# "Normalising" join probability of n events, by taking n-th root

I have a group of events which I guess you could call a compound events.

Each event is something like: $$A=A_1\cap A_2\cap...\cap A_{n_a}$$

I am estimating the probability of the over all event by assuming independence of the components

$$P(A) = \prod^{i=n_a}_{i=1} P(A_i)$$

I have another compound event (I actaully have untold thousands of them)

$$B=B_1\cap B_2\cap...\cap B_{n_b}$$

$$P(B) = \prod^{i=n_b}_{i=1} P(B_i)$$

But since $n_a \ne n_b$ comparing the probabilities between $A$ and $B$ didn't seem fair. Since it is significantly less likely for $n+1$ co-occurances than for $n$ And even worse when the difference in $n_a$ and $n_b$ is greater

So I took to using:

$$P^\prime(A) = \prod^{i=n_a}_{i=1} \left(P(A_i)\right)^{\frac{1}{n_a}}$$

$$P^\prime(B) = \prod^{i=n_b}_{i=1} \left(P(B_i)\right)^{\frac{1}{n_b}}$$

Which I now realize is equivalent to substituting for the probabilities of $A$ and $B$, the geometric means of the probabilities of their components.

Does this make sense to do? Is this an actual technique people use?

• You haven't said what you are trying to compare about A and B. Considering your specific application is it "unfair" to compare P(A) to P(B) simply because $n_b>n_a$?
– Hugh
Sep 12 '16 at 8:26

To expand on my comment about your specific application, here's a practical example about deciding if normalization is appropriate. The example isn't about probabilities but it makes a point about normalization in general.

Consider a school which has two children, one who is 10 years old and is 4ft tall, the other is 15 years old and is 5ft tall.

You want to compare their success as a basketball player by examining their height. There are two example applications for this.

1. You want to know who would be better for the school basketball team this year. The 15 year old kid is taller so he looks like a better player. Just like your probability example you might say "This is unfair, the 10 year old is shorter but he is tall for his age, we can't compare purely on height"; it's true that the 10 year old is tall for his age but that is irrelevant because you want a good basketball player for this year's school team.

2. A second comparison you might do is to guess which child has a better chance of becoming an NBA star is his life. Neither of them can become a professional basketball player immediately so their immediate heights are not a good measure of their performance. Now it is valid to say "height is an unfair comparison, the 10 year old is tall for his age". You could normalise their heights by dividing their height by their age (or something more sophisticated) and this can give you a better measure of who will be taller when they play at a professional level.

For your question about probabilities you might find that normalizing is a suitable way to compare the two probabilities, but if you just care about which event has a greater chance of occurring then it's not "unfair" on event $B$ at all.

• My particular question is about geometric mean, as a method for normalizing. Rather than "Is normalizing appropriate at all?" Sep 12 '16 at 9:34
• I'm demonstrating the need for you to examine your specific application. There must exist similar cases where geometric mean is a better or worse way to quantify the probability.
– Hugh
Sep 12 '16 at 12:23