Summary statistic for 'average' conditional probabilities

Question

I have the following scenario:

Let $X_i$ denote some event. Now I have the following calculated probabilities.

$P(X_1 \cap X_2 \cap \cdots \cap X_9 | X_{10})$

$P(X_1 \cap X_2 \cap \cdots \cap X_{8} \cap X_{10} | X_{9})$

$\vdots$

$P(X_2 \cap \cdots \cap X_{9} \cap X_{10} | X_{1})$

In words, the above expressions represent the (joint) probability of the 'remaining' events given that a particular event has occurred.

To calculate these probabilities, I know the underlying probability distribution that describes the entire system, i.e., $p(x_1, x_2, \cdots. x_{10})$, so for example, calculating the probability $P(X_1 \cap X_2 \cap \cdots \cap X_9 | X_{10})$ just simply requires me to find $\frac{P(X_1 \cap \cdots \cap X_{10})}{P(X_{10})}$ where both the denominator and numerator can be calculated by integrating over certain regions of the probability density function $p(x_1, x_2, \cdots. x_{10})$.

My question is, I wish to find one summary value that describes the 'average' probability of this system, e.g. say $P(X_1 \cap X_2 \cap \cdots \cap X_9 | X_{10})= 0.4$, $P(X_1 \cap X_2 \cap \cdots \cap X_{8} \cap X_{10} | X_{9})= 0.3$ etc, how can I "combine" these values $0.4$, $0.3$, etc into one value that describes the "average" probability of this system? My initial method is just to take the arithmetic average of each value, but that is mathematically incorrect since conditional probabilities aren't summable (except when they are conditioned on the same event). So are there any other measures/techniques I can use to somehow "combine" these single probabilities into "one" value? I.e., some kind of summary statistic that describes the average probability?

Answer 1

There is a bad-boy gang of 10 members. Your crew has 9 members, and you mostly prefer to be in the same bar with the bad-boy gang if your numbers match or theirs are lower. You have calculated what is the (different) probabilities that whenever a specific one of the bad boys is in a bar, all the other nine are there too.

One day, you are told that one member of the bad-boys gang was seen entering the bar that you and your crew want also to go to. But you are not told which one. Hmmm, what to do now? You need the conditional probability, in order to evaluate the situation and decide whether to take the chance or not. But you cannot have it. What can you have?

Well, what is the probability that BB-1 was actually the one that entered the bar ("BB" as in "Bad Boy" or "Brigitte Bardot")? Your crew is good with probabilities, you have calculated that too. And what is the probability that BB-2 was the one? etc

Since you do not know which Bad Boy has entered the bar, you resort to weight each conditional probability by multiplying it with the probability that the associated conditioning event (Bad Boy-X is in bar, that is) is the one that has actually happened. In other words you calculate

$$P(X_2 \cap \cdots \cap X_{9} \cap X_{10} | X_{1})\cdot P(X_1) +...+P(X_1 \cap X_2 \cap \cdots \cap X_9 | X_{10})\cdot P(X_{10})$$

$$=10\cdot P(X_1 \cap \cdots \cap X_{10})$$

Is the above weighting scheme a "weighted average"? Not necessarily, at least in the proper meaning of the term. What is the sample/probability space here? What are the elementary events?

Returning to the more practical issues, note that if your informer (too stoned for further details) had not tell you that one of the Bad Boys had entered the bar, the probability that all 10 of them would be there would have been $P(X_1 \cap \cdots \cap X_{10})$. Now that you know that one of them is there, the probability that they are all there has increased ten-fold... So this is mostly a layman's example of how probability calculus can help somebody avoiding some (possibly unwanted) bruises, rather than a proper mathematical answer to your question.