Occurrence certainty

Question

I think the best way to ask this question is with a hypothetical situation. Let's say I have a sensor that measures on average one occurrence every 3 days with a standard deviation of about 4. I want to move that sensor to a new unknown location. How many days can I leave the sensor in this new location until I'm confident that this location will not produce an occurrence? Like 6,7,8 days...?

In this plot I have included, the data is 93 days where integer occurrences are recorded. The average occurrence per day is 2.95. The std is 4.03. 0 is by far the most common value. In the data (which you cant see in the histogram) the longest stretch of 0 entries was 6 days. So if I move the sensor to some unknown location, how many days can it sit there reporting 0's until I'm confident that this unknown location doesn't experience occurrences

Answer 1

You seem to need to change the mean and standard deviation according to circumstances. So, it may be best to model waiting times between events as gamma distributed and round up to integers on account of daily data collection. This matches the story in the first part of your question.

(Notice that this is different from looking at the number of events per day, as in the last part with the histogram.)

Specifically, consider $\mathsf{Gamma}(\mathrm{shape}=9/16, \mathrm{rate} = 3/16),$ which has $\mu = 3, \sigma=4.$

Here is a summary of $1000$ such waiting times simulated in R:

set.seed(2022)
x = ceiling(rgamma(1000, 9/16, 3/16))
mean(x);  sd(x)
[1] 3.627
[1] 3.809512

This gamma distribution has probability about $0.1232$ of a wait exceeding 7 days between events, and probability $0.002$ of three consecutive waits exceeding 7 days. You could choose some such criterion to decide if events at a new site are too sparse to be of interest.

1-pgamma(7, 9/16, 3/16)
[1] 0.1231928
(1-pgamma(7, 9/16, 3/16))^3
[1] 0.00186963

Here is a histogram of my simulated sample of $1000$ waiting times.

hist(x, br=(0:31)-.5, col="skyblue2")
abline(v=7.5, col="red")

Waits of $7$ or more among the first $100$ waiting times (total of $341$ days) are designated by TRUEs below:

y = x[1:100]
sum(y)
[1] 341

y >= 7
  [1] FALSE FALSE FALSE FALSE FALSE FALSE  TRUE FALSE  TRUE FALSE
 [11] FALSE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE FALSE FALSE
 [21] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
 [31] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE  TRUE
 [41] FALSE  TRUE FALSE FALSE  TRUE  TRUE FALSE FALSE FALSE FALSE
 [51] FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE FALSE
 [61] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE  TRUE
 [71] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
 [81] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE  TRUE
 [91] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE