Why does Dini's theorem apply here?
When reading a textbook I encounter the following argument which I couldn't understand. The stochastic process $X$ below is assumed to be a Levy process, but we might not need this fact and might just take it as a stochastic process.
So the authors manage to prove that for any $n$ and $t in [0,c]$, where $c$ is a constant, the following inequality hold:
$$
lim_{s rightarrow t} P(|X_c - X_s| geq n) leq P(|X_c - X_t| geq n)
$$
Then they claim "by Dini's theorem" we can establish the following equality
$$
lim_{n rightarrow infty} sup_{0 leq t leq c}P(|X_c - X_t| geq n) = 0
$$
I cannot see why Dini's theorem apply here: I suppose that the authors are suggesting that we consider the sequence of functions $f_n(t) = P(|X_c - X_t| geq n)$ defined on the compact interval $[0,c]$. But they haven't established the continuity of these functions. I'm thereby confused and have been wondering if we don't need Dini's theorem to prove the desired equality?
Any hint will be greatly appreciated!
real-analysis probability-theory stochastic-processes
edited May 23 '17 at 17:47
saz
77.7k756120
asked May 23 '17 at 17:37
Dormire
479214
add a comment |
When reading a textbook I encounter the following argument which I couldn't understand. The stochastic process $X$ below is assumed to be a Levy process, but we might not need this fact and might just take it as a stochastic process.
So the authors manage to prove that for any $n$ and $t in [0,c]$, where $c$ is a constant, the following inequality hold:
$$
lim_{s rightarrow t} P(|X_c - X_s| geq n) leq P(|X_c - X_t| geq n)
$$
Then they claim "by Dini's theorem" we can establish the following equality
$$
lim_{n rightarrow infty} sup_{0 leq t leq c}P(|X_c - X_t| geq n) = 0
$$
I cannot see why Dini's theorem apply here: I suppose that the authors are suggesting that we consider the sequence of functions $f_n(t) = P(|X_c - X_t| geq n)$ defined on the compact interval $[0,c]$. But they haven't established the continuity of these functions. I'm thereby confused and have been wondering if we don't need Dini's theorem to prove the desired equality?
Any hint will be greatly appreciated!
real-analysis probability-theory stochastic-processes
edited May 23 '17 at 17:47
saz
77.7k756120
asked May 23 '17 at 17:37
Dormire
479214
add a comment |
When reading a textbook I encounter the following argument which I couldn't understand. The stochastic process $X$ below is assumed to be a Levy process, but we might not need this fact and might just take it as a stochastic process.
So the authors manage to prove that for any $n$ and $t in [0,c]$, where $c$ is a constant, the following inequality hold:
$$
lim_{s rightarrow t} P(|X_c - X_s| geq n) leq P(|X_c - X_t| geq n)
$$
Then they claim "by Dini's theorem" we can establish the following equality
$$
lim_{n rightarrow infty} sup_{0 leq t leq c}P(|X_c - X_t| geq n) = 0
$$
I cannot see why Dini's theorem apply here: I suppose that the authors are suggesting that we consider the sequence of functions $f_n(t) = P(|X_c - X_t| geq n)$ defined on the compact interval $[0,c]$. But they haven't established the continuity of these functions. I'm thereby confused and have been wondering if we don't need Dini's theorem to prove the desired equality?
Any hint will be greatly appreciated!
real-analysis probability-theory stochastic-processes
edited May 23 '17 at 17:47
saz
77.7k756120
asked May 23 '17 at 17:37
Dormire
479214
When reading a textbook I encounter the following argument which I couldn't understand. The stochastic process $X$ below is assumed to be a Levy process, but we might not need this fact and might just take it as a stochastic process.
So the authors manage to prove that for any $n$ and $t in [0,c]$, where $c$ is a constant, the following inequality hold:
$$
lim_{s rightarrow t} P(|X_c - X_s| geq n) leq P(|X_c - X_t| geq n)
$$
Then they claim "by Dini's theorem" we can establish the following equality
$$
lim_{n rightarrow infty} sup_{0 leq t leq c}P(|X_c - X_t| geq n) = 0
$$
I cannot see why Dini's theorem apply here: I suppose that the authors are suggesting that we consider the sequence of functions $f_n(t) = P(|X_c - X_t| geq n)$ defined on the compact interval $[0,c]$. But they haven't established the continuity of these functions. I'm thereby confused and have been wondering if we don't need Dini's theorem to prove the desired equality?
