Does mean-square differentiability imply mean-square continuity?
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We say that a random process $X_t$ is mean-square differentiable at time $t_0$ if there is a random process $X_{t_0}'$ such that
$$lim_{tto t_0} mathbb{E}Big[big(X_{t_0}' - frac{X_t - X_{t_0}}{t-t_0}big)^2Big]=0$$
We say that a random process $X_t$ is mean-square continuous at time $t_0$ if
$$lim_{tto t_0} mathbb{E}big[(X_{t} - X_{t_0})^2big]=0$$
From calculus, we know that differentiability implies continuity. Does the same statement hold in stochastic calculus?
More formally, suppose a process is mean-square differentiable for all $tin mathbb{R}$. Is this process mean-square continuous for all $tinmathbb{R}$?
stochastic-processes stochastic-calculus
asked Nov 15 at 19:11
sharkbait
254
add a comment |
up vote
0
down vote
favorite
We say that a random process $X_t$ is mean-square differentiable at time $t_0$ if there is a random process $X_{t_0}'$ such that
$$lim_{tto t_0} mathbb{E}Big[big(X_{t_0}' - frac{X_t - X_{t_0}}{t-t_0}big)^2Big]=0$$
We say that a random process $X_t$ is mean-square continuous at time $t_0$ if
$$lim_{tto t_0} mathbb{E}big[(X_{t} - X_{t_0})^2big]=0$$
From calculus, we know that differentiability implies continuity. Does the same statement hold in stochastic calculus?
More formally, suppose a process is mean-square differentiable for all $tin mathbb{R}$. Is this process mean-square continuous for all $tinmathbb{R}$?
stochastic-processes stochastic-calculus
asked Nov 15 at 19:11
sharkbait
254
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
We say that a random process $X_t$ is mean-square differentiable at time $t_0$ if there is a random process $X_{t_0}'$ such that
$$lim_{tto t_0} mathbb{E}Big[big(X_{t_0}' - frac{X_t - X_{t_0}}{t-t_0}big)^2Big]=0$$
We say that a random process $X_t$ is mean-square continuous at time $t_0$ if
$$lim_{tto t_0} mathbb{E}big[(X_{t} - X_{t_0})^2big]=0$$
From calculus, we know that differentiability implies continuity. Does the same statement hold in stochastic calculus?
More formally, suppose a process is mean-square differentiable for all $tin mathbb{R}$. Is this process mean-square continuous for all $tinmathbb{R}$?
stochastic-processes stochastic-calculus
asked Nov 15 at 19:11
sharkbait
254
We say that a random process $X_t$ is mean-square differentiable at time $t_0$ if there is a random process $X_{t_0}'$ such that
$$lim_{tto t_0} mathbb{E}Big[big(X_{t_0}' - frac{X_t - X_{t_0}}{t-t_0}big)^2Big]=0$$
We say that a random process $X_t$ is mean-square continuous at time $t_0$ if
$$lim_{tto t_0} mathbb{E}big[(X_{t} - X_{t_0})^2big]=0$$
From calculus, we know that differentiability implies continuity. Does the same statement hold in stochastic calculus?
More formally, suppose a process is mean-square differentiable for all $tin mathbb{R}$. Is this process mean-square continuous for all $tinmathbb{R}$?
stochastic-processes stochastic-calculus
stochastic-processes stochastic-calculus
asked Nov 15 at 19:11
sharkbait
254
asked Nov 15 at 19:11
sharkbait
254
asked Nov 15 at 19:11
sharkbait
254
asked Nov 15 at 19:11
sharkbait
254
asked Nov 15 at 19:11
sharkbait
254
254
add a comment |
add a comment |
1 Answer
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Yes, that's true.
Using the elementary inequality $$(x+y)^2 leq 2x^2+2y^2, qquad x,y in mathbb{R}$$ we find
$$begin{align*} (X_t-X_{t_0})^2 &= (t-t_0)^2 left( frac{X_t-X_{t_0}}{t-t_0} - X_{t_0}' + X_{t_0}' right)^2 \ &leq 2 (t-t_0)^2 left( frac{X_t-X_{t_0}}{t-t_0}-X_{t_0}' right)^2 + 2(t-t_0)^2 (X_{t_0}')^2. end{align*}$$
Taking expectations on both sides we get
$$begin{align*} mathbb{E}((X_t-X_{t_0})^2) &leq 2(t-t_0)^2 underbrace{mathbb{E} left[ left( frac{X_t-X_{t_0}}{t-t_0}-X_{t_0}' right)^2 right]}_{xrightarrow{t to t_0} 0} + 2 underbrace{(t-t_0)^2}_{xrightarrow{t to t_0} 0} mathbb{E}((X_{t_0}')^2) \ &xrightarrow{t to t_0} 0. end{align*}$$
answered Nov 15 at 19:32
saz
76.3k755117
Thank you! One note: from this proof, it looks like this should be true only when $X_{t_0}'$ is second order, since we require that $mathbb{E}[(X_{t_0}')^2]$ be finite. So this would apply when $X_t$ is WSS, but may not work otherwise.
