What is a white noise ? What is the derivate of the Brownian motion? [duplicate]












0












$begingroup$



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 ?










share|cite|improve this question









$endgroup$



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.























    0












    $begingroup$



    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 ?










    share|cite|improve this question









    $endgroup$



    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.





















      0












      0








      0





      $begingroup$



      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 ?










      share|cite|improve this question









      $endgroup$





      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      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.
























          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          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 $xi_t=dB_t$.






          share|cite|improve this answer









          $endgroup$




















            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            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 $xi_t=dB_t$.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              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 $xi_t=dB_t$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                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 $xi_t=dB_t$.






                share|cite|improve this answer









                $endgroup$



                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 $xi_t=dB_t$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 16 '18 at 10:29









                SurbSurb

                38k94375




                38k94375















                    Popular posts from this blog

                    Quarter-circle Tiles

                    build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

                    Mont Emei