{"id":10756,"date":"2011-04-06T12:31:45","date_gmt":"2011-04-06T16:31:45","guid":{"rendered":"http:\/\/www.smbtraining.com\/blog\/?p=10756"},"modified":"2021-05-06T15:44:35","modified_gmt":"2021-05-06T19:44:35","slug":"finding-an-edge-in-random-markets","status":"publish","type":"post","link":"https:\/\/www.smbtraining.com\/blog\/finding-an-edge-in-random-markets","title":{"rendered":"Finding an edge in random markets"},"content":{"rendered":"<p>As many of you know from reading my previous posts, one of the tools I use for thinking about markets is to imagine that markets are pure random walks. If this were true, there would be no way to make money trading. Readers often comment that good money management, trailing stops, psychology, etc could help, but the bottom line is if you do not have an edge in the market, you cannot make money. Period. A trailing stop is not an edge in itself (I am aware of the studies in a well-known trading book that show a profitable trading system with random entries and trailing stops. The key is that the markets tested were not random (they were real futures markets), and the exits were not random (they were a good trailing stop system). If markets were random you could not make money trading, and, if you are not convinced of that, you are missing a key piece of the intuition behind understanding price action.<\/p>\n<p>This comment from a reader this morning on an old post is a good chance to dig into the concept a little deeper:<\/p>\n<blockquote><p><em>\u2026 But it is as simple as this. If you take 10 long trades (let&#8217;s keep it at ten for the example) and you enter a random market at 1000. You know that 5 out of ten you reach 990 first and the other 5 you reach 1010 first. What you need to do is to cut your losses of those five that reach 990 at that level. The once that go to 1010 you let them run and you trail them in a way you are comfortable with. And yes, 2 of those 5 that reach 1010 might end up with a loss after all of let&#8217;s say 7 and 3. One of those 5 might result in only a win of 7. One might result in a win of 25&#8230; But the last one could be that big winner of 60!<br \/>\nSo in the end<br \/>\nTotal loss: 60<br \/>\nTotal profit: 92<br \/>\nNett profit: 32<\/em><\/p>\n<p><em>Everybody that works with a fixed stop loss AND take profit is doomed to fail. When long you should always get stopped out because prices are coming down and short vice versa. That way your profit is unlimited, your losses are limited and in case you are up your profits are protected.<br \/>\n<\/em><\/p><\/blockquote>\n<p>A useful way to think about this problem is to use something called a binomial tree. Imagine a tree that starts at 1,000 and then goes up or down +\/- 10 on the next step, with 50% probability of either step. This pattern repeats as it branches out from the starting point. We might draw it like this:<\/p>\n<p>So, let&#8217;s consider the reader&#8217;s scenario and break it down into three possibilities:<\/p>\n<ol>\n<li>The first step is down, to our stop of -10. This happens 50% of the time.<\/li>\n<li>The first step is up, to a profit of +10. We do not take profit at this level, but now we &#8220;reduce our risk&#8221; but raising our stop to 0. The probability of this first step up is 50%.<\/li>\n<\/ol>\n<ul>\n<li>The second step is down, for a loss of 0. This happens with 50% probability from the previous step. The overall probability of this happening is 50% * 50% = 25%.<\/li>\n<li>The second step is up, and we book a win of +20. This also has a 25% probability from the beginning.<\/li>\n<\/ul>\n<p style=\"text-align: justify;\">Remember, our expected value equation for n scenarios is: <em>Expected value = probability of scenario<sub>1<\/sub> * payoff of scenario<sub>1<\/sub> + \u2026. + probability of scenario<sub>n<\/sub> * payoff of scenario<sub>n<\/sub><\/em>. In this case, the math is: (50% * -10) + (25% * 0) + (25% * +20) = -5 + 0 + 5 = 0. Even with the trailing stop, this is a zero expectancy game. In the reader&#8217;s comment above he was assuming you could let the winners run, but, in a random market, you will find the probabilities shrink as the profit increases so that the zero expectancy game is maintained. This is an important intuition to build about trading systems and price action. You must have an edge to make money.<\/p>\n<p style=\"text-align: justify;\">I know many of you will protest that if you had more steps in the tree (I simplified so each step was +10) that it would be different. Work it out carefully on paper and you will see that it is not. The message is simple: you cannot create a positive expectancy situation without having some kind of edge, and you cannot have an edge (by definition) in a random market.<\/p>\n<p><a href=\"http:\/\/www.twitter.com\/AdamG_SMB\"><img decoding=\"async\" src=\"http:\/\/twitter-badges.s3.amazonaws.com\/follow_me-c.png\" alt=\"Follow AdamG_SMB on Twitter\" \/><\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Adam asks if traders can make money using trailing stops in random walk markets. <\/p>\n","protected":false},"author":388,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[943,1,1373,15],"tags":[1704,788,1488,1496,1434,1428,1546],"class_list":["post-10756","post","type-post","status-publish","format-standard","hentry","category-adam-grimess-blogs","category-general-comments-2","category-trading-lesson","category-trading_theory","tag-binomial-tree","tag-edge","tag-expectancy","tag-expected-value","tag-random","tag-random-walk","tag-trailing-stop","no-post-thumbnail"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.1.1 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Finding an edge in random markets | SMB Training<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" 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