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# Some basic math reminder

Discussion in 'Technical Analysis' started by kanonka, Feb 26, 2010.

1. ### kanonka

The reason why I decided to get to this is that I regularly see how people forget math basics and try to convince others based on wrong assumptions. For example, in (another yet) recent discussion about (another) "best indicator" the following false logic was used:

1) If indicator A gives you correct signal in 60% cases, and indicator B gives you correct signal in 60% cases, then combining them will give you correct signal in 0.6*0.6 = 36% cases.

Not to going into discussion whether to use indicators or not, I just want to remind basic math. The above statement would be true only if you enter into trade when ANY of the indicator fires, i.e. when OR logic is used. But, if you enter the trade only when BOTH indicators fire, i.e. when AND logic is used, then correct calculation will be:
2) (1 - (1-0.6)*(1-0.6)) = 84% of correct signal chance. In simple words, you filter bad signals with one filter, and then whatever went through, you filter with another filter.

I put together one example to illustrate that.
Suppose, we have a signal, represented as a segmented digit on an LCD screen, like this (I tried to "draw" digit "2"):
......
......|
|.....

Now, we have an indicator A, that can read if number of horizontal bars is higher/lower than some number, and indicator B, that can read if number of vertical bars is higher/lower than some number. Let's describe all possibilies:
digit 0 has 2hor bars + 4 vert bars
1 has 0h + 2v
2 has 3h + 2v
3 has 3h + 2v
4 has 1h + 3v
5 has 3h + 2v
6 has 3h + 3v
7 has 1h + 2v
8 has 3h + 4v
9 has 3h + 3v

Let's assume, that the best time to go into trade is when we see digit 3. We can't actually see it, but we can use our indicators. Let's make indicator A fire when number of horizontal bars is 3; indicator B fires when number of vertical bars is 2.

Now, what is probability of "good trade", if we rely on indicator A alone? There are 6 digits with 3 horizontal bars, so we will be right only one time out of 6, i.e. our "good trade" probability is 1/6 (~17%). What is probability of "good trade", if we rely on indicator B alone? There are 5 digits with 2 vertical bars, so we will be right only one time out of 5, i.e. our "good trade" probability is 1/5 (~20%).

Now, what happens if we use BOTH indicators A and B? According to bad logic and wrong math, we should multiply "good" probabilities, making it 1/30 (~3%). But according to normal math, there are only 3 digits that have BOTH 3 horizontal and 2 vertical bars, so our chances of good trade is 1/3 (~33%). The math is, as I described above:

(1 - (1 - 1/6) * (1 - 1/5)) = 1 - (5/6)* (4/5) = 1 - 20/30 = 1 - 2/3 = 1/3.

Again, I don't advocate pro or contra of using indicators, I just want to get math strait