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Is there a probability cut-off point?

e.g. if the probability of something keeps halving do you stop at a certain point and say the rest of the possibilities are negligible?

What do you do then?

Ignore the rest of the possibilities in all following calculations? e.g. percentages, averages, ratios, etc.

What is an event rate and is it related?

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I'm curious as to the context of your question. A brief foray into google seems to indicate that 'Event Rate' is associated with biostatistics/Evidence-Based Medicine. So, what context is this in? – Drew Christianson Dec 10 '11 at 21:10
@DrewChristianson I came across discussions involving event rate when Googling "probability cut-off rate" – Adam Lynch Dec 13 '11 at 15:39

If you're trying to do an exact calculation, then no, you should not ignore any of the possibilities. Of course, if you just want an approximate result, ignoring low-probability events may be justified, as long as the effect of those events on what you're trying to calculate is not disproportionately large.

For example, say you're playing a game where you toss a fair coin repeatedly. Every time the coin comes up heads, you win \$1; the first time it comes up tails, the game ends. Then your expected winnings from the game are $$\frac{\$1}2 + \frac{\$1}4 + \frac{\$1}8 + \frac{\$1}{16} + \dotsb = \sum_{k=1}^\infty \frac{\$1}{2^k} = \$1.$$

If you truncate this series after $n$ terms, the sum will not be exactly \$1. But it won't be off by more than $\$1/2^n$, so the difference becomes negligible (less that one cent) after only 7 terms.

However, let's say you're playing a different game, where you win \$0.01 on the first heads, \$0.02 on the second, \$0.04 on the third, \$0.08 on the fourth and so on. You might think that this game doesn't pay as well as the previous one, since the rewards are in cents instead of dollars, and since the game is very likely to end before they grow much larger. But if you calculate the expected winnings from playing this game, they are $$\frac{\$0.01}{2} + \frac{\$0.02}{4} + \frac{\$0.04}{8} + \dotsb = \sum_{k=1}^\infty \frac{2^{k-1}}{2^k}\cdot\$0.01 = \sum_{k=1}^\infty \frac{\$0.01}{2} = \infty.$$

This despite the fact that, if you only play up to $n$ rounds of this game, your expected winnings are only $n/2$ cents, and that the odds of your game lasting more than, say, 20 rounds are literally less than one in a million! However, if you do manage to toss 20 heads in a row, you'll already have won over \$10,000, and the prizes just keep going up from there.

The example I used above is known as the St. Petersburg paradox. You can find a lot of interesting discussion about the implications of this paradox, but the message to take home from this example is that you shouldn't neglect even extremely unlikely events if the effects of those events, if and when they do happen, is equally extreme.

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But if the probabilities keep having like I said then it never ends so how can you include all possibilities? e.g. if a grasshoppers jump is as long as half the distance between itself and its destination then it never gets there. – Adam Lynch Dec 13 '11 at 15:48
@AdamLynch: By summing the infinite series, like I did in the examples above. If you really can't do that, then yes, you need to figure out a point where it's safe to truncate the sum. Which typically involves (either blind faith or) estimating some upper bound for the error resulting from the truncation, which often again means summing an infinite series -- but hopefully a simpler one than the one you started with. – Ilmari Karonen Dec 13 '11 at 15:55

From the center for evidence based medicine, the event rate is defined as:

The proportion of patients in a group in whom the event is observed. Thus, if out of 100 patients, the event is observed in 27, the event rate is 0.27. Control event rate (CER) and experimental event rate (EER) are used to refer to this in control and experimental groups of patients, respectively. The patient expected event rate (PEER) refers to the rate of events we'd expect in a patient who received no treatment or conventional treatment.

Ilmari is completely correct that, in general, neglecting low probability events is a bad idea. However, how does this relate to the event rate? Take as an example some disease with probability of $p=.5$ of manifesting in any member of the population. Then, take a sample of the population of size $n=500$. How many people would we expect to have the disease? Under the strong assumption that the disease is an independent random process, the number of cases in the same is a binomial RV - lets call it $\mathbb{X}$. We use the usual form of expected value to find the expected number of cases:$$E\left[ \mathbb{X} \right] = \sum_{k=0}^{n} k \binom{n}{k} p^k (1-p)^{n-k} = np$$

Which essentially gives use the event rate for the sample. For the n and p given above, that expected event rate would be $.5*500=250$.

Now consider what would happen if you had some cut-off level m below which you equated the probability to 0. You'd still run the same equation from above, but only pick off terms with a probability greater than a certain value. Essentially, you'd be doing the calculation:$$E\left[ \mathbb{X} \right] = \sum_{k=0}^nk{n \choose k}p^k(1-p)^{n-k}H\left({n \choose k}p^k(1-p)^{n-k}-m\right)$$ Where H(x) is 0 if $x\leq 0$ and 1 otherwise (per Ilmari's suggestion).

For $m=\frac{1}{1000000}, E\left[ \mathbb{X} \right]= 249.9733$. For $m=\frac{1}{1000}, E\left[ \mathbb{X} \right]= 233.184.$ If n were larger, the underestimations would be more significant.

The upshot is that it is rarely a good idea to neglect probabilities close to 0, regardless of bound. Effects might not be present in small samples or with very small bounds, but they will be there. Use the fullest precision possible, whenever possible, at least while working out a problem. Round only at the end if the situation demands it.

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Nice example, but your numbers seem a bit off. (I get $np = .5 \cdot 1000 = 500$, not $250$.) Also, wouldn't "neglecting events with probability less than $m$" correspond to $$\sum_{k=0}^nk{n \choose k}p^k(1-p)^{n-k}H\left({n \choose k}p^k(1-p)^{n-k}-m\right),$$ where $H(x)$ is $0$ if $x<0$ and $1$ otherwise? – Ilmari Karonen Dec 12 '11 at 20:11
Nice catch, I was playing around with sample size and proportion to get the effect I wanted for the example and it looks like I forgot to change the numbers all the way through. My intention was n=500, p=.5, I think everything matches now. You're right on the summand too, mine underestimates the probabilities by a bit. – Drew Christianson Dec 12 '11 at 22:21
In theory, Drew and Ilmari are correct. In practice, there's not much point in worrying about events with extremely low probabilities. For example, the kinetic theory of gases says there is a nonzero probability that all the oxygen molecules in the room will move away from me, causing me to suffocate. But I think I'm pretty safe in calling that negligible. – Robert Israel Dec 13 '11 at 19:11

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