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#1
Feb1914, 10:00 PM

P: 1

Hello, I have been searching around for a topic of math that I am not familiar with. I'm hoping somebody can point me in the right direction.
This has to do with Compound growth and decay when the two are combined. This occurs in trading(ie stocks,options,futures,etc.) And I am hoping I can find an accurate formula that describes the nature of this idea. The main Idea is that when compound growth and decay are added together, they tend to go towards zero. Ex) Starting with $100, risk is 10% and reward is 10%. The chance of a win is 50%, chance of a loss is 50%. On the first trade we win: $100 + $10(10%) = $110. On the next trade we lose: $110  $11(10%) = $99. As you can see, one win of 10% and one loss of 10% does not result in the starting balance, but tends to get closer to zero. Also, the order of these wins/losses is not important. I have not been able to find a formula that explains this phenomena, or even what this concept is called. 


#2
Feb1914, 10:34 PM

P: 81

I don't know how this concept is called; it sounds like something in economics or financial mathematics to me. Having said that, I believe your problem is a purely probabilistic one. Here is an explanation: The chances of winning or losing is 50%50%. If you win, your new balance is 110% of your previous balance. If you lose, your new balance is 90% of your previous balance. Consider the following combination as an example: 1.win, 2.lose, 3.win, 4.win. Then your new balance is 110%*90%*110%*110%*(initial balance). Thus you can see why the order does not matter, for suppose the order is changed, you would still have the same balance because multiplication is commutative and associative. Now to see why in the end your final balance would tend towards zero, let the number of trades be T and let the number of wins be W. Since the chances of winning or losing is 50%50%, in the limit T→∞ we have W/T = 0.5 . Thus, in the limit T→∞ the number of wins and losses will become equal. Therefore, to calculate your new balance, each time you multiply your initial balance by 110%, you would also have to multiply the new balance by 90% (and you would have to do that T/2 times): new balance = lim (110%*90%)^{T/2}*(initial balance) = lim 0.99^{T/2} * (initial balance)T→∞ T→∞ = 0, as 0.99 < 1. Thus, you can see why your balance will eventually tend to zero. To summarize, we've explained (i) why the order of win/loss do not make a difference, and (ii) why your balance will eventually tend to $0. We made use of the commutative and associative laws of multiplication, simple probability theory, and limits. I hope this helps. 


#3
Feb1914, 11:11 PM

Engineering
Sci Advisor
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Thanks
P: 6,967

I don't know what "finance experts" call it, but I don't think it is anything more complicated than ##(1+x)(1x) = 1  x^2##.
There is a old classic "are you financially clueless" question that says "if you are gambling and you win 40% and then lose 30%, have you made a profit?" The answer is "no". 1.4 x 0.7 = 0.98. 


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