Any hint will be greatly appreciated!
real-analysis probability-theory stochastic-processes
real-analysis probability-theory stochastic-processes
edited May 23 '17 at 17:47
saz
77.7k756120
asked May 23 '17 at 17:37
Dormire
479214
edited May 23 '17 at 17:47
saz
77.7k756120
asked May 23 '17 at 17:37
Dormire
479214
edited May 23 '17 at 17:47
saz
77.7k756120
edited May 23 '17 at 17:47
saz
77.7k756120
edited May 23 '17 at 17:47
saz
77.7k756120
77.7k756120
asked May 23 '17 at 17:37
Dormire
479214
asked May 23 '17 at 17:37
Dormire
479214
asked May 23 '17 at 17:37
Dormire
479214
479214
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
The inequality
$$lim_{s to t} mathbb{P}(|X_c-X_s| geq n) leq mathbb{P}(|X_c-X_t| geq n)$$
shows that
$$f_n(t) := mathbb{P}(|X_t-X_c| geq n)$$
is upper semicontinuous; therefore the following version of Dini's theorem is applicable:
Let $(f_n)_{n in mathbb{N}}$ be a sequence of upper semicontinuous functions such that $f_n geq f_{n+1}$ for all $n geq 1$. If there exists a continuous function $f$ such that $f_n(x) to f(x)$ for all $x$, then $f_n to f$ uniformly on compact sets.
answered May 23 '17 at 17:47
saz
77.7k756120
Thanks for the hint! Going to through the proof I realize that semi-continuity is really enough:-)
– Dormire
May 23 '17 at 21:01
@Dormire You are welcome
– saz
May 24 '17 at 4:23
add a comment |
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The inequality
$$lim_{s to t} mathbb{P}(|X_c-X_s| geq n) leq mathbb{P}(|X_c-X_t| geq n)$$
shows that
$$f_n(t) := mathbb{P}(|X_t-X_c| geq n)$$
is upper semicontinuous; therefore the following version of Dini's theorem is applicable:
Let $(f_n)_{n in mathbb{N}}$ be a sequence of upper semicontinuous functions such that $f_n geq f_{n+1}$ for all $n geq 1$. If there exists a continuous function $f$ such that $f_n(x) to f(x)$ for all $x$, then $f_n to f$ uniformly on compact sets.
answered May 23 '17 at 17:47
saz
77.7k756120
Thanks for the hint! Going to through the proof I realize that semi-continuity is really enough:-)
– Dormire
May 23 '17 at 21:01
@Dormire You are welcome
– saz
May 24 '17 at 4:23
add a comment |
The inequality
$$lim_{s to t} mathbb{P}(|X_c-X_s| geq n) leq mathbb{P}(|X_c-X_t| geq n)$$
shows that
$$f_n(t) := mathbb{P}(|X_t-X_c| geq n)$$
is upper semicontinuous; therefore the following version of Dini's theorem is applicable:
Let $(f_n)_{n in mathbb{N}}$ be a sequence of upper semicontinuous functions such that $f_n geq f_{n+1}$ for all $n geq 1$. If there exists a continuous function $f$ such that $f_n(x) to f(x)$ for all $x$, then $f_n to f$ uniformly on compact sets.
answered May 23 '17 at 17:47
saz
77.7k756120
Thanks for the hint! Going to through the proof I realize that semi-continuity is really enough:-)
– Dormire
May 23 '17 at 21:01
@Dormire You are welcome
– saz
May 24 '17 at 4:23
add a comment |
The inequality
$$lim_{s to t} mathbb{P}(|X_c-X_s| geq n) leq mathbb{P}(|X_c-X_t| geq n)$$
shows that
$$f_n(t) := mathbb{P}(|X_t-X_c| geq n)$$
is upper semicontinuous; therefore the following version of Dini's theorem is applicable:
Let $(f_n)_{n in mathbb{N}}$ be a sequence of upper semicontinuous functions such that $f_n geq f_{n+1}$ for all $n geq 1$. If there exists a continuous function $f$ such that $f_n(x) to f(x)$ for all $x$, then $f_n to f$ uniformly on compact sets.
answered May 23 '17 at 17:47
saz
77.7k756120
The inequality
$$lim_{s to t} mathbb{P}(|X_c-X_s| geq n) leq mathbb{P}(|X_c-X_t| geq n)$$
shows that
$$f_n(t) := mathbb{P}(|X_t-X_c| geq n)$$
is upper semicontinuous; therefore the following version of Dini's theorem is applicable:
Let $(f_n)_{n in mathbb{N}}$ be a sequence of upper semicontinuous functions such that $f_n geq f_{n+1}$ for all $n geq 1$. If there exists a continuous function $f$ such that $f_n(x) to f(x)$ for all $x$, then $f_n to f$ uniformly on compact sets.
answered May 23 '17 at 17:47
saz
77.7k756120
edited May 23 '17 at 19:32
answered May 23 '17 at 17:47
saz
77.7k756120
answered May 23 '17 at 17:47
saz
77.7k756120
answered May 23 '17 at 17:47
saz
77.7k756120
77.7k756120
Thanks for the hint! Going to through the proof I realize that semi-continuity is really enough:-)
– Dormire
May 23 '17 at 21:01
@Dormire You are welcome
– saz
May 24 '17 at 4:23
add a comment |
Thanks for the hint! Going to through the proof I realize that semi-continuity is really enough:-)
– Dormire
May 23 '17 at 21:01
@Dormire You are welcome
– saz
May 24 '17 at 4:23
Thanks for the hint! Going to through the proof I realize that semi-continuity is really enough:-)
– Dormire
May 23 '17 at 21:01
Thanks for the hint! Going to through the proof I realize that semi-continuity is really enough:-)
– Dormire
May 23 '17 at 21:01
@Dormire You are welcome
– saz
May 24 '17 at 4:23
@Dormire You are welcome
– saz
May 24 '17 at 4:23
add a comment |
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