Sampling from a population so that the sample has a target mean/median.

Hi all,

if there a direct way to randomly collect samples from an existing population multiple times so that each time the sample has a target mean or median for a given metric?

Example:

Dataset = 400 observations X 10 metrics

I want to sample N times (with replaecement) 20 observations so that the mean or median for each 20-observation sample of metric 1 equals a target value and then calculate the mean for the other 9 metrics for each of these 20-observation samples.

Is there a way to do such sampling in Matlab?

Thank you!

[Torsten]

What do you mean with "for a given metric" ? So mean and median are not calculated in the normal way ?

[Antonios Asiminas]

Apologies for confusion.

What I mean is: Sampling from Dataset(:,1) and get indexes for the elements of each sample (idx) so that each sample has mean(Dataset(idx, 1)) == target, and calculate mean(Dataset(idx, 2)), mean(Dataset(idx, 3)),... mean(Dataset(idx, 10)). Repeat that N times.

The "or median" was cover the posibility there is a solution for this problem with median target rather than mean.

I thought trying a while loop and sample, check if the mean of the sample is the target (or close enough) and then calculate the other metrics means. This is not a nice a certainly not a fast solution though...

I hope this makes more sense now.

[Image Analyst]

Not really. How would you compute idx? And I'm not sure what N is when you say that you need to get the 10 means N times.

I'm thinking what you really want is like what Bruno said where you'd get a list of indexes that match a target or are in a target range, and then get the means for the other columns, so like

rowsOfInterest = Dataset(:, 1) == target;
theMeans = mean(Dataset(rowsOfInterest, :), 1);

That would give you the mean of all columns but only for those rows where column 1 is your target value.

Of course you could also use grpstats() or groupsummary() or splitapply() to do the same thing.

Answer 1

[Bruno Luong]

This is for "mean" (EDIT code for fixing BUG)

% Generate 1e6 Dummy Data test
A=10+sum(30*rand(1e6,3),2);
% number of subsamples to be draw from A
m = 1000;
targetmean=35; % taget mean

meanA = mean(A);
if targetmean > meanA
    dir = 'ascend';
else
    dir = 'descend';
end
As = sort(A, dir);
A1 = As(1);
Aend = As(end);
t = (As-A1)/(Aend-A1);
nA = length(A);

% if you give non-attainable targetmean, you get error here or NaN p
EspFun = @(p) sum(As.*t.^p) / sum(t.^p);
p = fzero(@(p) EspFun(p)-targetmean, [0 1000]);
if isnan(p)
    error('target mean not possible with this formulation')
end
idx = ceil(nA*rand(1,m).^(1/(p+1)));
Asubsample = As(idx);
% Check
mean(Asubsample)

ans = 34.5036

[Bruno Luong]

If you want non replacement draw you have to set m value larger than the subsample cardinal

desiredsubsamplecardinal = 100;
m = ceil(1.1*desiredsubsamplecardinal);

then later do

Asubsample = As(unique(idx));
if length(Asubsample) >= desiredsubsamplecardinal
    Asubsample = Asubsample(1:desiredsubsamplecardinal);
else
    ... retry
end

[Bruno Luong]

This is for "median" (EDIT code for fixing BUG)

% Generate 1e6 Dummy Data test
A=10+sum(30*rand(1e6,3),2);
% number of subsamples to be draw from A
m = 1000;
targetmedian=35; % taget edian

medianA = median(A);
nA = length(A);
if medianA > targetmedian
    dir = 'ascend';
    thalf = sum(A <= targetmedian) / nA;
else
    dir = 'descend';
    thalf = sum(A >= targetmedian) / nA;
end
As = sort(A, dir);
p = max(log(thalf)/log(0.5),0);
idx = ceil(nA*rand(1,m).^p);
Asubsample = As(idx);

% Check
median(Asubsample)

ans = 35.5359

Answer 2

[Image Analyst]

No, not in general. The target mean may not exist in the population. If my target mean weight is 10 kg, and my population is the weights of elephants, I cannot sample the weights of the elephants and ever get a weight of 10 Kg.

