Okay, I understand now. Those perpendiculars are no longer perpendiculars when you change the scale. That's a good point. You are right, it has to be scale-invariant.
Risk (pain) vs Reward (gain)
- Thread starter nonlinear5
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Okay, I understand now. Those perpendiculars are no longer perpendiculars when you change the scale. That's a good point. You are right, it has to be scale-invariant.
Yes you got it. I don't know how to prove it changes the point of maximum perpendicular distance between the line and the curve, but I suspect it may have some effect. Probably not too much though.
I am thinking perhaps this revised version makes more sense (see below). The "spirit" is still the same, which is to maximize the distance between the red identity line and the risk/reward curve. The change is that instead of the maximum North-West distance, this version uses the maximum North-South distance. This version should be scale-invariant. Please let me know if you find otherwise.
The optimal point occurs when the instantaneous slope of the thick blue line becomes smaller than the slope of of the red line. This is equivalent to the point where the thin tangent line to the thick blue curve becomes parallel to the red identity line.
My next step is to quantify it, and run a test against a real trade distribution to see what it produces.
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As you said since you have no facts, "a bracket trader will consistently make bad decisions at bad times..." is a non factual statement, next you be saying Easter Bunny exists and you love wearing Bunny outfit at bedtime.....
I can't think of any trader than I know of that trades for a living consistently makes bad decisions at any time as all have a sequence of what to do when in there Trading Plan. I think most that have bracketing into their trading is not some newbie off the street as it is much more complex of doing stocks and/or futures and hedging as risk management.
Maybe you don't have that in your Trading Plan and don't have clue of having multiple sequences? I think you stay in the game long enough, you are always ready and have answer for any situation. Best way to see if one is correct or not is one doing back testing, never know, your trading could improve much better.
perhaps, but everything I have to show for myself is the result of some lucky trade that just happened to trend longer and harder than anybody thought possible, and if i was just sitting there with a bracket target I would have never even known.
I am thinking perhaps this revised version makes more sense (see below). The "spirit" is still the same, which is to maximize the distance between the red identity line and the risk/reward curve. The change is that instead of the maximum North-West distance, this version uses the maximum North-South distance. This version should be scale-invariant. Please let me know if you find otherwise.
The optimal point occurs when the instantaneous slope of the thick blue line becomes smaller than the slope of of the red line. This is equivalent to the point where the thin tangent line to the thick blue curve becomes parallel to the red identity line.
My next step is to quantify it, and run a test against a real trade distribution to see what it produces.
Looks good and should be scale-invariant concerning scales of axis. Though, you will still have the conundrum of scale of rates, ie. "if I halve risk I will gain more reward relative to that huge decrease of risk". And you could always decide that the current "max reward" point is unrealistic/unwanted, and move it more to the left, changing things once more.
So you want to call the optimal point where the blue tangent slope go below the linear interpolation slope. Could be interesting since there is a max reward in play that practically bounds the graph. Seems like a logical way to go about it if you accept the premises. Though what you get, is really just some kind of average between "infinite" reward per risk, and a max reward that's not that tempting relative to the risk increase. Eg. 1.25 vs 0.3 risk is over 4 times the risk or money allocation.
Is this the same point as the max distance between the curve and line? That difference would be something like (just so we speak about the same thing): ln(n(x+1)) - ax.
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Okay, here is the quantification. I'll try to be very precise and detailed, to avoid any ambiguities, so that the results can be reproduced by anyone who wishes to try.
Let R be a series of trade returns:
R: {r(1), r(2), r(3), ..., r(N)},
where N is the number of trades,
r(i) is the return on the ith trade
Each trade return r is:
r = (S - B) / B,
where B is the price at which an instrument was bought,
and S is the price at which an instrument was sold.
(we'll ignore the commissions for simplicity).
For example, if we bought at $100 and sold at $110, then
r = (S - B) / B = (110 - 100) / 100 = 0.1
Notice that this formula is exactly the same for short and long trades. For example, if we sold short at $110, and bought back at $100, then this trade return r is
r = (S - B) / B = (110 - 100) / 100 = 0.1
Next, we define leveraged cumulative log return CR(L) as the sum up all individual leveraged returns in R as follows:
CR(L) = log(1 + L * r(1)) + log(1 + L * r(2)) + ... + log(1 + L * r(N)) [formula 1]
where log() is the natural logarithm,
L is leverage, L >= 0.
The reason for using log-returns instead of regular returns is that it's computationally more efficient. The result is the same as multiplying the returns, which gives us the total compounded return. The resulting quantity CR(L) is our terminal gain, given leverage L.
