The opinions expressed in these forums do not represent those of C2, and any discussion of profit/loss
is not indicative of future performance or success.
There is a substantial risk of loss in trading. You should therefore carefully consider
whether such trading is suitable for you in light of your financial condition. You should read,
understand, and consider the Risk Disclosure Statement that is provided by your broker
before you consider trading. Most people who trade lose money.
Your S&P 500 example only shows that it uses no hard stop, nothing more.
Now take any system that does use a hard stop and you will notice that you won’t be able to vary the number of contracts (what you call leverage).
In other words you won’t be able to demonstrate that “high” leverage will deteriorate the performance of a system, because as soon as the hard stop and the risk per trade are known, it becomes impossible to change the number of contracts to trade (“leverage”).
Both were fixed at a hard stop of $300 loss. The stops are set at entry but partially decided based on how many contracts are traded. You act as though they are changing mid trade. If you select number of points first such as ten you have the same issue. You lose more if you have more leverage and the trade goes against you.
If you say S&P is at 3000 and my stop is going to be at 2990 then a strategy using more contracts (more leverage) will lose more than a strategy using fewer. The cumulative max loss then goes up by $50 for each contract that is purchased.
This shows both methods. First method starts with what max loss in dollars the trader wants, $300 aka 3% of NLV. The second method starts with we price the trader wants to be stopped out at for the S&P. I used 10 points because you referenced it.
With both methods we look at 1 and 5 contracts (varying amounts of leverage).
No, your hard stop was based on the 3% maximum loss. In other words it is a meaningless stop with no technical analysis logic behind it. We need both a hard stop (based on sound analysis) AND a maximum loss per position to trade at the optimal and safe level.
No, we don’t select anything, the system itself must dictate the location (size) of the hard stop, not the trader. Then you add the second parameter (maximum loss per trade) and you get the correct number of contracts to trade.
You cannot change that number. If you do you are INDEED using more leverage, because now you have decided to use a much bigger stop. In other words now you are no longer trading, you are gambling.
Assume you enter a trade at a specific price on MES such as $3862.00. Your system using its methodology selects a stop price. Let’s just assume the stop price is 10 points below the entry. So you enter at $3862.00 and you have a stop set at $3852.00. Would a position of 1 contract or 5 contracts have a larger potential loss?
Nothing you said there is wrong. If you keep the “hard stop” of 3852.00 the risk will increase above 3% if you buy more than 6 contracts (increase leverage) and it will decrease below 3% if you buy fewer than 6 contracts (decrease leverage) while still keeping the same stop price.
The other thing to consider too is how probable it is for a stop to be hit. Obviously a stop closer to the current price is more likely to be hit. A stop with a NLV max loss of 3% that is only 10 points away is much more likely to be hit than a stop with a NLV max loss of 3% that is 100 points away. Trading more contracts forces you to have tighter stop prices assuming you are keeping the same target NLV max loss.
Well, only a serious backtest can show you the optimal hard stop to use, as well as the optimal % of risk to apply to each position (assuming your system has clearly defined trading rules).
In general trend following systems usually need large hard stops, while reversals systems need tighter stops (due to their low risk high reward ratio).
So if the S&P is at 4,000 do you not think there is objectively a higher probability that it will cross the value of 3,990 compared to crossing the value of 3,900? I feel like you do but your previous comment seems to imply you don’t.
Yes of course, the market has a higher probability of moving 10 points higher first versus moving 100 points lower first.
But again, the trader needs to backtest all kinds of stops to find the optimal one. A lot of seemingly logical trading ideas (like a small stop is better than a large stop, or vice versa) fail big time when you backtest them.
Oh I agree. An more leverage means you need a tighter stop to have the same per trade max loss. In other words, the risk goes up with more leverage. If not in terms of max loss by adjusted stops, in terms of likelihood of the stop being hit.
Yes, but the trader must never mess with the hard stop (make it smaller than the original), just to maintain the 3% risk per trade (leverage).
If the trader tampers with the original hard stop, the performance of the system will quickly deteriorate. He can even turn his winning system into a losing system in no time.
If you follow the trend (with a moving average for instance) and the price just whipsaws back and forth across the moving average line, you will be stopped out many many times, sometimes on day one, so you need a bigger stop (like a volatility stop based on ATR for example *) to avoid the “noise” and the inevitable pullbacks.
On the other hand, if you trade reversals you are essentially buying near the low of the move (or shorting near the high of the move) so your stop is much much smaller.
Yep, you are right. A lot of small losses and small number of the large profits when you catch the trend. Low percent of profit trades, large profit/loss trade ratio. Typical trend following.
cutting the losses and letting profits run
And opposite on the mean reversal. Buy right at the low/top is kind of a goal, typically you buy earlier hoping for reversal right now. Not happens. Then you allow price to go against you, hoping that it will revert back. Large stops, small take profits and 90% of profitable trades.
that risks are higher with the larger leverage.
