the fundamental risk/reward question

T

traderkay

This has been discussed before in a slightly different form but that's ok. Let's suppose the market is random, ie. the chances of it going up 1 point at any given moment are 1/2, the chances of going down 1 point are 1/2. Let's also suppose that the market moves in 1 point increments up and down. Now what happens if we enter and set a 1 point stop loss and a 3 point target. What is our mathematical expectation? This question mathematically is not that simple, or I'd solve it:) Basically we have to look at all the possible market steps: For example, lets say we're long. Market goes up with prob. of 1/2, then it again has two options, up or down, let's say it goes up again, we're up two points, but then it might go down again going to +1, then it might go to +2 again then +1, +0 and -1 stopping us out. And at every 'junction' there's two events with the 1/2 probabilities. The basic question is what is the probability of us getting to +3 without getting to -1 first? What is the probability of getting to -1 without getting to +3 first? Not obvious at all. But I'm positive this can be mathematically solved, as I've seen a similar problem solved. That problem involved a drunkard standing one step away from the cliff. The Prob of a step toward the cliff is 1/3, the Prob of step away from the cliff is 2/3. What is the probability of the drunkard falling off the cliff? Believe me this is a complicated problem. But I'm sure there's people on ET qualified to solve it.
P.S. If there's even the slightest auto-correlation between moves, ie. it's more probable to go up if we went up on the last movement, vica versa for down, then any moron can get a positive expectation no matter what setup he uses.
 
Quote from traderkay:

Let's suppose the market is random, ie. the chances of it going up 1 point at any given moment are 1/2, the chances of going down 1 point are 1/2. Let's also suppose that the market moves in 1 point increments up and down. Now what happens if we enter and set a 1 point stop loss and a 3 point target. What is our mathematical expectation?
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Unless I've missed something, you have answered the question yourself. Our mathematical expectation is zero, and remains so no matter what stop loss or target we set.
 
not sure whether i get what you mean .. but if you want the probabilities just apply the path rule and multiply the single probabilities
 
I think you're talking about a monte carlo simulation. You would figure the expectation for each path (chance of this path occuring * result) and then adding them.

I could be way off base.. I barely remember my combinatorics/discrete class.

regards,

laz


Quote from traderkay:

the possible market steps: For example, lets say we're long. Market goes up with prob. of 1/2, then it again has two options, up or down, let's say it goes up again, we're up two points, but then it might go down again going to +1, then it might go to +2 again then +1, +0 and -1 stopping us out. And at every 'junction' there's two events with the 1/2 probabilities. The basic question is what is the probability of us getting to +3 without getting to -1 first? What is the probability of getting to -1 without getting to +3 first? Not obvious at all.
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well expectancy would equal:

= 3*Prob(Going +3 before going -1)-1*Prob(Going -1 before going +3)

Now, we just have to figure out the probabilities. Need some skills for that.
 
The problem is not too hard if we take it one step at a time and realize how the problem changes with each step.

In the first step the market will either move down 1 point or up 1 point. If it moves down 1 point, we hit the stop loss and exit with a 1 point loss. This situation of being stopped out on the first step occurs 50% of time. The other 50% of the time, the first step sees the price increase by 1.

Now for the twist that helps us solve the problem. Consider the situation of having an open profit of 1 point after that first favorable step. Having gained a point we are now only 2 points away from our profit target and 2 points away from our stop loss point. Thus we have a symmetric situation for the chance of wandering either up 2 more points or down 2 points. Given the 50:50 probabilities of the steps, the chance of either move is 50%.

In total we have a 50% chance of losing 1 point from an immediate stop-out, a 50%*50% chance of gaining a point and then gaining two points for a net gain of 3, and a 50%*50% chance of gaining a point and then losing two points for a net loss of 1. The expectation is:

Expectation = (0.5)*-1 + (0.5*0.5)*3 + (0.5*0.5)*-1
Expectation = (0.75)*-1 + (0.25)*3
Expectation = 0

Of course, we did not need to do all this work to realize that the expectation was 0. As Mr Subliminal said, the fact that each step in the market has an expectation of 0 and that each step is independent guarantees that we will have an expectation of 0 for this trading rule or any other trading rule.

Happy (nonrandom) trading to all,
Traden4Alpha
 
Quote from Traden4Alpha:


Of course, we did not need to do all this work to realize that the expectation was 0. As Mr Subliminal said, the fact that each step in the market has an expectation of 0 and that each step is independent guarantees that we will have an expectation of 0 for this trading rule or any other trading rule.
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your explanation is very intuitive. But I remembered that Livermmore (spelling) said from his book that during old times brokers used this trick to make money. So I just did some algebra for fun.

Assume you have infinite time horizon (infinite movements allowed as long as neither stop nor target hit).
m = expectation of this problem
p = expectation of starting from +1 to hit either -1 or +3
q = expectation of starting from +2 to hit either -1 or +3

Then,
m = .5 * ( p - 1)
p = .5 * (q + m)
q = .5 * (3 +p)

Those three equations yield:
m = 0
p = 1
q = 2
 
Quote from wan2BTrader:


your explanation is very intuitive. But I remembered that Livermmore (spelling) said from his book that during old times brokers used this trick to make money. So I just did some algebra for fun.

Assume you have infinite time horizon (infinite movements allowed as long as neither stop nor target hit).
m = expectation of this problem
p = expectation of starting from +1 to hit either -1 or +3
q = expectation of starting from +2 to hit either -1 or +3

Then,
m = .5 * ( p - 1)
p = .5 * (q + m)
q = .5 * (3 +p)

Those three equations yield:
m = 0
p = 1
q = 2
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Don't have time to check your math, but the expectation will be zero regardless of where you start from. The logic T4A outlined applies the same way. It's a random process.
 
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