What is a white noise ? What is the derivate of the Brownian motion? [duplicate]
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This question already has an answer here:
What is “white noise” and how is it related to the Brownian motion?
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Could someone explain me what is a whit noise ? In my course it's written that it's the derivate of a Brownian motion, but how can it be the derivative of something that doesn't exist ?
probability brownian-motion noise
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marked as duplicate by Did, user10354138, KReiser, Eric Wofsey, Andrew Dec 22 '18 at 6:16
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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$begingroup$
This question already has an answer here:
What is “white noise” and how is it related to the Brownian motion?
2 answers
Could someone explain me what is a whit noise ? In my course it's written that it's the derivate of a Brownian motion, but how can it be the derivative of something that doesn't exist ?
probability brownian-motion noise
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marked as duplicate by Did, user10354138, KReiser, Eric Wofsey, Andrew Dec 22 '18 at 6:16
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
What is “white noise” and how is it related to the Brownian motion?
2 answers
Could someone explain me what is a whit noise ? In my course it's written that it's the derivate of a Brownian motion, but how can it be the derivative of something that doesn't exist ?
probability brownian-motion noise
$endgroup$
This question already has an answer here:
What is “white noise” and how is it related to the Brownian motion?
2 answers
Could someone explain me what is a whit noise ? In my course it's written that it's the derivate of a Brownian motion, but how can it be the derivative of something that doesn't exist ?
This question already has an answer here:
What is “white noise” and how is it related to the Brownian motion?
2 answers
probability brownian-motion noise
probability brownian-motion noise
asked Dec 16 '18 at 10:10
user623855user623855
1507
1507
marked as duplicate by Did, user10354138, KReiser, Eric Wofsey, Andrew Dec 22 '18 at 6:16
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Did, user10354138, KReiser, Eric Wofsey, Andrew Dec 22 '18 at 6:16
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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1 Answer
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Take a derivable function $f:mathbb Rlongrightarrow mathbb R$. The definition of the derivative function is $$f'(x)=lim_{hto 0}frac{f(x+h)-f(x)}{h}.tag{$D_1$}$$
The thing is, if $f$ is derivable and $f'$ is integrable (in Lebesgue sense), then $$f(x)=f(0)+int_0^x f'(t)dt.tag{$D_2$}$$
In fact, you can use $(D_2)$ as a more general definiton of the derivative. I.e, $f$ is derivable if there is $g$ integrable s.t. $$f(x)=f(0)+int_0^x g(t)dt.$$
One can prove that $g$ is unique in the sense that if $h$ is s.t. $$f(x)=f(0)+int_0^x h(t)dt$$
then $h=g$ a.e. An other thing is if $f$ is $mathcal C^1$ in then is derivable in $(D_2)$ sense. If $f$ is derivable in $(D_1)$ sense I would says that is also derivable in $(D_2)$ but I'm not completely sure of that (I'll check).
Now, we can do almost the same construction with stochastic process. If $$X_t=X_0+int_0^t f(Y_t,t)dt+int_0^t g(Z_t,t)dB_t,$$ then we write formally the derivative of $X_t$ as $$dX_t=f(Y_t,t)dt+g(Z_t,t)dB_t.$$
Now, we have that $$B_t=int_0^t dB_t.$$ So what we call a white noise is simply $xi_t=dB_t$.
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add a comment |
1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take a derivable function $f:mathbb Rlongrightarrow mathbb R$. The definition of the derivative function is $$f'(x)=lim_{hto 0}frac{f(x+h)-f(x)}{h}.tag{$D_1$}$$
The thing is, if $f$ is derivable and $f'$ is integrable (in Lebesgue sense), then $$f(x)=f(0)+int_0^x f'(t)dt.tag{$D_2$}$$
In fact, you can use $(D_2)$ as a more general definiton of the derivative. I.e, $f$ is derivable if there is $g$ integrable s.t. $$f(x)=f(0)+int_0^x g(t)dt.$$
One can prove that $g$ is unique in the sense that if $h$ is s.t. $$f(x)=f(0)+int_0^x h(t)dt$$
then $h=g$ a.e. An other thing is if $f$ is $mathcal C^1$ in then is derivable in $(D_2)$ sense. If $f$ is derivable in $(D_1)$ sense I would says that is also derivable in $(D_2)$ but I'm not completely sure of that (I'll check).
Now, we can do almost the same construction with stochastic process. If $$X_t=X_0+int_0^t f(Y_t,t)dt+int_0^t g(Z_t,t)dB_t,$$ then we write formally the derivative of $X_t$ as $$dX_t=f(Y_t,t)dt+g(Z_t,t)dB_t.$$
Now, we have that $$B_t=int_0^t dB_t.$$ So what we call a white noise is simply $xi_t=dB_t$.
