Probability & Statistics/Probability
Thankyou for your time & would appreciate an answer to a statement in a book I am reading regarding trading as I do not know how the answer is worked out. Perhaps you could explain the workings for me which would be greatly appreciated. How does the author arrive at a 65% chance? I have no problems with arriving at the average or standard deviation.
Lets say your expectancy
is 0.33R, but your standard
deviation is 3R. What this means
is that even though your average
gain after 20 trades should total
about 6.6R, you only have about
a 65% chance of being profitable
after 20 trades because of the
Thankingyou in anticipation
The statement is explaining how mean and standard deviation are both
important factors in determining the probability that one sees a particular type of outcome (in this case, a profit).
If the expectation (mean) is 0.33R, this just means that the average outcome is 0.33R.
Having a positive mean does not
guarantee that you are likely to make a profit. For example, let's say you choose one of six numbers at random, and the possible outcomes are -1, -2, -3, -5, -7, or +20 outcome. Although the average is (20-1-2-3-5-7)/6 = 0.33R, you only have a 1/6 chance of picking a positive number.
However, under a normal distribution
, a positive mean will guarantee a likelihood (better than 50% at least) that the outcome is positive, because half of all outcomes are higher than the mean (in the case you have, another 15% is also between 0 and the mean for a total of 65%).
For more on the normal distribution, see Wikipedia
In a normal distribution there will still be some chance that the outcome is negative, and if the mean is close to zero (compared to the standard deviation), that chance will be significant.
For the expectancy (mean) equal to μ and standard deviation σ², the probability that you are above some threshold T can be computed. For example, the probability of being higher than T=μ is exactly 1/2. The probability of being higher than T=μ-σ is about 84%. The probability of being higher than T=μ-3σ is over 99%.
So if μ = 0.33R and σ² = 0.03, it is a virtual certainty that the outcome is positive (about 97%).
If μ = 0.33R and σ² = 0.3, then it is somewhat likely to be positive (73%).
And in the case given, where σ²=3, it is about 57% likely to be positive.
For more about this, see Wikipedia
Now, that's only true for a single transaction -- not twenty of them.
It's important that if you have multiple normal distributions, and they add up, then the μ and σ² values add up. For two, it might be:
X1 is distributed by μ1 and σ1
X2 is distributed by μ2 and σ2
X3 is distributed by μ and σ
⇒ μ = μ1 + μ2, σ² = σ1² + σ2²
So for twenty identical trades, the new values of μ and σ are:
μ = 20*0.33 = 6.6
σ² = 20*3 = 60
Notice that this means σ is only changed by √(20) factor, not 20.
So you can then use calculus to compute an integral from x=0 to x=∞ of 1/(σ√(2π) ∫ e^( (x-μ)^2 / 2σ² ) dx with the values of μ and σ above to get 80.2. This is actually way higher than 65%, and I think your book may have a typo or error of some type.
For more on computing such things, see Wikipedia
So, what is interesting to observe is this: Say you have n trades (n=1 and n=20 we have done, and seen 57% and 80% as probability of a positive outcome).
Well, if you have n=100 trades, it should be even higher than 80% right? The more trades you do, the closer you get to the average -- meaning you are less and less likely to dip below zero. This is because the mean is multiplied by n (which was 20 above), while the variance σ is only multiplied by √(n).