How to deal with 'cut-off' selection bias/sampling bias? (truncated distribution) In short
When measuring an outcome with a normal distribution, but whos mean is below the detection threshold, can you still make statements about differences between populations?
Example
Say I only pick up outcomes that are above a certain value. In one population it happens to be the 90th percentile for instance, so I only measure 10% of the outcomes. So In the image below I only pick up outcomes to the right of the vertical bar.
Then in another population I do the same measurement and get different outcomes. For both populations I know the mean is below my threshold but for neither I know how far below it is or what the means and variances are.

My question
Can I still make a sensible comparison between these groups in any way? If my samples would be evenly distribubted over the possible values of the outcome I would just have done a t-test.
My thoughts so far
Will the mean I measure still be higher if the actual mean is higher? (I think if the variance is equal then yes, but I might be wrong)
And what if theres not an exact cutoff point but my sampling probability is more like a sigmoidal function with its center at a certain value? The chance I measure an outcome is dependent my measurement so probably there is a 'sampling distribution' which I think should have the shape of the integral of a normal distribution. Maybe when taking this into account I can deduce something more about the true distribution in the population other than just my measurements which are mostly in the tail of the actual distribution.
I have about 20 measured outcomes per population, if I plot its histogram can I deduce something about the actual distribution?
 A: In the case of censoring, one sensible comparison could be performing quantile regression and comparing both populations at a higher percentile. Say you have 60% of your data below your detection threshold, you could still compare both populations at the 75th percentile. Here's a useful paper about this approach:
@Article{22769433,
AUTHOR = {Eilers, Paul and Roder, Esther and Savelkoul, Huub and van Wijk, Roy},
TITLE = {Quantile regression for the statistical analysis of immunological data with many non-detects},
JOURNAL = {BMC Immunology},
VOLUME = {13},
YEAR = {2012},
NUMBER = {1},
PAGES = {37},
URL = {http://www.biomedcentral.com/1471-2172/13/37},
DOI = {10.1186/1471-2172-13-37},
PubMedID = {22769433}}

According to your last comments, you are dealing with truncation. In that case, if you know the percentile of truncation you still can use quantile regression to compare both populations at a higher percentile. However, you'll need a large sample size for this aproach to work. With a threshold at 80th percentile, 200 samples per group should provide reasonable estimates. Let's see an example:
a<-rlnorm(1000)  #Population A
at<-na.omit(ifelse(a>2.32, a, NA))  #Truncation at 80th percentile, this is your sample
at_imp<-c(rep(min(at), length(at)*4), at)    #We have 20% of values, so we "fill" the other 80% with our minimum

b<-rlnorm(1000, 1)   #Population B
bt<-na.omit(ifelse(b>6.31, b, NA))  #Truncation at 80th percentile, this is your sample
bt_imp<-c(rep(min(bt), length(bt)*4), bt) #We have 20% of values, so we "fill" the other 80% with our minimum

truncated_data<-data.frame(values=c(at_imp, bt_imp), group=c(rep("A", length(at_imp)), rep("B", length(bt_imp))))
original_data<-data.frame(values=c(a, b), group=c(rep("A", 1000), rep("B", 1000)))

#Comparison of results
library(quantreg)
fit1<-rq(values ~ group, truncated_data, tau=0.9)  #Quantile regression for 90th percentile
summary(fit1)  #Results for truncated data

fit2<-rq(values ~ group, original_data, tau=0.9)
summary(fit2)  #Results for original data

If you don't know the percentile of truncation but know the raw value, you could try truncated regression, but with such a high degree of truncation I dont think you would get reasonable estimates.
