Consider the simplfied math. model for asset price (it is nevertheless quite practical for specific situations see "PS" part below) assume price "p(n)" at moment "n" is equal to N(0,1) - i.i.d - independent Gaussians.

Assume we can possess not more than one asset in any moment. I.e. if buy it we can keep it or sell it, but cannot buy another until we sold the previous one.

Our profit is accumulation of differences between the prices we bought and sold.

**Informal Question** What are the "best" trading strategies ?

**Mathematically rigorous Question 1** What is the strategy which will maximize expectation value of the profit for trading time n=1...N ?

**Conjecture (YES/NO)** Is it true that the answer to the question above is given by the following simple strategy if the price is greater than zero - sell, if less than zero buy.
See MatLab code for details.

**Variations on the question**

Profit here is random variable, and what means "best" random variable is ambiguous. The simplest version is to take expectation as in question above, but more profound measures of quality can be something like E(profit)/std(profit), I mean taking "risk" into account.

Here is MatLab code for simple strategy - buy if price < threshold1, sell if price > threshold2. Results of simulation suggests that best choice is threshold1=threshold2=0. It motivates the conjecture above.

`enter code here`

function profit = TradeStrategy2(threshold1, threshold2 )

```
len = 1e6;
p = randn(1,len);
flagBought = 0; profit = 0;
for k=1:len
if (p(k) < threshold1 ) && (flagBought == 0)
pSave = p(k);
flagBought = 1;
elseif (p(k) > threshold2 ) && (flagBought == 1)
profit = profit + (p(k)-pSave) ;
flagBought = 0;
end;
end;
fprintf(1,'Strategy 2, threshold1 =, %f, threshold2 =, %f profit/len = , %f, \n ', threshold1, threshold2, profit/len );
profit = profit/len;
```

end

## Simulation results:

Strategy 2, threshold1 =, -0.000000, threshold2 =, 0.000000 profit/len = , 0.398951,

Strategy 2, threshold1 =, -0.100000, threshold2 =, 0.100000 profit/len = , 0.395956,

Strategy 2, threshold1 =, -0.100000, threshold2 =, 1.000000 profit/len = , 0.281029,

Strategy 2, threshold1 =, -1.000000, threshold2 =, 0.100000 profit/len = , 0.281722,

Strategy 2, threshold1 =, -1.000000, threshold2 =, 1.000000 profit/len = , 0.242517,

Strategy 2, threshold1 =, -3.000000, threshold2 =, 3.000000 profit/len = , 0.004250,

Similar questions can be asked for arbitrary random process. I guess, that the general answer should also be known, however I have not yet received answer at

PS

Motivation for considering such simplified model of the price is the following. Consider the "trend" part of some real price, the simplest model for the trend is linear trend: p(t) = At + B + noise. The most simple situation is to take A=0. So we almost come to our model. Please notice that choice of "B" does not affect the optimal stategy, so we can assume it to be zero.

However, physiologically, it is better to assume it to be B = million or billion - very big number. It will explain the second assumption of the model - you cannot buy two assets at one time - that it is just because your resources are limited - you have 1 million (or billion), but not two.