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## Sampling some elements without replacement out of big population, based on a probability distribution

Asked by Clarisha Nijman

### Clarisha Nijman (view profile)

on 20 Oct 2018
Latest activity Commented on by Clarisha Nijman

### Clarisha Nijman (view profile)

on 22 Oct 2018
Accepted Answer by Bruno Luong

### Bruno Luong (view profile)

Hello,
Given population U=1:20, The associated probability series is related to the importance of each element: p=[0.01 0.05 0.11 0.08 0.02 0.02 0.01 0.055 0.1 0.2 0.05 0.05 0.01 0.03 0.025 0.06 0.04 0.05 0.01 0.02] Based on this probability distribution I want to sample 10 numbers from population U without replacement (one number should be selected at most once).
The best codes I could composed is this one, but it is a code that results in sampling with replacement U = 1:20; p=[0.01 0.05 0.11 0.08 0.02 0.02 0.01 0.055 0.1 0.2 0.05 0.05 0.01 0.03 0.025 0.06 0.04 0.05 0.01 0.02]; n = 10; c = cumsum([0,p(:).']); c = c/c(end); [~,i] = histc(rand(1,n),c); r = v(i); % map to v values
A next code I tried is U = 1:20; p=[0.01 0.05 0.11 0.08 0.02 0.02 0.01 0.055 0.1 0.2 0.05 0.05 0.01 0.03 0.025 0.06 0.04 0.05 0.01 0.02]; %to get the cumulative distribution in reversed order q=flip(p); r=cumsum(q); s=[0 r];
n = 10;%number of columns that should be selected X = zeros(n,1);%initiate list to save selected columns
%draw 10 uniform numbers, each one represents an arbitrary cumulative distribution x = rand(n,1);
%link the cdf value depending on its range to the right element of the populatione for j=1:size(U,2)-1 X(x>=s(j) & x<s(j+1))=20-j; end
But this code gives me problems at the last line and I do not know how to cover the situation when cum probabiliy equals one.
Can somebody give me some feedback/suggestion?
Thank you in advance!

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## 2 Answers

Answer by Bruno Luong

### Bruno Luong (view profile)

on 20 Oct 2018
Edited by Bruno Luong

### Bruno Luong (view profile)

on 20 Oct 2018
Accepted Answer

Your question is mathematically inconsistent. Take a very simple example, a population of 2: [1,2], and you want to generate 1 with probability 3/4 and 2 with probability 1/4.

If you draw 2 samples without replacement you either get [1,2] or [2,1], so the prescribed probabilities of 3/4 and 1/4 never meet.

Now you could simulate the real-life drawing by a loop:

```U = 1:20;
p=[0.01 0.05 0.11 0.08 0.02 0.02 0.01 ...
0.055 0.1 0.2 0.05 0.05 0.01 0.03 ...
0.025 0.06 0.04 0.05 0.01 0.02];
n = 10;
nset = 10000;
```
```% without replacement
ik = zeros(nset,n);
for i=1:nset
pk = p(:).';
for k=1:n
c = cumsum([0,pk]);
c = c/c(end);
[~,d] = histc(rand,c);
ik(k) = d;
pk(d) = 0;
end
end
wor = U(ik);
```

This give one set of drawing n element without replacement. Repeat it as long as you like. But there is no warranty the appearance is the prescribed P.

For example the element #10 has prescribed probability of 0.2, however it can appears at mots once in when drawing a sequence of 10 without replacement, so the probability the element #10 appears can never goes above 1/10 = 0.1, which is not 0.2 you want it to be. (The histogram plot in the subsequent comment bellow also confirms this, look at bin #10)

Bruno Luong

### Bruno Luong (view profile)

on 20 Oct 2018
If you run this code with the above data, you'll see the probability without replacement is distorted, it get equalized among elements (brown on the left). This is not the case with replacement (blue on the left) U = 1:20;
p=[0.01 0.05 0.11 0.08 0.02 0.02 0.01 ...
0.055 0.1 0.2 0.05 0.05 0.01 0.03 ...
0.025 0.06 0.04 0.05 0.01 0.02];
n = 10;
nset = 10000;
% without replacement
ik = zeros(nset,n);
for i=1:nset
pk = p(:).';
for k=1:n
c = cumsum([0,pk]);
c = c/c(end);
[~,d] = histc(rand,c);
ik(i,k) = d;
pk(d) = 0;
end
end
wor = U(ik);
% with replacement
c = cumsum([0,p]);
c = c/c(end);
[~,d] = histc(rand(nset,n),c);
wr = U(d);
close all
subplot(1,2,1);
h2 = histogram(wr(:),20);
hold on
h1 = histogram(wor(:),20);
legend([h1,h2],'Without replacement','with replacement');
subplot(1,2,2);
bar(p)
title('prescribed probability');

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Answer by Clarisha Nijman

### Clarisha Nijman (view profile)

on 20 Oct 2018

Bruno, the answer is very very clear, thanks a lot!

Clarisha Nijman

### Clarisha Nijman (view profile)

on 21 Oct 2018
Aha, I see, that dimension is the problem. Tnx a lot!!!
Bruno Luong

### Bruno Luong (view profile)

on 21 Oct 2018
Also you might incorporate the last edited version of algorithm without KEEP but set P(D) = 0. It will be faster.
Clarisha Nijman

### Clarisha Nijman (view profile)

on 22 Oct 2018
Tnx a lot Bruno!!!|

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