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Assume such a "random walk with reset" X(t) is defined based on Wiener process (GBM)

X(t + dt) =  
   0, by probability lambda * dt; or, 
   X(t) + dW, by probability (1 - lambda * dt)

, where lambda is a constant, and

dW = W(t+dt) - W(t), 

while W is a Wiener process (GBM).

What is the differential form for X? I'd like to have something like

dX = ( 1 - lambda * dt) * dW + lambda * dt * ....

so that I could use this with Ito's lemma to do some calculus.

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The random walk that is in your mind must have jump discontinuities, which is not allowed in Ito process. So it should be a Levy process. I know not much about this subject, though... – Sangchul Lee Mar 10 '12 at 13:47
@sos440 , thanks for the comments, however for such X, dX is dependent on X(t), so it's not i.i.d. Then it's even not a "random process", other than Levy.... I'm a bit lost here... – athos Mar 10 '12 at 15:20
What is GBM? I assume BM is Brownian Motion, but what is the G? – Nate Eldredge Mar 10 '12 at 16:43
@Nate: GBM is usually a geometric Brownian motion: $$ dX_t = \mu X_tdt+\sigma X_t dW_t. $$ Though it's not a Wiener process, so I'm a bit confused about OP. – Ilya Mar 10 '12 at 21:40
@Athos: Let me consider the case when $W_t = B_t$ is just a Brownian motion and $N_t$ is a Poisson process with an intensity $\lambda$, then the process $$ \mathrm dX_t = -X_t\cdot \mathrm dN_t + \mathrm dB_t $$ should satisfy your conditions. Informally speaking, when the Poisson process does not have a jump in $[t,t+dt)$ (which is of probability $1-\lambda \cdot\mathrm dt$) then $X_t$ evolves like a Brownian motion, while at a moment of jump - it jumps to the $0$ since $$ -X_t\cdot \mathrm dN_t = -X_t \cdot \Delta N_t $$ where $\Delta N_t = N_t - N_{t-}$ – Ilya Mar 10 '12 at 21:45
up vote 2 down vote accepted

Ok, let me expand my comment into an answer. Let $W_t$ be either a geometric Brownian motion, or a Brownian motion, or any Ito process - it does not matter for our case, we only will use the continuity properties of the process.

The idea is to use the Poisson process in order to describe jumps, namely let $N_t$ be the Poisson process with a given intensity $\lambda$, so the probability of jump for $N_t$ in the interval $[t,t+dt]$ is roughly $\lambda\cdot dt$.

The question now is: the jumps of a Poisson process is always equal to $1$ whenever it happens, so how to use it for your problem? The good thing about a Poisson process it that it has piecewise-constant paths, which are of finite variatino with probability one so the integral is easy to define: for any function $f:\mathbb R_{+}\to\mathbb R$ which has limits from the left $$ \int\limits_0^t f(t)dN_t = \sum\limits_{s\leq t}f(s-)\Delta N_s \tag{1} $$ where $f(s -) = \lim\limits_{r\uparrow s}f(r)$ and $\Delta N_s = N_s - N_{s-}$. Note that a Poisson process has finitely many jumps in each interval $[0,t]$ with probability $1$, so the summation in $(1)$ is finite.

If we manage to make $X_t$ jump to zero at each moment $N_t$ has a jump, we solve the problem. How should we derive an equation, though? Well, suppose that $N_t$ has jumps at times $$ t_0<t_1<\dots<t_n<\dots $$ and consider $t\in [t_{i},t_{i+1})$ - there the process $X_t$ behaves as $W_t$ since there are no jumps: $$ X_t = (X_{t_i}-W_{t_i})+W_t\quad\text{ for }t\in[t_i,t_{i+1}).\tag{2} $$

If we consider a jump at time $t_{i+1}$ downwards, of the absolute value $X_{t_{i+1}}$ it will make everything work. On the other hand, whenever you work with integrals w.r.t. the Poisson process you have to consider limits from the left as in $(1)$. In our case it is ok, since any of processes $W_t$ we considered above has continuous paths and so $W_{t_{i+1}-} = W_{t_i}$. Clearly, $-X_{t-}\cdot\Delta N_t$ is what we need since it makes jumps only together $N_t$ makes them and these jumps are of an appropriate value and direction.

As a result, the process defined by $$ dX_t = - X_{t-}dN_t + dW_t \tag{3} $$ solves your problem. The explicit formula for such process is given in $(2)$.

For the stochastic calculus of such process, a very nice reference is "Stochastic Calculus for Finance II" by S. Shreve, Chapter 11.

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is the X defined in (3) a semi-martingale? – athos Jun 3 '12 at 4:04
@athos: would you first fix OP as requested? – Ilya Jun 3 '12 at 17:08

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