– sharkbait
Nov 15 at 19:39
@sharkbait Actually $mathbb{E}((X_{t_0}')^2)<infty$ follows from $$mathbb{E} left[ left( frac{X_t-X_{t_0}}{t-t_0} - X_{t_0}' right)^2 right]< infty$$ for $t$ close to $t_0$ and the fact that $(X_t)_t$ is square-integrable.
– saz
Nov 15 at 19:48
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Yes, that's true.
Using the elementary inequality $$(x+y)^2 leq 2x^2+2y^2, qquad x,y in mathbb{R}$$ we find
$$begin{align*} (X_t-X_{t_0})^2 &= (t-t_0)^2 left( frac{X_t-X_{t_0}}{t-t_0} - X_{t_0}' + X_{t_0}' right)^2 \ &leq 2 (t-t_0)^2 left( frac{X_t-X_{t_0}}{t-t_0}-X_{t_0}' right)^2 + 2(t-t_0)^2 (X_{t_0}')^2. end{align*}$$
Taking expectations on both sides we get
$$begin{align*} mathbb{E}((X_t-X_{t_0})^2) &leq 2(t-t_0)^2 underbrace{mathbb{E} left[ left( frac{X_t-X_{t_0}}{t-t_0}-X_{t_0}' right)^2 right]}_{xrightarrow{t to t_0} 0} + 2 underbrace{(t-t_0)^2}_{xrightarrow{t to t_0} 0} mathbb{E}((X_{t_0}')^2) \ &xrightarrow{t to t_0} 0. end{align*}$$
answered Nov 15 at 19:32
saz
76.3k755117
Thank you! One note: from this proof, it looks like this should be true only when $X_{t_0}'$ is second order, since we require that $mathbb{E}[(X_{t_0}')^2]$ be finite. So this would apply when $X_t$ is WSS, but may not work otherwise.
– sharkbait
Nov 15 at 19:39
@sharkbait Actually $mathbb{E}((X_{t_0}')^2)<infty$ follows from $$mathbb{E} left[ left( frac{X_t-X_{t_0}}{t-t_0} - X_{t_0}' right)^2 right]< infty$$ for $t$ close to $t_0$ and the fact that $(X_t)_t$ is square-integrable.
– saz
Nov 15 at 19:48
add a comment |
up vote
1
down vote
accepted
Yes, that's true.
Using the elementary inequality $$(x+y)^2 leq 2x^2+2y^2, qquad x,y in mathbb{R}$$ we find
$$begin{align*} (X_t-X_{t_0})^2 &= (t-t_0)^2 left( frac{X_t-X_{t_0}}{t-t_0} - X_{t_0}' + X_{t_0}' right)^2 \ &leq 2 (t-t_0)^2 left( frac{X_t-X_{t_0}}{t-t_0}-X_{t_0}' right)^2 + 2(t-t_0)^2 (X_{t_0}')^2. end{align*}$$
Taking expectations on both sides we get
$$begin{align*} mathbb{E}((X_t-X_{t_0})^2) &leq 2(t-t_0)^2 underbrace{mathbb{E} left[ left( frac{X_t-X_{t_0}}{t-t_0}-X_{t_0}' right)^2 right]}_{xrightarrow{t to t_0} 0} + 2 underbrace{(t-t_0)^2}_{xrightarrow{t to t_0} 0} mathbb{E}((X_{t_0}')^2) \ &xrightarrow{t to t_0} 0. end{align*}$$
answered Nov 15 at 19:32
saz
76.3k755117
Thank you! One note: from this proof, it looks like this should be true only when $X_{t_0}'$ is second order, since we require that $mathbb{E}[(X_{t_0}')^2]$ be finite. So this would apply when $X_t$ is WSS, but may not work otherwise.
– sharkbait
Nov 15 at 19:39
@sharkbait Actually $mathbb{E}((X_{t_0}')^2)<infty$ follows from $$mathbb{E} left[ left( frac{X_t-X_{t_0}}{t-t_0} - X_{t_0}' right)^2 right]< infty$$ for $t$ close to $t_0$ and the fact that $(X_t)_t$ is square-integrable.
– saz
Nov 15 at 19:48
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Yes, that's true.
