# Wiener process with a random mean [closed]

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.

## closed as not a real question by Did, Jim, rschwieb, Andreas Caranti, Asaf Karagila♦Mar 9 '13 at 22:00

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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
• @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". – Ilya Feb 20 '13 at 13:07
• 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. – Ilya Feb 20 '13 at 16:20
• 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

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.