 4 at 6:08
saz
73.9k553113
Please observe that the range space of the cadlag functions is discrete and countable.
– yadaddy
Sep 4 at 6:11
@yadaddy Sorry, I missed that; see my edited answer.
– saz
Sep 4 at 15:10
Thank you. I see that $x_n(lambda_n(tau)) = x(tau)$ for all $n geq N$, but I don't see why $tau_n = lambda_n(tau)$. The time shifts $lambda_n$ that are given to us by $J_1$-convergence need not be those that map the jump time point $tau$ of $x$ to the jump timepoints $tau_n$ of $x_n$.
– yadaddy
Sep 5 at 7:41
1
@yadaddy For $s in [0,tau)$, we have $x(s)=x(0)$, and it follows from $(1)$ that $x_n(s)=x(0)$ for any $s in [0,lambda_n(tau))$. At $s=lambda_n(tau)$, it follows again from (1) that $x_n$ jumps and the new position is exactly $x(tau)$. If this makes things easier for you, you may note that $lambda_n$ is strictly increasing (hence invertible), and then $(1)$ becomes $x_n(s) = x(lambda_n^-1(s))$. By looking at the "trajectory" of $x$ you can recover all the information about $x_n$.
– saz
Sep 5 at 7:52
Yes, that helped.
– yadaddy
Sep 5 at 8:13
 |Â
show 1 more comment
up vote
2
down vote
accepted
Lemma: If $x_n to x$ in $J_1$ topology and $x$ is not constant, then $tau_n to tau$ and $x_n(tau_n) to x(tau)$.
Proof: Since $x_n to x$ in $J_1$ we can find a sequence $(lambda_n)_n in mathbbN$ of continuous strictly increasing functions $lambda_n: [0,1] to [0,1]$ with $lambda_n(0)=0$, $lambda_n(1)=1$ such that $$|lambda_n-textid|_infty to 0 qquad |x_n circ lambda_n-x|_infty to 0.$$ In particular, there exists $N in mathbbN$ such that $$|x_n circ lambda_n-x|_infty < 1 quad textfor all $n geq N$,$$ and since the functions are taking values in $mathbbN$, this already implies $$ forall n geq N , forall t in [0,1]: quad x_n(lambda_n(t))= x(t). tag1$$ As $x$ is right-continuous, non-constant and takes values in $mathbbN$, we know that the first jump time $tau$ of $x$ satisfies $$tau = inft in (0,1]: x(0) neq x(t) in (0,1].$$ It follows from $(1)$ that $$tau_n = lambda_n(tau)$$ and $$x_n(tau_n) = x_n(lambda_n(tau)) = x(tau)$$ for all $n geq N$. In particular, $x_n(lambda_n(tau)) to x(tau)$ and $$tau_n =lambda_n(tau) xrightarrown to infty tau.$$
answered Sep 4 at 6:08
saz
73.9k553113
Please observe that the range space of the cadlag functions is discrete and countable.
– yadaddy
Sep 4 at 6:11
@yadaddy Sorry, I missed that; see my edited answer.
– saz
Sep 4 at 15:10
Thank you. I see that $x_n(lambda_n(tau)) = x(tau)$ for all $n geq N$, but I don't see why $tau_n = lambda_n(tau)$. The time shifts $lambda_n$ that are given to us by $J_1$-convergence need not be those that map the jump time point $tau$ of $x$ to the jump timepoints $tau_n$ of $x_n$.
– yadaddy
Sep 5 at 7:41
1
@yadaddy For $s in [0,tau)$, we have $x(s)=x(0)$, and it follows from $(1)$ that $x_n(s)=x(0)$ for any $s in [0,lambda_n(tau))$. At $s=lambda_n(tau)$, it follows again from (1) that $x_n$ jumps and the new position is exactly $x(tau)$. If this makes things easier for you, you may note that $lambda_n$ is strictly increasing (hence invertible), and then $(1)$ becomes $x_n(s) = x(lambda_n^-1(s))$. By looking at the "trajectory" of $x$ you can recover all the information about $x_n$.
– saz
Sep 5 at 7:52
Yes, that helped.