Using the elementary inequality $$(x+y)^2 leq 2x^2+2y^2, qquad x,y in mathbb{R}$$ we find
$$begin{align*} (X_t-X_{t_0})^2 &= (t-t_0)^2 left( frac{X_t-X_{t_0}}{t-t_0} - X_{t_0}' + X_{t_0}' right)^2 \ &leq 2 (t-t_0)^2 left( frac{X_t-X_{t_0}}{t-t_0}-X_{t_0}' right)^2 + 2(t-t_0)^2 (X_{t_0}')^2. end{align*}$$
Taking expectations on both sides we get
$$begin{align*} mathbb{E}((X_t-X_{t_0})^2) &leq 2(t-t_0)^2 underbrace{mathbb{E} left[ left( frac{X_t-X_{t_0}}{t-t_0}-X_{t_0}' right)^2 right]}_{xrightarrow{t to t_0} 0} + 2 underbrace{(t-t_0)^2}_{xrightarrow{t to t_0} 0} mathbb{E}((X_{t_0}')^2) \ &xrightarrow{t to t_0} 0. end{align*}$$
answered Nov 15 at 19:32
saz
76.3k755117
Yes, that's true.
Using the elementary inequality $$(x+y)^2 leq 2x^2+2y^2, qquad x,y in mathbb{R}$$ we find
$$begin{align*} (X_t-X_{t_0})^2 &= (t-t_0)^2 left( frac{X_t-X_{t_0}}{t-t_0} - X_{t_0}' + X_{t_0}' right)^2 \ &leq 2 (t-t_0)^2 left( frac{X_t-X_{t_0}}{t-t_0}-X_{t_0}' right)^2 + 2(t-t_0)^2 (X_{t_0}')^2. end{align*}$$
Taking expectations on both sides we get
$$begin{align*} mathbb{E}((X_t-X_{t_0})^2) &leq 2(t-t_0)^2 underbrace{mathbb{E} left[ left( frac{X_t-X_{t_0}}{t-t_0}-X_{t_0}' right)^2 right]}_{xrightarrow{t to t_0} 0} + 2 underbrace{(t-t_0)^2}_{xrightarrow{t to t_0} 0} mathbb{E}((X_{t_0}')^2) \ &xrightarrow{t to t_0} 0. end{align*}$$
answered Nov 15 at 19:32
saz
76.3k755117
answered Nov 15 at 19:32
saz
76.3k755117
answered Nov 15 at 19:32
saz
76.3k755117
answered Nov 15 at 19:32
saz
76.3k755117
76.3k755117
Thank you! One note: from this proof, it looks like this should be true only when $X_{t_0}'$ is second order, since we require that $mathbb{E}[(X_{t_0}')^2]$ be finite. So this would apply when $X_t$ is WSS, but may not work otherwise.
– sharkbait
Nov 15 at 19:39
@sharkbait Actually $mathbb{E}((X_{t_0}')^2)<infty$ follows from $$mathbb{E} left[ left( frac{X_t-X_{t_0}}{t-t_0} - X_{t_0}' right)^2 right]< infty$$ for $t$ close to $t_0$ and the fact that $(X_t)_t$ is square-integrable.
– saz
Nov 15 at 19:48
add a comment |
Thank you! One note: from this proof, it looks like this should be true only when $X_{t_0}'$ is second order, since we require that $mathbb{E}[(X_{t_0}')^2]$ be finite. So this would apply when $X_t$ is WSS, but may not work otherwise.
– sharkbait
Nov 15 at 19:39
@sharkbait Actually $mathbb{E}((X_{t_0}')^2)<infty$ follows from $$mathbb{E} left[ left( frac{X_t-X_{t_0}}{t-t_0} - X_{t_0}' right)^2 right]< infty$$ for $t$ close to $t_0$ and the fact that $(X_t)_t$ is square-integrable.
– saz
Nov 15 at 19:48
Thank you! One note: from this proof, it looks like this should be true only when $X_{t_0}'$ is second order, since we require that $mathbb{E}[(X_{t_0}')^2]$ be finite. So this would apply when $X_t$ is WSS, but may not work otherwise.
– sharkbait
Nov 15 at 19:39
Thank you! One note: from this proof, it looks like this should be true only when $X_{t_0}'$ is second order, since we require that $mathbb{E}[(X_{t_0}')^2]$ be finite. So this would apply when $X_t$ is WSS, but may not work otherwise.
– sharkbait
Nov 15 at 19:39
@sharkbait Actually $mathbb{E}((X_{t_0}')^2)<infty$ follows from $$mathbb{E} left[ left( frac{X_t-X_{t_0}}{t-t_0} - X_{t_0}' right)^2 right]< infty$$ for $t$ close to $t_0$ and the fact that $(X_t)_t$ is square-integrable.
– saz
Nov 15 at 19:48
@sharkbait Actually $mathbb{E}((X_{t_0}')^2)<infty$ follows from $$mathbb{E} left[ left( frac{X_t-X_{t_0}}{t-t_0} - X_{t_0}' right)^2 right]< infty$$ for $t$ close to $t_0$ and the fact that $(X_t)_t$ is square-integrable.
– saz
Nov 15 at 19:48
add a comment |
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