$endgroup$
add a comment |
$begingroup$
Take a derivable function $f:mathbb Rlongrightarrow mathbb R$. The definition of the derivative function is $$f'(x)=lim_{hto 0}frac{f(x+h)-f(x)}{h}.tag{$D_1$}$$
The thing is, if $f$ is derivable and $f'$ is integrable (in Lebesgue sense), then $$f(x)=f(0)+int_0^x f'(t)dt.tag{$D_2$}$$
In fact, you can use $(D_2)$ as a more general definiton of the derivative. I.e, $f$ is derivable if there is $g$ integrable s.t. $$f(x)=f(0)+int_0^x g(t)dt.$$
One can prove that $g$ is unique in the sense that if $h$ is s.t. $$f(x)=f(0)+int_0^x h(t)dt$$
then $h=g$ a.e. An other thing is if $f$ is $mathcal C^1$ in then is derivable in $(D_2)$ sense. If $f$ is derivable in $(D_1)$ sense I would says that is also derivable in $(D_2)$ but I'm not completely sure of that (I'll check).
Now, we can do almost the same construction with stochastic process. If $$X_t=X_0+int_0^t f(Y_t,t)dt+int_0^t g(Z_t,t)dB_t,$$ then we write formally the derivative of $X_t$ as $$dX_t=f(Y_t,t)dt+g(Z_t,t)dB_t.$$
Now, we have that $$B_t=int_0^t dB_t.$$ So what we call a white noise is simply $xi_t=dB_t$.
$endgroup$
add a comment |
$begingroup$
Take a derivable function $f:mathbb Rlongrightarrow mathbb R$. The definition of the derivative function is $$f'(x)=lim_{hto 0}frac{f(x+h)-f(x)}{h}.tag{$D_1$}$$
The thing is, if $f$ is derivable and $f'$ is integrable (in Lebesgue sense), then $$f(x)=f(0)+int_0^x f'(t)dt.tag{$D_2$}$$
In fact, you can use $(D_2)$ as a more general definiton of the derivative. I.e, $f$ is derivable if there is $g$ integrable s.t. $$f(x)=f(0)+int_0^x g(t)dt.$$
One can prove that $g$ is unique in the sense that if $h$ is s.t. $$f(x)=f(0)+int_0^x h(t)dt$$
then $h=g$ a.e. An other thing is if $f$ is $mathcal C^1$ in then is derivable in $(D_2)$ sense. If $f$ is derivable in $(D_1)$ sense I would says that is also derivable in $(D_2)$ but I'm not completely sure of that (I'll check).
Now, we can do almost the same construction with stochastic process. If $$X_t=X_0+int_0^t f(Y_t,t)dt+int_0^t g(Z_t,t)dB_t,$$ then we write formally the derivative of $X_t$ as $$dX_t=f(Y_t,t)dt+g(Z_t,t)dB_t.$$
Now, we have that $$B_t=int_0^t dB_t.$$ So what we call a white noise is simply $xi_t=dB_t$.
$endgroup$
Take a derivable function $f:mathbb Rlongrightarrow mathbb R$. The definition of the derivative function is $$f'(x)=lim_{hto 0}frac{f(x+h)-f(x)}{h}.tag{$D_1$}$$
The thing is, if $f$ is derivable and $f'$ is integrable (in Lebesgue sense), then $$f(x)=f(0)+int_0^x f'(t)dt.tag{$D_2$}$$
In fact, you can use $(D_2)$ as a more general definiton of the derivative. I.e, $f$ is derivable if there is $g$ integrable s.t. $$f(x)=f(0)+int_0^x g(t)dt.$$
One can prove that $g$ is unique in the sense that if $h$ is s.t. $$f(x)=f(0)+int_0^x h(t)dt$$
then $h=g$ a.e. An other thing is if $f$ is $mathcal C^1$ in then is derivable in $(D_2)$ sense. If $f$ is derivable in $(D_1)$ sense I would says that is also derivable in $(D_2)$ but I'm not completely sure of that (I'll check).
Now, we can do almost the same construction with stochastic process. If $$X_t=X_0+int_0^t f(Y_t,t)dt+int_0^t g(Z_t,t)dB_t,$$ then we write formally the derivative of $X_t$ as $$dX_t=f(Y_t,t)dt+g(Z_t,t)dB_t.$$
Now, we have that $$B_t=int_0^t dB_t.$$ So what we call a white noise is simply $xi_t=dB_t$.
answered Dec 16 '18 at 10:29
SurbSurb
38k94375
38k94375
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