– yadaddy
Sep 5 at 8:13
 |Â
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Lemma: If $x_n to x$ in $J_1$ topology and $x$ is not constant, then $tau_n to tau$ and $x_n(tau_n) to x(tau)$.
Proof: Since $x_n to x$ in $J_1$ we can find a sequence $(lambda_n)_n in mathbbN$ of continuous strictly increasing functions $lambda_n: [0,1] to [0,1]$ with $lambda_n(0)=0$, $lambda_n(1)=1$ such that $$|lambda_n-textid|_infty to 0 qquad |x_n circ lambda_n-x|_infty to 0.$$ In particular, there exists $N in mathbbN$ such that $$|x_n circ lambda_n-x|_infty < 1 quad textfor all $n geq N$,$$ and since the functions are taking values in $mathbbN$, this already implies $$ forall n geq N , forall t in [0,1]: quad x_n(lambda_n(t))= x(t). tag1$$ As $x$ is right-continuous, non-constant and takes values in $mathbbN$, we know that the first jump time $tau$ of $x$ satisfies $$tau = inft in (0,1]: x(0) neq x(t) in (0,1].$$ It follows from $(1)$ that $$tau_n = lambda_n(tau)$$ and $$x_n(tau_n) = x_n(lambda_n(tau)) = x(tau)$$ for all $n geq N$. In particular, $x_n(lambda_n(tau)) to x(tau)$ and $$tau_n =lambda_n(tau) xrightarrown to infty tau.$$
answered Sep 4 at 6:08
saz
73.9k553113
Lemma: If $x_n to x$ in $J_1$ topology and $x$ is not constant, then $tau_n to tau$ and $x_n(tau_n) to x(tau)$.
Proof: Since $x_n to x$ in $J_1$ we can find a sequence $(lambda_n)_n in mathbbN$ of continuous strictly increasing functions $lambda_n: [0,1] to [0,1]$ with $lambda_n(0)=0$, $lambda_n(1)=1$ such that $$|lambda_n-textid|_infty to 0 qquad |x_n circ lambda_n-x|_infty to 0.$$ In particular, there exists $N in mathbbN$ such that $$|x_n circ lambda_n-x|_infty < 1 quad textfor all $n geq N$,$$ and since the functions are taking values in $mathbbN$, this already implies $$ forall n geq N , forall t in [0,1]: quad x_n(lambda_n(t))= x(t). tag1$$ As $x$ is right-continuous, non-constant and takes values in $mathbbN$, we know that the first jump time $tau$ of $x$ satisfies $$tau = inft in (0,1]: x(0) neq x(t) in (0,1].$$ It follows from $(1)$ that $$tau_n = lambda_n(tau)$$ and $$x_n(tau_n) = x_n(lambda_n(tau)) = x(tau)$$ for all $n geq N$. In particular, $x_n(lambda_n(tau)) to x(tau)$ and $$tau_n =lambda_n(tau) xrightarrown to infty tau.$$
answered Sep 4 at 6:08
saz
73.9k553113
edited Sep 4 at 15:10
answered Sep 4 at 6:08
saz
73.9k553113
answered Sep 4 at 6:08
saz
73.9k553113
answered Sep 4 at 6:08
saz
73.9k553113
73.9k553113
Please observe that the range space of the cadlag functions is discrete and countable.
– yadaddy
Sep 4 at 6:11
@yadaddy Sorry, I missed that; see my edited answer.
– saz
Sep 4 at 15:10
Thank you. I see that $x_n(lambda_n(tau)) = x(tau)$ for all $n geq N$, but I don't see why $tau_n = lambda_n(tau)$. The time shifts $lambda_n$ that are given to us by $J_1$-convergence need not be those that map the jump time point $tau$ of $x$ to the jump timepoints $tau_n$ of $x_n$.
– yadaddy
Sep 5 at 7:41
1
@yadaddy For $s in [0,tau)$, we have $x(s)=x(0)$, and it follows from $(1)$ that $x_n(s)=x(0)$ for any $s in [0,lambda_n(tau))$. At $s=lambda_n(tau)$, it follows again from (1) that $x_n$ jumps and the new position is exactly $x(tau)$. If this makes things easier for you, you may note that $lambda_n$ is strictly increasing (hence invertible), and then $(1)$ becomes $x_n(s) = x(lambda_n^-1(s))$. By looking at the "trajectory" of $x$ you can recover all the information about $x_n$.
