We know that prob argument in sample is used to assign a probability of weights.
For example,
table(sample(1:4, 1e6, replace = TRUE, prob = c(0.2, 0.4, 0.3, 0.1)))/1e6
# 1 2 3 4
#0.2 0.4 0.3 0.1
table(sample(1:4, 1e6, replace = TRUE, prob = c(0.2, 0.4, 0.3, 0.1)))/1e6
# 1 2 3 4
#0.200 0.400 0.299 0.100
In this example, the sum of probability is exactly 1 (0.2 + 0.4 + 0.3 + 0.1), hence it gives the expected ratio but what if the probability does not sum to 1? What output would it give? I thought it would result in an error but it gives some value.
When the probability sums up to more than 1.
table(sample(1:4, 1e6, replace = TRUE, prob = c(0.2, 0.5, 0.5, 0.1)))/1e6
# 1 2 3 4
#0.1544 0.3839 0.3848 0.0768
table(sample(1:4, 1e6, replace = TRUE, prob = c(0.2, 0.5, 0.5, 0.1)))/1e6
# 1 2 3 4
#0.1544 0.3842 0.3848 0.0767
When the probability sums up to less than 1
table(sample(1:4, 1e6, replace = TRUE, prob = c(0.1, 0.1, 0.5, 0.1)))/1e6
# 1 2 3 4
#0.124 0.125 0.625 0.125
table(sample(1:4, 1e6, replace = TRUE, prob = c(0.1, 0.1, 0.5, 0.1)))/1e6
# 1 2 3 4
#0.125 0.125 0.625 0.125
As we can see, running multiple times gives the output which is not equal to prob but the results are not random as well. How are the numbers distributed in this case? Where is it documented?
I tried searching on the internet but didn't find any relevant information. I looked through the documentation at ?sample which has
The optional prob argument can be used to give a vector of weights for obtaining the elements of the vector being sampled. They need not sum to one, but they should be non-negative and not all zero. If replace is true, Walker's alias method (Ripley, 1987) is used when there are more than 200 reasonably probable values: this gives results incompatible with those from R < 2.2.0.
So it says that the prob argument need not sum to 1 but doesn't tell what is expected when it doesn't sum to 1? I am not sure if I am missing any part of the documentation. Does anybody have any idea?