Corrections to probability calculation

This document will discuss corrections to the document on track-hit association using IST pads and strips located here.

G. Van Buren - BNL
23 Jan 2007
24 Jan 2007 (modified)

We start with the equation:

and define an effective area of the ellipse with radii equivalent to the resolutions:

yielding

First term

The first term in the numerator is meant to be the probability of finding the correct hit when there is only one hit. This probability (ignoring inefficiencies in actually reconstructing the hit) is 1.0. So this should read instead:

and we can even incorporate it into the numerator sum now:

This is a rather small correction. Red is before and blue is after the correction in this plot:

Sum limits

Next, the summation limit of n=4 becomes an issue as the hit density (and therefore the probabilities to get more hits) grows. We find that this is probably not an issue for occupancies up to 2.0 hits/cm^2, but can be set to a limit of n=100 to insure accuracy out well beyond our expected range of occupancies:

Again, before (red) and after (blue) this correction:

Per Track

I believe that this sum is being taken on a per detector unit basis. We are asking the question (A), "What is the probability that I will properly associate a hit in this detector element given that there is at least one hit?" I believe we should instead ask the question on a per track basis (B), "What is the probability that I will properly match the hit in this detector given that I have a track passing through it?" This has a non-neligible effect on the answer. I have a few justifications for it:

The per track argument is necessary to arrive at the formula used in the simpler case of strips only:

or

To get that result we ask the simple question, "How many hit candidates do I have for association with a track?" The existence of the track implies at least one hit. So we want to know the number of hits in the effective area, and our correct hit will be one out of those hits. If we do not ask this question per track, we find simply that we expect:

and

This equation is in fact well behaved as ρ goes to zero, but I think it isn't what we want...
If instead we ask this question per track, we multiply each possibile scenario of number of hits (n) by the number of tracks that will experience that number (also n), and then correct for this in the denominator:

and

This matches the other study, which has also been verified by Monte Carlo. If we are incorrect here, then I think we must be incorrect in the other study.
My (perhaps tragic?) logic: Let us ask the performance for correctly reconstructing D0s. We do not answer this question by asking the average of (the average number of D0s reconstructed per event with at least one D0). Instead, we ask how many of all the D0s that we threw were reconstructed. If having two D0s in an event is correlated with a worse reconstruction rate, we must (I believe) include both D0s in our performance question.
At this point in time, I think question (A) is malformed. We want to know how a detector performs, but in order to do so we must make a cut on having at least one hit in the detector unit. Question (B) does not suffer from this because we are asking the question for each track, not each detector unit.

This is not a simple topic to fully appreciate. Pressing forward with this I arrive at:

and the before/after corrections (red/blue) plot is as follows:

Combinatorics

I thought I disagreed with the document in question on this point, but I no longer do. The equation so far assumes that for the case of n hits in the unit, the probability of associating the correct hit is

This is essentially the same as the following equation where A_σ has decreased by a factor of the pad length over the strip length and ρ is replaced by (n-1)/A_p:

For a while, I thought this was wrong. I figured that since we know there are n hits in the pad, with n-1 hits not from the track in question, I could calculate the probability of properly matching the track by calculating the probability that none of the other candidates are found within the effective area. I proceeded assuming that the probability for any one hit to be in the effective area is p=A_σ/A_p, and the probability that it is not is q=(1-(A_σ/A_p)). The equations then become:

This notably reduces the correct hit association probability. Why exactly this is not right, I believe, has to do with the fact that even if you get other hits in the effective area, you still have some probability to choose the right one. So we must determine the number of expected hits and find the probability of getting it correct (as we have done before) via 1/N_hits.

My first attempt at calculating that for the binomial probability distribution with n candidates (binomial 'events' or 'samples') where N (the number of binomial successes) is greater than zero gives, interestingly, the same result as for the Poisson distribution! (note that the N in brackets is not the same as the n on the right hand side of the equation, sorry for any confusion; n is an upper bound on N)

If this is correct, it means that it is unimportant whether a binomial or Poissonian probability distribution is used for this calculation. As some affirmation, Howard's Monte Carlo simulation seems to agree with this calculation.

However, it is not clear to me why we should use this formula for N_hits in this instance. I used this before because I wanted to calculate this quantity per track, which meant putting an extra n in both the numerator and denominator. However, this particular probability sum for P_c takes that factor into account outside of this particular N_hits calculation. So I think it is wrong to do it twice.

If I instead assume that N_hits = 1 + B, where the 1 is from the track of interest, and B is from the background n-1 hits, then I come to yet another happy coincidence when I take B to be the expected number of hits from a binomial distribution with (n-1) samples because this works out to be the same value for N_hits:

Whatever the reason, this formula seems correct!

Naive intuition

I have to admit that going into this, I expected the pads to buy nothing: yes, the effective area for correct hit association decreases by a significant factor (20 for some examples of IST configurations), but this is countered by the ghost hits which will increase the hit density by the same factor. I had naively expected this to wash out.

I am still trying to understand why this is not so. The pads+strips do improve performance slightly. It has been suggested that this is a subtlety of edge effects, but I have not included such effects in my calculations, and yet the performance has improved!