Next, we apply [formula 1] to all leverage levels, from L=0 to some arbitrary high leverage. Since CR(L) is convex, it will peak sooner or later. Now, we have a special value, max[CR(L)], which corresponds to the maximum gain, and the corresponding Kelly leverage KL, which is the level of L where CR(L) peaks. This is the red point on the graph in my previous post.
Now, we can identify the thick red "identity line", which goes from [0,0] to [KL, CR(KL)]. The slope of that line is simply:
slope = CR(KL) / KL
With that, we can identify every point of the red line as:
CR(L) = L * slope
So, now we have a value of every point of the red line, and a value of every point of the blue line. With that, we can identify leverage L where the difference between the blue line and the red line reaches its largest value. That is, we want to maximize this delta D with respect to L:
D = CR(L) - (L * slope)
By maximizing D, we find what we were looking for: the green point on the graph, which corresponds to the "optimal leverage" OL.
Now, let's do a sanity test by computing KL and OL for a well studied example, from
https://en.wikipedia.org/wiki/Kelly_criterion
There is a game where you have a 60% chance of winning, and 40% chance of losing. If you win, you get 100% of your bet (in addition to getting your bet back). If you lose, you lose your bet. How much do you bet? The wikipedia article says, "bet 20% of your bankroll at each opportunity".
Let's verify this by using the numerical methodology above (rather than the analytical solution provided by wikipedia).
Here is the (Java) code:
Code:
/**
* @author nonlinear5
*/
public class OptimalLeverageCalculator {
private final List<Double> tradeReturns;
private OptimalLeverageCalculator(List<Double> tradeReturns) {
this.tradeReturns = tradeReturns;
}
public static void main(String[] args) {
List<Double> tradeReturns = new ArrayList<>();
tradeReturns.add(1.0);
tradeReturns.add(1.0);
tradeReturns.add(1.0);
tradeReturns.add(1.0);
tradeReturns.add(1.0);
tradeReturns.add(1.0);
tradeReturns.add(-1.0);
tradeReturns.add(-1.0);
tradeReturns.add(-1.0);
tradeReturns.add(-1.0);
OptimalLeverageCalculator optimalLeverageCalculator = new OptimalLeverageCalculator(tradeReturns);
optimalLeverageCalculator.evaluateOptimalLeverage();
}
private double getGain(double leverage) {
double gain = 0;
for (double tradeReturn : tradeReturns) {
double leveragedReturn = leverage * tradeReturn;
double logLeveragedReturn = leveragedReturn > -1 ? Math.log1p(leveragedReturn) : Double.NEGATIVE_INFINITY;
gain += logLeveragedReturn;
}
return gain;
}
private void evaluateOptimalLeverage() {
double maxGain = 0, maxLeverage = 0, optimalLeverage = 0, deltaLeverage = 0.01;
for (double leverage = 0; leverage <= 2; leverage += deltaLeverage) {
double gain = getGain(leverage);
if (gain >= maxGain) {
maxGain = gain;
maxLeverage = leverage;
} else {
break;
}
}
double slope = maxGain / maxLeverage;
double maxDistance = 0;
for (double leverage = 0; leverage <= maxLeverage; leverage += deltaLeverage) {
double gain = getGain(leverage);
double identityGain = leverage * slope;
double distance = gain - identityGain;
if (distance >= maxDistance) {
maxDistance = distance;
optimalLeverage = leverage;
} else {
break;
}
}
System.out.println("Kelly leverage: " + maxLeverage);
System.out.println("Optimal leverage: " + optimalLeverage);
}
}The output of running this code is:
Kelly leverage: 0.2
Optimal leverage: 0.1
So, my proposed numerical method agrees fully with the wikipedia's analytical solution (Kelly leverage = 0.2), and it also calculates the green point (optimal leverage = 0.1), which is, logically enough, between L=0 and L=KL.
I welcome you to use this numerical solution, and to apply it to any given return distribution. I'd be interested to see if the results make sense for a wide variety of return distributions.
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Correct.
Yes, what we are looking for is the max distance between the thick blue curve and the straight red identity line. That would be:
Max[CR(L) - (L * slope)] with respect to L.
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I won't say I am lucky, it more like I have done enough back testing that shows I have to be willing to give back huge open profits to make the homerun profits, except now I can hedge those open profits so the loss won't be as great, I do have a healthy target where I take off half between $4800-10k, but am trying to get over 55% of nine year range. So doing options only way I can manage risk.
As they say: no pain no gain. 