Asked by mutah
on 29 Aug 2014

Hi all

is the code below represent inverse cdf method?

y is pdf of any distribution

cdf_y = cumsum(y);

sum_y = sum(y);

for j = 1:N

randx = sum_y*rand();

i = 1;

while cdf_y(i) < randx

i = i + 1;

end

f(j) = x(i); end

please give me explain

Answer by Roger Stafford
on 1 Sep 2014

Edited by Roger Stafford
on 1 Sep 2014

Accepted Answer

In the code you show, Mutah, strictly speaking the y vector is not a pdf - a probability distribution function. It is a pmf - a probability mass function. Actually the author of the code has allowed y to be merely proportional to these probabilities, so y is proportional to a pmf. That is because it is dealing with a discrete set of values, not a continuous distribution. For each i, y(i) must be proportional to the probability of a random variable, f, being equal to the corresponding x(i).

With that understanding, the code is correct and the N-length array, f, will have a statistical distribution in accordance with the y values. The reason the author finds the cumsum(y) and sum(y) is that cumsum(y) is the cumulative sum of the y values and therefore proportional to the cumulative probability that is required of the result, and sum(y) would be the proportionality constant. If on the i-th trip through the for-loop, we have 0 < randx <= cdf_y(1), the while loop will not execute at all and you will get i = 1 which will insert x(1) into f. If cdf_y(1) < randx <= cdf_y(2), the while loop will make just one trip and i will emerge equal to 2, which inserts x(2) into f. This continues to be true all the way to the case cdf_y(end-1) < randx < cdf_y(end) which would then put x(end) into f. Therefore this has the effect of finding the inverse function of cdf_y, that is for each cumulative probability value, randx, it provides the appropriate x(i) value.

However, it must be said that this code is not very efficient. It does not take advantage of the fact that the cdf_y values are an ascending sequence of values. It should not be necessary to possibly have to run through the entire cdf_y sequence with the while loop seeking the point where each randx fits into the correct interval of cdf_y values. The following code uses the matlab function 'histc' which does take advantage of the ordering in cdf_f and accordingly should be much faster code for large x and y arrays.

cdf_y = cumsum(y);

cdf_y = cdf_y/cdf_y(end);

[~,ix] = histc(rand(N,1),[0;cdf_y]);

f = x(ix);

I have assumed here that y and x are column vectors. To understand why this works, you should make a careful study of the 'histc' function at:

http://www.mathworks.com/help/matlab/ref/histc.html

Roger Stafford
on 2 Sep 2014

mutah
on 15 Sep 2014

Error using histc

Edge vector must be monotonically non-decreasing.

Roger Stafford
on 15 Sep 2014

"y is pdf of any distribution" The only way I can see that your edge vector failed to be monotonically non-decreasing is that one of your 'y' values was negative. However, the pdf of distributions should always be non-negative.

You should be able to check that for yourself. Find the point where cdf_y decreases and check the corresponding y value.

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Answer by Image Analyst
on 29 Aug 2014

Image Analyst
on 29 Aug 2014

mutah
on 29 Aug 2014

what about this

N=10000

for j = 1:N

randx = sum_y*rand();

i = 1;

while cdf_y(i) < randx

i = i + 1;

end

f(j) = x(i);

end

Image Analyst
on 31 Aug 2014

Well, like I said, the while (and line before and two lines after) could be replaced:

i = find(cdf_y > randx, 1, 'first')

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## José-Luis (view profile)

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