– saz
Sep 5 at 7:52
Yes, that helped.
– yadaddy
Sep 5 at 8:13
 |Â
show 1 more comment
Please observe that the range space of the cadlag functions is discrete and countable.
– yadaddy
Sep 4 at 6:11
@yadaddy Sorry, I missed that; see my edited answer.
– saz
Sep 4 at 15:10
Thank you. I see that $x_n(lambda_n(tau)) = x(tau)$ for all $n geq N$, but I don't see why $tau_n = lambda_n(tau)$. The time shifts $lambda_n$ that are given to us by $J_1$-convergence need not be those that map the jump time point $tau$ of $x$ to the jump timepoints $tau_n$ of $x_n$.
– yadaddy
Sep 5 at 7:41
1
@yadaddy For $s in [0,tau)$, we have $x(s)=x(0)$, and it follows from $(1)$ that $x_n(s)=x(0)$ for any $s in [0,lambda_n(tau))$. At $s=lambda_n(tau)$, it follows again from (1) that $x_n$ jumps and the new position is exactly $x(tau)$. If this makes things easier for you, you may note that $lambda_n$ is strictly increasing (hence invertible), and then $(1)$ becomes $x_n(s) = x(lambda_n^-1(s))$. By looking at the "trajectory" of $x$ you can recover all the information about $x_n$.
– saz
Sep 5 at 7:52
Yes, that helped.
– yadaddy
Sep 5 at 8:13
Please observe that the range space of the cadlag functions is discrete and countable.
– yadaddy
Sep 4 at 6:11
Please observe that the range space of the cadlag functions is discrete and countable.
– yadaddy
Sep 4 at 6:11
@yadaddy Sorry, I missed that; see my edited answer.
– saz
Sep 4 at 15:10
@yadaddy Sorry, I missed that; see my edited answer.
– saz
Sep 4 at 15:10
Thank you. I see that $x_n(lambda_n(tau)) = x(tau)$ for all $n geq N$, but I don't see why $tau_n = lambda_n(tau)$. The time shifts $lambda_n$ that are given to us by $J_1$-convergence need not be those that map the jump time point $tau$ of $x$ to the jump timepoints $tau_n$ of $x_n$.
– yadaddy
Sep 5 at 7:41
Thank you. I see that $x_n(lambda_n(tau)) = x(tau)$ for all $n geq N$, but I don't see why $tau_n = lambda_n(tau)$. The time shifts $lambda_n$ that are given to us by $J_1$-convergence need not be those that map the jump time point $tau$ of $x$ to the jump timepoints $tau_n$ of $x_n$.
– yadaddy
Sep 5 at 7:41
1
1
@yadaddy For $s in [0,tau)$, we have $x(s)=x(0)$, and it follows from $(1)$ that $x_n(s)=x(0)$ for any $s in [0,lambda_n(tau))$. At $s=lambda_n(tau)$, it follows again from (1) that $x_n$ jumps and the new position is exactly $x(tau)$. If this makes things easier for you, you may note that $lambda_n$ is strictly increasing (hence invertible), and then $(1)$ becomes $x_n(s) = x(lambda_n^-1(s))$. By looking at the "trajectory" of $x$ you can recover all the information about $x_n$.
– saz
Sep 5 at 7:52
@yadaddy For $s in [0,tau)$, we have $x(s)=x(0)$, and it follows from $(1)$ that $x_n(s)=x(0)$ for any $s in [0,lambda_n(tau))$. At $s=lambda_n(tau)$, it follows again from (1) that $x_n$ jumps and the new position is exactly $x(tau)$. If this makes things easier for you, you may note that $lambda_n$ is strictly increasing (hence invertible), and then $(1)$ becomes $x_n(s) = x(lambda_n^-1(s))$. By looking at the "trajectory" of $x$ you can recover all the information about $x_n$.
– saz
Sep 5 at 7:52
Yes, that helped.
– yadaddy
Sep 5 at 8:13
Yes, that helped.
– yadaddy
Sep 5 at 8:13
 |Â
show 1 more comment
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