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Exact requirements for interchanging mean square derivatives with limits and expectations
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I've recently took an applied course in stochastic processes, and I'm far from satisfied with how we dealt with mean square derivatives.
Firstly, we defined a mean square derivative of a process $mathbb X(t)$ with an underlying probability space $(mathbb R, Sigma, mathbb P)$ as a process denoted by $mathbb X'(t)$ such that
$$displaystylelim_{hto 0}mathbb Eleft[left(frac{mathbb X(t+h)-mathbb X(t)}{h} - mathbb X'(t)right)^2right] = 0.$$
In the exercises, we seemed to replace $mathbb X'(t)$ with a difference quotient and a limit in the following fashion:
$$mathbb E[f(mathbb X'(t))] = mathbb Eleft[displaystylelim_{hto 0} fleft(frac{mathbb X(t+h)-mathbb X(t)}{h}right)right] = displaystylelim_{hto 0} mathbb Eleft[ fleft(frac{mathbb X(t+h)-mathbb X(t)}{h}right)right]$$
for sufficiently "nice" functions $f$.
How this worked was absolutely not explained and just ad-hoced away. When can you actually do this sort of thing?
probability-theory measure-theory stochastic-processes stochastic-calculus
asked Jan 18 at 20:27
Markus KlyverMarkus Klyver
392314
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add a comment |
$begingroup$
I've recently took an applied course in stochastic processes, and I'm far from satisfied with how we dealt with mean square derivatives.
Firstly, we defined a mean square derivative of a process $mathbb X(t)$ with an underlying probability space $(mathbb R, Sigma, mathbb P)$ as a process denoted by $mathbb X'(t)$ such that
$$displaystylelim_{hto 0}mathbb Eleft[left(frac{mathbb X(t+h)-mathbb X(t)}{h} - mathbb X'(t)right)^2right] = 0.$$
In the exercises, we seemed to replace $mathbb X'(t)$ with a difference quotient and a limit in the following fashion:
$$mathbb E[f(mathbb X'(t))] = mathbb Eleft[displaystylelim_{hto 0} fleft(frac{mathbb X(t+h)-mathbb X(t)}{h}right)right] = displaystylelim_{hto 0} mathbb Eleft[ fleft(frac{mathbb X(t+h)-mathbb X(t)}{h}right)right]$$
for sufficiently "nice" functions $f$.
How this worked was absolutely not explained and just ad-hoced away. When can you actually do this sort of thing?
probability-theory measure-theory stochastic-processes stochastic-calculus
asked Jan 18 at 20:27
Markus KlyverMarkus Klyver
392314
$endgroup$
$begingroup$
Your statement has an obvious error. It should have $frac{X(t+h)-X(t)}{h}$ in several places.
$endgroup$
– herb steinberg
Jan 18 at 20:34
add a comment |
$begingroup$
I've recently took an applied course in stochastic processes, and I'm far from satisfied with how we dealt with mean square derivatives.
Firstly, we defined a mean square derivative of a process $mathbb X(t)$ with an underlying probability space $(mathbb R, Sigma, mathbb P)$ as a process denoted by $mathbb X'(t)$ such that
$$displaystylelim_{hto 0}mathbb Eleft[left(frac{mathbb X(t+h)-mathbb X(t)}{h} - mathbb X'(t)right)^2right] = 0.$$
In the exercises, we seemed to replace $mathbb X'(t)$ with a difference quotient and a limit in the following fashion:
$$mathbb E[f(mathbb X'(t))] = mathbb Eleft[displaystylelim_{hto 0} fleft(frac{mathbb X(t+h)-mathbb X(t)}{h}right)right] = displaystylelim_{hto 0} mathbb Eleft[ fleft(frac{mathbb X(t+h)-mathbb X(t)}{h}right)right]$$
for sufficiently "nice" functions $f$.
How this worked was absolutely not explained and just ad-hoced away. When can you actually do this sort of thing?
probability-theory measure-theory stochastic-processes stochastic-calculus
asked Jan 18 at 20:27
Markus KlyverMarkus Klyver
392314
$endgroup$
I've recently took an applied course in stochastic processes, and I'm far from satisfied with how we dealt with mean square derivatives.
Firstly, we defined a mean square derivative of a process $mathbb X(t)$ with an underlying probability space $(mathbb R, Sigma, mathbb P)$ as a process denoted by $mathbb X'(t)$ such that
$$displaystylelim_{hto 0}mathbb Eleft[left(frac{mathbb X(t+h)-mathbb X(t)}{h} - mathbb X'(t)right)^2right] = 0.$$
In the exercises, we seemed to replace $mathbb X'(t)$ with a difference quotient and a limit in the following fashion:
$$mathbb E[f(mathbb X'(t))] = mathbb Eleft[displaystylelim_{hto 0} fleft(frac{mathbb X(t+h)-mathbb X(t)}{h}right)right] = displaystylelim_{hto 0} mathbb Eleft[ fleft(frac{mathbb X(t+h)-mathbb X(t)}{h}right)right]$$
for sufficiently "nice" functions $f$.
How this worked was absolutely not explained and just ad-hoced away. When can you actually do this sort of thing?
probability-theory measure-theory stochastic-processes stochastic-calculus
probability-theory measure-theory stochastic-processes stochastic-calculus
asked Jan 18 at 20:27
Markus KlyverMarkus Klyver
392314
asked Jan 18 at 20:27
Markus KlyverMarkus Klyver
392314
edited Jan 18 at 20:35
Markus Klyver
asked Jan 18 at 20:27
Markus KlyverMarkus Klyver
392314
asked Jan 18 at 20:27
Markus KlyverMarkus Klyver
392314
asked Jan 18 at 20:27
Markus KlyverMarkus Klyver
392314
392314
$begingroup$
Your statement has an obvious error. It should have $frac{X(t+h)-X(t)}{h}$ in several places.
$endgroup$
– herb steinberg
Jan 18 at 20:34
add a comment |
$begingroup$
Your statement has an obvious error. It should have $frac{X(t+h)-X(t)}{h}$ in several places.
$endgroup$
– herb steinberg
Jan 18 at 20:34
$begingroup$
Your statement has an obvious error. It should have $frac{X(t+h)-X(t)}{h}$ in several places.
$endgroup$
– herb steinberg
Jan 18 at 20:34
$begingroup$
Your statement has an obvious error. It should have $frac{X(t+h)-X(t)}{h}$ in several places.
$endgroup$
– herb steinberg
Jan 18 at 20:34
add a comment |
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$begingroup$
Your statement has an obvious error. It should have $frac{X(t+h)-X(t)}{h}$ in several places.
$endgroup$
– herb steinberg
Jan 18 at 20:34