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Simple problem: Gain after n trials given success rate and profibility 
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#1
Jan2812, 12:25 PM

P: 3

I'm stuck on what I'm sure is a very simple problem.
I'm trying to calculate predicted output for my share trading model. For example, for a strategy the probability of it winning on each trade may be 0.4 and hence a loss is 0.6. But for each trade it wins, my account balance is increased by 5% and each loss it loses 2%. How do I calculate how much I would have won or lost after n amount of trades? Can you break it down to on average each trade its a win or loss of x% so I can just use a simple compound interest formula to calculate expected profit after n amount of trades? 


#2
Jan2812, 05:38 PM

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#3
Jan2812, 06:01 PM

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#4
Jan2812, 06:29 PM

P: 2,501

Simple problem: Gain after n trials given success rate and profibility
EDIT: Note that for the ten trades I described above, I did not add the compounding which would be quite small. N(t)= 1.00 exp{(.008)10}= 1.08329 vs 1.08000 without compounding. Note I'm just working from the fixed assumptions of the model described and not saying anything about actual trading in stocks. 


#5
Jan2912, 05:56 AM

P: 3

Yes its from a trading account, so if I start with $5000 and I have a win and earn 5% then I end up with $5250, then If I win again I win 5% of the $5250 ($5512.5) and so on...
I'm confused with the formula because lets say I start with 1000. and my probability of winning or losing is 50/50. and each win or loss results in a 10% gain or loss of my account. If I win first I end up with 1100, then I lose I end up with 990. Of if I lose first I end up with 900 and then if I win I end up with 990. (same result) But if I put these values into the formula I end up with a value of 1000. 


#6
Jan2912, 11:20 AM

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#7
Jan2912, 11:37 AM

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Depending on whether you define the random variable in terms of percent of your current stake or percent of your initial stake, you get different problems to solve. To consider it as a percentage of your current stake, let's get rid of the idea of "percentage" completely , since it is a source of ambiguity in practical discussions: Define the random variable X as: P(X = 1.05) = 0.4 P(X = 0.98) = 0.6 Let {X1,X2,.... Xn} be an independent sequence of n random variables, each with the same distribution as X. Let the random variable R equal the product (K)(X1)(X2)...(Xn) where K is you initial stake. The term "gain" is also ambiguous. (e.g. is its R  K? R/K ? (RK)/K etc.). Let's look at the expected value E(R) of R. E(R) =K E( (X1)(X2)...Xn)). Since the Xi are independent random variables, you can compute this as E(R) =K E(X1) E(X2)...E(Xn) = K (E(X))^n 


#8
Jan2912, 03:13 PM

P: 2,501

I'm sure I'm missing something obvious. Could you show how to apply your formula for say 11 trades using P(X)=0.4, X=1.05, and Y=0.98? 


#9
Jan2912, 07:13 PM

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It's expected value is (1.05)(0.4) + (0.98)(0.6). So I'd raise that quantity to the 11 th power to get the expected multiplier for 11 trades. 


#10
Jan3012, 03:45 AM

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#11
Jan3012, 10:15 AM

P: 3

So are you saying.
If I have a 0.5 chance winning and hence 05. losing, and each win or loss is +10% or 10% respectively. I would write it as... P(X = 1.1) = 0.5 P(X = 0.9) = 0.5 and that after 5 trades I would have ((1.1)(0.5) + (0.9)(0.5))^5=1 Does this mean on average after 5 trades with those odds I would end up with the same amount of money? 


#12
Jan3012, 10:56 AM

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#13
Jan3012, 10:59 AM

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In any practical application, there will be some minimum stake that the trader needs in order to make a trade. The above formula doesn't account for the possibility of "ruin" when the traders stake falls below that minimum. 


#14
Jan3012, 05:50 PM

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