what called this method of generation random sample

2 visualizzazioni (ultimi 30 giorni)
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

Risposta accettata

Roger Stafford
Roger Stafford il 1 Set 2014
Modificato: Roger Stafford il 1 Set 2014
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
  5 Commenti
mutah
mutah il 15 Set 2014
Error using histc
Edge vector must be monotonically non-decreasing.
Roger Stafford
Roger Stafford il 15 Set 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.

Accedi per commentare.

Più risposte (1)

Image Analyst
Image Analyst il 29 Ago 2014
Not sure exactly what you mean and how specific your question is to that exact code, but in general the process of generating a bunch of random numbers and running an "experiment" N times (in a loop like you did) is called a "Monte Carlo Simulation". For example, check out the attached Monty Hall Problem that I coded up as a Monte Carlo simulation.
  10 Commenti
mutah
mutah il 29 Ago 2014
Modificato: mutah il 29 Ago 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
Image Analyst il 31 Ago 2014
Well, like I said, the while (and line before and two lines after) could be replaced:
i = find(cdf_y > randx, 1, 'first')

Accedi per commentare.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by