Preferentially Sampling Based upon Value Size

Question

So, this is something I think I'm complicating far too much but it also has some of my other colleagues stumped as well.

I've got a set of areas represented by polygons and I've got a column in the dataframe holding their areas. The distribution of areas is heavily right skewed. Essentially I want to randomly sample them based upon a distribution of sampling probabilities that is inversely proportional to their area. Rescaling the values to between zero and one (using the {​​​​​​​​x-min(x)}​​​​​​​​/{​​​​​​​​max(x)-min(x)}​​​​​​​​ method) and subtracting them from 1 would seem to be the intuitive approach, but this would simply mean that the smallest are almost always the one sampled.

I'd like a flatter (but not uniform!) right-skewed distribution of sampling probabilities across the values, but I am unsure on how to do this while taking the area values into account. I don't think stratifying them is what I am looking for either as that would introduce arbitrary bounds on the probability allocations.

Reproducible code below with the item of interest (the vector of probabilities) given by prob_vector. That is, how to generate prob_vector given the above scenario and desired outcomes?

# Data
n= 500
df <- data.frame("ID" = 1:n,"AREA" = replicate(n,sum(rexp(n=8,rate=0.1))))

# Generate the sampling probability somehow based upon the AREA values with smaller areas having higher sample probability::
prob_vector <- ??????

# Sampling:
s <- sample(df$ID, size=1, prob=prob_vector)```

Answer 1

There is no one best solution for this question as a wide range of probability vectors is possible. You can add any kind of curvature and slope. In this small script, I simulated an extremely right skewed distribution of areas (0-100 units) and you can define and directly visualize any probability vector you want.

area.dist = rgamma(1000,1,3)*40
area.dist[area.dist>100]=100
hist(area.dist,main="Probability functions")

area = seq(0,100,0.1)
prob_vector1 = 1-(area-min(area))/(max(area)-min(area))  ## linear
prob_vector2 = .8-(.6*(area-min(area))/(max(area)-min(area))) ## low slope
prob_vector3 = 1/(1+((area-min(area))/(max(area)-min(area))))**4  ## strong curve
prob_vector4 = .4/(.4+((area-min(area))/(max(area)-min(area))))  ## low curve
legend("topright",c("linear","low slope","strong curve","low curve"), col = c("red","green","blue","orange"),lwd=1)


lines(area,prob_vector1*500,col="red")
lines(area,prob_vector2*500,col="green")
lines(area,prob_vector3*500,col="blue")
lines(area,prob_vector4*500,col="orange")

The output is:

The red line is your solution, the other ones are adjustments to make it weaker. Just change numbers in the probability function until you get one that fits your expectations.