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I have found this kind of stochastic process $$ dX=dW-{\rm sgn}(dW)dt. $$ What would the probability distribution be for $X$ assuming that the distribution for ${\rm sgn}(dW)$ is a Bernoulli with $p=\frac{1}{2}$? Thanks.

Jon
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    The process $t\mapsto\mathrm{sgn}(\mathrm dW_t)$ does not exist. – Did Feb 17 '13 at 13:32
  • @did: Sorry, I meant the sign of the displacements in the Brownian motion. What is the right expression for this signum random process? Thanks. – Jon Feb 17 '13 at 13:40
  • This is something that was already explained: on every interval $(s,t)$ there exists as many points $s\lt s_1\lt t_1\lt\cdots\lt s_n\lt t_n\lt t$ as one wants such that $W_{s_1}\lt W_{t_1}\gt W_{s_2}\lt\cdots\gt W_{s_n}\lt W_{t_n}$, hence the process $\mathrm{sgn}(\mathrm dW)$ does not exist. – Did Feb 17 '13 at 13:51
  • @Did: I agree of course but you should read better my previous comment. I asked as I should write this , changing the notation. Please refer to http://math.stackexchange.com/questions/98479/probability-distribution-of-sign-changes-in-brownian-motion. – Jon Feb 17 '13 at 15:43
  • @Did: Of course, Matlab does not care about your beautiful non-existence proof and just compute it:nstep = 5000000; dt = 50; t=0:dt/nstep:dt; t=t(1:length(t)-1); B = normrnd(0,sqrt(dt/nstep),1,nstep); dB=zeros(1,nstep); dS=zeros(1,nstep); dB = cumsum(B); dS = sign(dB); dA = dB./dS; – Jon Feb 17 '13 at 15:53
  • A way out of NARQ here would be to transcribe mathematically the discrete algorithm this Matlab program corresponds to, and to ask a question about the mathematical properties of the result of this algorithm. – Did Feb 17 '13 at 18:20
  • @Did: I agree. But I did not want to resurrect a buried corpse. This was not in my mind. I have a process with a random mean. What is the probability distribution? How can I reformulate my question to get this answer? – Jon Feb 17 '13 at 18:30
  • Anything not clear? – SBF Feb 20 '13 at 09:10
  • @Ilya: Of course, thanks for the answer but the problem here is similar to the proposal of Dirac "function". When it was put forward nobody believed it. – Jon Feb 20 '13 at 11:46
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    @Jon: 1. You are welcome, but you may guess that even just typing an answer takes some effort, so if you have an opinion about it in future please fell free leave it by yourself without waiting to be contacted by the author. $$

    $$ 2. I showed that limit of your MATLAB manipulations is $0$ a.s. and thus whatever you defined is just trivial whenever exists. $$

    $$ 3. There is point of "believing" in some math object. It is either defined formally, or not. Dirac $\delta$ is defined formally. Moreover, it is useful, so... $$

    $$ 4. ... don't overestimate your "proposals".

     – SBF Feb 20 '13 at 13:07
  • @Ilya: 1) I have been on this group for more than one year now and I know usages. But, of course, if an answer does not seem to fit what I am asking for, I am not forced to vote or accept it. Thanks anyway. 2) I know that this has null measure in a Ito sense so you are not saying nothing new to me. 3) Of course, a posteriori one can say this happily. When Dirac firstly proposed it there was no such consensus. 4) Here there is a longstanding debate that involved also Did (Didier Piau) last year. But I do not want to wake up the beast again. So, feel free to do your search. – Jon Feb 20 '13 at 15:00
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    By no means I tried forcing you to upvote my answer, but clearly my comment 1. can be read in any way one has in his mind. Rather, I meant that it's certainly an unwelcomed behavior not to react on answers given to your question, and it's pity that you didn't get that. $$

    $$ You are asking to help you reformulating the question about the stochastic process with a random mean. Such a reformulation shall be based on examples/motivation. The only example provided is trivial and doesn't give any insight in the concept of random mean. Try providing more examples. So far -1 for ill-posed question.

     – SBF Feb 20 '13 at 16:20
  • @Ilya: -1 for missing of understanding. – Jon Feb 20 '13 at 20:24
  • Debate? What debate? There is/was never a debate (similarly, there is no question here). Fooling yourself once again... – Did Feb 27 '13 at 08:07
  • @Did: I cannot appreciate your arrogance. Your "beautiful" proof clashes against numerical evidence. Take some fun trying to understand it. I think that in the end of the story there will be just one fool and I avoid to name it. – Jon Feb 27 '13 at 08:54
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    Well, this is kind of reassuring, in a way: you are as unable to listen to other mathematicians' (in this case, @Ilya's) explanations (especially the To elaborate on your MATLAB algorithm paragraph) as to mine. As usual, you can cry and yell everything you want, as long as no definition of the process $\mathrm{sgn}(\mathrm dW)$ is provided, the question makes no sense. – Did Feb 27 '13 at 15:34
  • @Did: Don't worry, be happy. – Jon Feb 27 '13 at 17:36
  • Jon, don't you mean sgn(W(s)) rather than sgn(dW(s)) – Nick Mar 02 '13 at 19:03
  • @Nick: Matlab code accounts for the sign of displacements. This can also be taken one by one by hand. It is possible that a notational matter could have generated much ado about nothing. – Jon Mar 02 '13 at 19:16

1 Answers1

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You have to be careful, when dealing with SDEs, as their "intuitive" meaning can be misleading. Let us even for a moment forget about the diffusive part since whenever the meaningful solution $X$ of the SDE you wrote exists, it shall satisfy $$ X_t = X_0 + \int_0^t \mathrm dW_s - \int_0^t \mathrm{sgn}(\mathrm dW_s)\; \mathrm ds = X_0 + W_t - \int_0^t \mathrm{sgn}(\mathrm dW_s)\; \mathrm ds $$ and thus we can focus on $X_t - W_t$ that satisfy $$ \mathrm d(W_t-X_t) = \mathrm{sgn}(\mathrm dW_t)\;\mathrm dt. $$

Let us consider a more general formal equation $$ \mathrm dY_t = \mathrm{sgn}(\mathrm df_t)\;\mathrm dt \tag{1} $$ where $f:[0,\infty)\to\Bbb R$ is some continuous function, not necessary a trajectory of a Brownian motion. The only meaningful solution of such equation shall be $$ Y_t = Y_0 + \int_0^t \mathrm{sgn}(\mathrm df_s)\mathrm ds. $$ It would be naturally to assume that $\mathrm{sgn}(\mathrm df_s)$ is some function of the variable $s$, as we have to integrate this function. However, even if $f$ is a smooth function, its differential is not a function of $s$ only, it is also as function of $\mathrm ds$: $$ \mathrm df_s:= f'(s)\;\mathrm ds $$ and in the end $(1)$ makes a very little sense even in case $f$ is a smooth function. A better-defined equation is $$ \mathrm dZ_t = \mathrm{sgn}(f'_t)\; \mathrm dt $$ which indeed has a well-defined solution in case $f$ is smooth, and seems to be something that you are looking for. However, $f$ has to satisfy some regularity assumption in such case. As an example, if $\mathrm{sgn} f_t=:\xi_t$ is a collection of iid random variables that are $\pm1$ equiprobably, than it is even a non-trivial question whether $$ \{t\in [0,1]:\xi_t = 1\} $$ is a measurable set which is often needed for the integrability.

To elaborate on your MATLAB algorithm, and the comment by Did, let us focus on the interval $[0,1]$ and consider your integral sums $$ S_n = \sum_{k=0}^{n-1}\mathrm{sgn}(\Delta W_k)\cdot \Delta t_k $$ where $\Delta t_k = t_{k+1} - t_k$, $\Delta W_k = W(t_{k+1}) - W(t_k)$ and $$ 0 = t_0<t_1<\dots<t_n = 1 $$ is a partition of $[0,1]$. Suppose you allow for uniform partitions only, then $$ S_n = \frac1n\sum_{k=0}^{n-1}\xi_k $$ where $\xi_k = \pm1$ equiprobably, and thus by LLN you have $S_n \to \mathsf E\xi_0 = 0$ a.s. I leave non-uniform partition case to you.

SBF
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