Your opinion please - page 4
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(sith voice) Your diagram is weak old man.
On my diagram time goes left to right. A net long position is blue. A net short position is red. The net position size is shown mid way along the line. Green nodes are wins where the position is closed. Up is a higher price meaning a win for a long position.
My diagram is simplified to neglect parallel paths that nevertheless end up at the same point. There are actually an infinite number of routes to C3, even sticking to the grid. The calculations then become summations of infinite series or finding the solution to a recurrence relation (this latter one is sadly beyond my mathematical ability).
Assuming a random walk, the chance of reaching C1 is 50%, neglecting the spread. The chance of reaching C2 is 25%, neglecting the spread. Your homework question is what is the chance of reaching C3?
I'll like to entertain your diagram but I don't understand what it's doing. If the chances of flipping a Heads on a Coin is 50%. Or the chances of landing on Red vs Black is 50%, or the chances of moving Up-1-Grid vs moving Down-1-Grid is 50%. This has No relevance at the time of the Hedge. Simply because At-the-time-of-the-Hedge, your independent event has passed. IE - We already know the results of the 1st coin toss, spin or price move.
It's precisely at this point where you're wrong in assuming that there's less chance of the price continuing down vs up. If you flip a coin 10 times and get 10 Heads, the probability of Heads coming up again on the next toss is still 50:50. Therefore your Start point on your graph as it relates to this conversation is Not-Static, but rather always changing based on the New-Results.
Plus bringing you back to your Original Example again. If you Buy, Close your order on Stop-Loss, And then Re-Buy again at the original price. Whats to stop Price from going down again? IE- your 50:50 chance. Of course, it goes down again and close you out at a Stop-Loss again. And re-bounds again. Staying true to your system will cost you dearly in a Side-Ways market. <-- Again just food for taught :)
It's precisely at this point where you're wrong in assuming that there's less chance of the price continuing down vs up. If you flip a coin 10 times and get 10 Heads, the probability of Heads coming up again on the next toss is still 50:50. Therefore your Start point on your graph as it relates to this conversation is Not-Static, but rather always changing based on the New-Results.
Of course, but we are placing a martingale trade so we have to start counting from when we start putting on money. Yes the chance is always 50:50 for red or black. That isn't the question I am asking.
NOTE: I have edited the image in my earlier post as it was wrong :-(
In the absence of spread, and with a random market, the profit factor from random trades will be 1.00 regardless of the stop loss and take profit levels. You can use this idea to calculate the win rate. So if the stop loss is 50 pips and the take profit is 50 pips (neglecting the spread) then the win rate is obviously 50%. Now if you set the take profit at 20 pips and the stop loss at 40 pips (neglecting the spread) the win rate is easy to calculate. And that is pretty much all I was asking as the homework exercise (which now relates to node n2 and is comparing 3 up to 2 down).
I've made it obvious I'm not prepared to engage into what if probabilities. If someone comes along and can show mathematically that no matter which dance price made a hedge option will always be worse then I'm all ears. Until then I'm done with this thread. It was nice chatting..... chow. Ps> "I wouldn't hold my breath" :)
I'm with dabbler on this one. There is always a sequence of non-hedged trading actions which have a comparable effect on market position and account equity as a series of hedged orders. As dabbler says, the non-hedged actions avoid a swap cost, and avoid exposure to the spread suddenly widening. The advantages of the hedged route are that it's easier to code in MT4, and it's easier to assess the current state of the grid strategy from the MT4 open-order list and chart. Plus the fact that it makes you feel warm and fuzzy provided that you're prepared to ignore your account equity and only look at your account balance.
Going back to the diagram from 2012.02.11 13:54 and the subsequent explication, "I'll just Close Buy at Point-B and Place a Sell at Point-B" seems to be wrong. The equivalent series of non-hedged orders appears to be: open a buy at point A; close the buy at point B - but don't place a further order; place a buy for 2n lots at point C. In the original diagram you have no net position between points B and C, and closing the sell at C is then equivalent to going long. If I understand the note at point C correctly, the non-hedged equivalent is opening a buy, in terms of that being equivalent to closing the hedged sell, and also opening a second buy.
I'm with dabbler on this one. There is always a sequence of non-hedged trading actions which have a comparable effect on market position and account equity as a series of hedged orders. As dabbler says, the non-hedged actions avoid a swap cost, and avoid exposure to the spread suddenly widening. The advantages of the hedged route are that it's easier to code in MT4, and it's easier to assess the current state of the grid strategy from the MT4 open-order list and chart. Plus the fact that it makes you feel warm and fuzzy provided that you're prepared to ignore your account equity and only look at your account balance.
In celebration of the fact that it's exactly 2 years since I last participated in one of these discussions about hedging (https://www.mql5.com/en/forum/117708/page6) ...
I'm with dabbler on this one. There is always a sequence of non-hedged trading actions which have a comparable effect on market position and account equity as a series of hedged orders. As dabbler says, the non-hedged actions avoid a swap cost, and avoid exposure to the spread suddenly widening. The advantages of the hedged route are that it's easier to code in MT4, and it's easier to assess the current state of the grid strategy from the MT4 open-order list and chart. Plus the fact that it makes you feel warm and fuzzy provided that you're prepared to ignore your account equity and only look at your account balance.
