My approach to the market has become very probability based. I look at specific occurrences (such as 2 SD volume moves or price rotations) and look at the probability that something else is likely to happen. For example, and this is a very common stat used by traders, if the market has a 2 SD price rotation, there is an 80% chance the high/ low of that rotation will be retested or broken. I have accumulated a quite few of these “stats”. Sometimes more that one will be relevant at one time. My question is, how do I correctly combine these statistics if they occur simultaneously? I was looking over all my CFA probability notes and I cannot say I am sure how to approach this question.
I have created this made up situation below (all the stats are made up except the impulsive move). There are 2 stats for price to head lower and one that suggest the opposite is true. What is the probability that price will tag yesterday’s low considering all 3 stats?
So they must be conditional probabilities. If you can define the constituent mutually exclusive events (with probability %s adding to 100), then it becomes easy to calculate price based on those probabilities.
Second, to answer your question, your analysis is historical based, that is “when A has happened, there has been a probability P of B happening”. You can find the conditional probability of event C using the same principle - just count the times A or B has happened, then count the frequency of C given A or B. You should be able to construct a probability tree starting with any event A, B or C.
I don’t fully understand the context of the problem and your charts, but I’ll just comment on the math of it. As others have said, it sounds like you have a few conditional probabilities of the type “Probability that a stock price will go up given Event A” = p%. So I’ll label this P(X|A)=p% where X is the event that stock price goes up and A is the catalyst event. So basically you’ve accumulated some data like P(X|A)=p% and P(X|B)=q% and would like to “combine” them and find out P(X|A&B)=? where A&B is A and B occurring at the same time. All you need to do is apply Bayes’ rule and make a few assumptions.
You need the denominator: P(A&B)=P(A&B|X)*P(X)+P(A&B|X^)*P(X^) where X^ is the opposite of even X, e.g. the stock price goes down.
Then P(A&B)=P(A&B|X)*P(X)+P(A&B|X^)*P(X^)=P(A|X)*P(B|X)*P(X)+P(A|X^)*P(B|X^)*P(X^) {b}.
A couple of clarifications: since X^ is the complimentary event to X, P(X^)=1-P(X), P(X^|A)=1-P(X|A), etc. Also, P(A|X)=P(X|A)*P(A)/P(X), etc. So using these, and plugging {b} into {a}, you can express P(X|A&B) entirely in terms of P(X|A), P(X|B), P(X), P(A) and P(B). If you don’t have any prior views on P(X), P(A) and P(B), you can always start with 50%.
This is just combining two conditional probabilities from two events A and B. You can generalize this to any number of events and get P(X|A1&A2&A3&…&An) by following the same steps and it’s not much more complicated than this.
Thanks everyone for the input. It looks like Ohai was headed in the same direction that Mobius kindly spelled out for me. I had initially considered Bayes but quickly became concerned I would not apply it correctly. I will look into this…
I have no idea what “5 min volume spike on down candle” means, but it sounds like 2SD move in a particular direction (as opposed to two-sigma move in either direction). If that is the case, assuming normal distribution the probability will be 4.5%/2. But why normal - is this assumption reasonable? If you don’t know the distribution, use Chebyshev’s inequality - for any distribution, the probability of a k-sigma move in a particular direction would be less than (1/2)*(1/k^2). In your case for k=2 that’s 12.5%
My “2SD stats” are collected this way. The chart represents a few hundred days of data (only a couple days shown). On the bottom the bars are volume/ 5min candle. Projected onto those bars is profile distribution. The pink line represents 95.4%. I look at at least 200 samples of mixed up and down candles with >2SD volume. So, if that makes sense, the probability of a volume spike occurring (either up or down) is 4.5%, yes? The way I compiled the statistic was I would call it a success if the low of a down candles was retested/ passed or if the high of a up candle was retested/ passed. I did not do separate studies on up and down candles.