If the targets are definitely in your population then you can't guarantee any mean if you randomly pick some samples. You may however get close to the population mean. If you want a specific mean I think you'd have to sort your population and then sample around the target mean, but then you're no longer doing it randomly.

[Antonios Asiminas]

I see what you're saying. The mean is definitely in the population.

What I will do it then decide on a range of value (target mean +/- a tolerance) and then using a while loop sample randomly and if the mean of metric of interest is within the range count up, and do that until I reach the required interations.

Thank you for the super fast response!

[Image Analyst]

What is the use case? Why would you ever want to get a value that is just some value you forced upon it, rather than a value that is true?

Let's say you had uniformly distributed data between 1000 and 2000. Ok, maybe you can do some manipulations to get a value of 1349 or 1842 or 1065, but what's the point of that? It really doesn't mean anything - it's just some value you decided upon in advance without any regard to what your data actually are!

[Bruno Luong]

@Image Analyst "What is the use case?"

Well I can imagine to do some stats, such as "what a population of 20 years old average think about ecology?" We then draw a subsample of 20 years of everage on the available database of and do some statistic on it. How you draw a subsample of 20 years of everage from the database? Well this is exactly the question asked.

[Antonios Asiminas]

Thank you again for the suggestions, discussion and code.

The use case is the following: I have several observations (100s) of a population for two time points. I have calculated X metrics for each observation. When I calculate the statistics for each metric/time point, I get a change (subtle but robust) in a few metrics between timepoints.

Now the question I have is: Could the changes in the majority of metrics secondary to the change in one target metric?

To address this question I am planning to get the mean from my first timepoint for the target metric and sample from the second timepoint many times a smaller population (20-30) with the same mean as the baseline for the target metric, and then calculate the means for the remaining metrics of interest.

If changes in the target metric are primary then the distribution of means for the different metrics are not going to be different between timepoints.

I suspect I can reach a similar conclusion by just correlating the target metric and each of the other metrics?

A

[Bruno Luong]

@Antonios Asiminas Frankly you should accept my answer with code if it does what you want rather than @Image Analyst who simply tell more or less it's not possible to carry out the task, which is wrong (since my code does that).

Still don't know what you meant by "metric" (is it a universal math term?) by never mind.

If you want (e.g. 5) many group of 20, just generate with m = 100, then randomly split in 5 x 20. They must have a same theoreticll mean/median.

If my code doesn't do what you want please tell why.

[Antonios Asiminas]

Thank you very much! By metric I mean a calculation that describes an aspect of each sample; If for example my samples a pebbles at a beach, I can calculate a few metrics to describe these pebbles (weight, shape, colour etc). In my case I have cells.

Thank you again.

AA

[Image Analyst]

@Antonios Asiminas you say "Now the question I have is: Could the changes in the majority of metrics secondary to the change in one target metric? "

So couldn't you use Principal Components Analysis (PCA) to determine the the percentage of variation explained by the various metrics in the two time-point populations? For example at time point 1 (say, Summer) you could find out that 95% of the variation is due to weight differences between the pebbles and 5% is due to color differences. And at time point 2 (say, winter) you might find out that 85% of the variation is due to weight differences between the pebbles and 15% is due to color differences. PCA will tell you exactly how important each metric is.

[Beorn Nijenhuis]

@Bruno Luong This was a helpful script for me. I had a cohort of n=280 samples of peoples ages (15<age < 75) and needed a subsample of 60 with a mean of 50. It worked. I have a question though: When I run this code I notice a tolerance in the output. With n=280 the dolerance was ±7 years approximatly. How is this tolerance calculated so I can write this in to the methods of my paper?

[Bruno Luong]

I don't know how it is calculate, but it's due to the fact that the sample is too sparse. My method assume it is well approximated a uniform distrubution; so it works well when one have a big sample of data.