Going back to the diagram from 2012.02.11 13:54 and the subsequent explication, "I'll just Close Buy at Point-B and Place a Sell at Point-B" seems to be wrong. The equivalent series of non-hedged orders appears to be: open a buy at point A; close the buy at point B - but don't place a further order; place a buy for 2n lots at point C. In the original diagram you have no net position between points B and C, and closing the sell at C is then equivalent to going long. If I understand the note at point C correctly, the non-hedged equivalent is opening a buy, in terms of that being equivalent to closing the hedged sell, and also opening a second buy.
You and I both know that I'm aware of that thread, and for a while I give it the benefit of the doubt. And I already taught about placing a 2x size at Point-C. If you place a 2x size at point C the the hedger will place a 3x size at Point-C. And no matter what you try to do the Hedger will always have you by +1 because he didn't close his original position.
This is exactly why I mentioned the lot size of 0.1 and also the reason why I opened another position and also why I mention the both that Hedger and the Stop-Loss guy have the same Information at Point-C. I did all this not with just Dabber in mind but with JJC and Gordon in mind should you revisit the topic.
I did all this not with just Dabber in mind but with JJC and Gordon in mind should you revisit the topic.
The sad thing is that I was trying to help you out with your maths, especially with respect to your earlier thread
https://www.mql5.com/en/forum/137889
whilst at the same time you thought you were trying to help me :-)
[...] And no matter what you try to do the Hedger will always have you by +1 because he didn't close his original position.
I'm getting quite puzzled now. As far as I can see, the diagram at 2012.02.11 13:54 is saying that the hedged orders are as follows:
* Open a buy at point A
* Open a sell at point B
* Close the sell at point C, and also open another buy order
* Therefore, make a profit once the price rises again towards point A.
The sequence of profitability is therefore going to be as follows:
* At point B, the system is $x in loss
* From point B to point C, the loss is fixed because an increasing loss on the buy order from A is offset by an increasing profit on the sell order from B. The net difference remains a loss of $x
* At point C the sell is closed. This realises a profit of $x on the account balance, and leaves a loss of 2 x $x on the floating P/L. Account equity remains in loss by $x compared to outset. A new buy is opened at this point.
* If the price then returns to point A, then the original buy order is at break-even; there's a banked profit from the B sell order of $x, and the buy order from C is 2 x $x in profit. Total profit is 3 x $x.
All I'm saying is that for this particular sequence of hedged orders there is - and there will always be - an equivalent sequence of non-hedged orders with the same effect on account equity and market exposure:
* Open a buy at point A
* Close the buy at point B, crystallizing a loss of $x
* Open a buy for 2 x the lot size at point C.
* When the price returns to point A, the buy order from C is 4 x $x in profit and there's the previous loss of 1 x $x, again leading to net profit of 3 x $x.
(EDIT, for further confirmation: if the price then rises a grid position above A, then the hedged system has open profit of $x on the buy from A, banked profit of $x on the sell from B, and open profit of 3 x $x from the buy at C, for a total of 5 x $x. Similarly, the non-hedged orders have a banked loss of $x from the buy at A and open profit of 6 x $x from the buy at C, again for a total of 5 x $x.)
The single biggest point here is in relation to C: if you have an open buy order and an open sell order, and you then close the sell, then it's equivalent to having no open orders and then opening a buy. In either case your net position goes from flat to +n lots. The more general point is that there's always a sequence of non-hedged orders corresponding to any sequence of hedged orders. And what Dabbler's saying is that if B and C happen over different days then you're making a loss on the swap which isn't the case with the non-hedged orders.
... And what Dabbler's saying is that if B and C happen over different days then you're making a loss on the swap which isn't the case with the non-hedged orders.
Agreed. But I am also saying that you also pay extra spread.
Assume lotsize of 1 initially and cost the position in terms of spread x lotsize for simplicity
First trade costs 1x spread. That fails.
So now add double size, costing 2x spread.
Suppose it wins. I have paid 3 x spread in total.
Do it the other way.
First trade costs 1x spread. That fails.
Close it and reverse, costing 1x spread.
Suppose it wins. I have paid 2 x spread in total.
Dabbler does not like paying spread unnecessarily :-)
* Therefore, make a profit once the price rises again towards point A. I don't think so. You make a profit once price reaches Point-B, I have that outlined in the diagram as Thanks to the Sell. Sure you can come up with a Non-Hedge equivalent in Hind-Sight 20/20. But when you're sitting at Point-B or C without your crystal ball, then there's no equal.
Do it the other way. First trade costs 1x spread. That fails. Close it and reverse, costing 1x spread. Suppose it wins. I have paid 2 x spread in total. Suppose that lose. You've cost your self the spreads + the distance of the stop-loss. :) I just killed the bird in the back of my hands.
Sitting here and saying well umm looking at your diagram... umm I'll just Sell-All at point A and Buy-All at point C will give me the best chance is totally folly.