# Can grid strategy make profit on a random walk?

I've seen this thread, but it's a little too advanced for me. I haven't studied finance, just recently had some experience with grid strategy bots on cryptocurrency exchanges (in future markets), and become interested in this variant of grid strategy:

Suppose the price of an asset is a random walk in the following manner: It either changes +a% or -b% each second, with equal 1/2 to 1/2 probabalities, and additionally we have (1+0.01a)*(1-0.01b)=1. For simplicity, let's call u=(1+0.01a) and d=(1-0.01b).

I have 1$at t=0, and the price is p=p0. At this moment, I set an infinite number of buy and sell orders. Buy orders: B1: 1/2 at p0*d^1 B2: 1/4 at p0*d^2 B3: 1/8 at p0*d^3 . . . Sell orders: S1: 1/2 at p0*u^1 S1: 1/4 at p0*u^2 S1: 1/8 at p0*u^3 . . . At t=1, I set the take-profit at p=p0, and wait until some t=t', when the price is p=p0 again. At that moment I have slightly more than 1$. I repeat the same process again, but this time, the size of all orders are multiplied by the tiny increase I had in my balance.

I can't see where this strategy (on such asset with exactly that random description) could go wrong. So I could go on and calculate the expected duration between zero crossings, and that gives very slow but exponentially growing profits?

I think I must have some or many errors in here, it doesn't seem right. Please point them out for me.

Thanks a lot.

Update: I thought I might need some basics, so I asked another question.

• A random walk is not mean reversing. Oct 7 at 10:44
• Right, but what can be said about the distribution of length of time between two consecutive zero (or any other points) crossings? Can we then calculate the expected time for 100 zero crossings? Oct 7 at 11:50