# Testing equivalence of two censored distributions

If we observe two censored distributions, where all observations above a cutoff are set at the value of the cutoff, how can we test whether the observable distributions suggest that the two censored distributions come from the same true distribution?

We could imagine this for income data where incomes above a certain level are reported in a form like '\\$250k/year and greater.' Or we could imagine data on campaign contributions where people can only donate $$\X$$, but some probably would have donated more in the absence of the cap.

For example:

d1 <- rnorm(n = 1000, sd = 5)

d2 <- rnorm(n = 1000, sd = 5)

d1 <- ifelse(d1>5,5,d1)

d2 <- ifelse(d2>10,10,d2)

par(mfrow=c(1,2))
hist(d1, xlim = c(-20, 20), ylim = c(0,200))
hist(d2, xlim = c(-20, 20), ylim = c(0,200))


• If it's recorded as "250+", it is censored, rather than truncated. – Glen_b Oct 30 '18 at 3:46
• Truncated would be the case where you somehow didn't even see the ones that were above your threshold. [For example, imagine a fishing net that only catches fish above a certain size; that's left-truncated] – Glen_b Oct 30 '18 at 21:25
• @jbowman I believe that the censoring here is happening at different points, so I don't think that a KS test is appropriate as the samples are, by definition, not from the same distribution. OP is asking about the true distribution from which the censored observations arose. – adityar Nov 13 '18 at 10:16
• We are assuming that you know the censoring points, right? If so, you just take the less censored distribution, censor it to match the more censored distribution, and then use whatever test for equality of distributions you like. – Bill Nov 13 '18 at 14:11
• @Bill, while that is a reasonable way to get an answer, you're losing some information. For example, just before censoring, the distribution is sloping downward in the first image, so it's a reasonable guess that this will continue moving forward, thus adding more certainty that the second image is produced from the same distribution. – adityar Nov 13 '18 at 15:10