I thought this was an interesting problem. Although I haven't went so far as to find a general form; I believe it's not too difficult to do.

There are a few (argumentatively non-practical assumptions) that need to be made, however.

1) Each one unit time step corresponds to one R unit up or down only (no flat).

2) You exit only when -1R units below the origin is hit, or a trailing stop is hit (multiple of R). It would make more sense to exit at the peak if it was known in advance, but for now we'll use the trailing stop exit with a fixed maximum peak.

From there just draw a decision tree that bifurcates to the right. Each time a position must be exited marks a leaf on the DT that represents the probability and payoff of that leaf.

Let's use an example:

Suppose our peak value is +2R and the trailing stop is -1R unit from any value in time, and always exit at -1R below origin if it is hit.

There are two possible branches to start:

1) 1/2*(1R)

1/2*(-1R)

It's easy to see that the expectation at the first possible step is .5R-.5R=0

2) The next step is a bit trickier, you know that 50% of all possible trades must result in -1R (from the first step assumption) so that node is fixed at 1/2*(-1R). However, the top node may continue to +2R with 1/2^n or (1/2*1/2=) 1/4 chance,

and 0R with 1/4 chance (simply step up one unit or down one unit). Notice again, that here you have a terminating node at 1/4*0R , so that node becomes fixed for all time because it was a -1R retrace (you will always exit when this node is hit.

3) This leaves two fixed nodes 1/2*(-1R) + 1/4*(0R), and these nodes must account for 1/2+1/4 of all trades (on average), leaving only 1/4 of possible trades left. Well the next possible trade can only be one: which is from +2R to +1R (it can't go higher because the max is fixed and no lower value can creep up because it was terminated at the last fixed node). Therefore, you have 100%*(1/4)*(+1R).

The path possibilities are closed, so you have net expectation:

1/2*(-R)+ 1/4*0R+ 1/4*1R= -1/4*R, which is not zero.

* If your criteria was exit at -1R from origin, -1R trailing, or exit at max when hit, the expectation result would be:

1/2*(-R)+1/4*(0*R)+1/4*(2*R) = 0

You can easily extend this concept to any number of maximum values (where max value is equal to max step value), and trailing stop values as integer multiples of R. As you go farther out the positive edge begins to win. It can't be too difficult to extend it to a compact general solution as well.

All that being said, it makes a lot more sense to empirically just write down just a few observed samples of trade entry and exit to get a better feel for the true statistics (after about 30 or so they should give at least some good feeling regardless of all the other variables you mention, the trade difference as a multiple of R is the only factor neccessary to record), and even better simulate via monte carlo boxcar methods, with many window samples.

P.S. I thought about this during bed, but looked back and saw your original question was without a trailing stop. The same concept applies, only now there will only be two terminating exit values which are -1R and +Rmax, just add all the possible paths and prob*payoffs of the branches until your exit criteria is met (i.e. the max value is hit by any of the paths = termination, the rest of the paths will eventually end at -1R termination).

There are a few (argumentatively non-practical assumptions) that need to be made, however.

1) Each one unit time step corresponds to one R unit up or down only (no flat).

2) You exit only when -1R units below the origin is hit, or a trailing stop is hit (multiple of R). It would make more sense to exit at the peak if it was known in advance, but for now we'll use the trailing stop exit with a fixed maximum peak.

From there just draw a decision tree that bifurcates to the right. Each time a position must be exited marks a leaf on the DT that represents the probability and payoff of that leaf.

Let's use an example:

Suppose our peak value is +2R and the trailing stop is -1R unit from any value in time, and always exit at -1R below origin if it is hit.

There are two possible branches to start:

1) 1/2*(1R)

1/2*(-1R)

It's easy to see that the expectation at the first possible step is .5R-.5R=0

2) The next step is a bit trickier, you know that 50% of all possible trades must result in -1R (from the first step assumption) so that node is fixed at 1/2*(-1R). However, the top node may continue to +2R with 1/2^n or (1/2*1/2=) 1/4 chance,

and 0R with 1/4 chance (simply step up one unit or down one unit). Notice again, that here you have a terminating node at 1/4*0R , so that node becomes fixed for all time because it was a -1R retrace (you will always exit when this node is hit.

3) This leaves two fixed nodes 1/2*(-1R) + 1/4*(0R), and these nodes must account for 1/2+1/4 of all trades (on average), leaving only 1/4 of possible trades left. Well the next possible trade can only be one: which is from +2R to +1R (it can't go higher because the max is fixed and no lower value can creep up because it was terminated at the last fixed node). Therefore, you have 100%*(1/4)*(+1R).

The path possibilities are closed, so you have net expectation:

1/2*(-R)+ 1/4*0R+ 1/4*1R= -1/4*R, which is not zero.

* If your criteria was exit at -1R from origin, -1R trailing, or exit at max when hit, the expectation result would be:

1/2*(-R)+1/4*(0*R)+1/4*(2*R) = 0

You can easily extend this concept to any number of maximum values (where max value is equal to max step value), and trailing stop values as integer multiples of R. As you go farther out the positive edge begins to win. It can't be too difficult to extend it to a compact general solution as well.

All that being said, it makes a lot more sense to empirically just write down just a few observed samples of trade entry and exit to get a better feel for the true statistics (after about 30 or so they should give at least some good feeling regardless of all the other variables you mention, the trade difference as a multiple of R is the only factor neccessary to record), and even better simulate via monte carlo boxcar methods, with many window samples.

P.S. I thought about this during bed, but looked back and saw your original question was without a trailing stop. The same concept applies, only now there will only be two terminating exit values which are -1R and +Rmax, just add all the possible paths and prob*payoffs of the branches until your exit criteria is met (i.e. the max value is hit by any of the paths = termination, the rest of the paths will eventually end at -1R termination).

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