Truncated Gaussian

Generate a pseudo-random vector X drawn from the truncated Gaussian distribution

Description

% function X = TruncatedGaussian(sigma, range)
% X = TruncatedGaussian(sigma, range, n)

Generate a pseudo-random vector X of size n, X are drawn from the truncated Gaussian distribution in a RANGE braket; and satisfies std(X)=sigma.

RANGE is of the form [left,right] defining the braket where X belongs. For a scalar input RANGE, the braket is [-RANGE,RANGE].

If input SIGMA is negative, X will be forced to have the same "shape" of distribution function than the unbounded Gaussian with standard deviation -SIGMA: N(0,-SIGMA). It is similar to calling RANDN and throw away values ouside RANGE. In this case, the standard deviation of the truncated Gaussian will be different than -SIGMA. The *effective* mean and the effective standard deviation can be obtained by calling:
 
 [X meaneffective sigmaeffective] = TruncatedGaussian(...)

http://en.wikipedia.org/wiki/Truncated_normal_distribution

• Acknowledgements: This submission has inspired the following:; Truncated multivariate normal
• MATLAB release: MATLAB 7.8 (R2009a)
• Other requirements: Probably work on most MATLAB version No statistics Toolbox is required

[summersyu Yu]

How to change the mean?

[Bruno Luong]

To change the mean, you might try this:

shift = something
X = shift + TruncatedGaussian(sigma, range-shift, ...)

[Ze Dagher]

Bruno,

your code has one problem. Enter this command and see what happens:
Y = TruncatedGaussian(-1,[10 20],[1 1000]);
All the values returned are 20s. I know that 10 to 20 is a very unlikely range with a distribution of N(0,1) but still, I am just showing here that I noticed that your code returns wrong values sometimes.

Another (possibly related) problem is when you try and generate a distribution with non-zero mean with the range being a bit far from the mean. The histogram of the returned values does not match a trusted method I used. I am not sure what this is about.

[Bruno Luong]

Thanks, Ze I made some correction to the code. The new version is available.

[Ze Dagher]

Thanks Bruno. This turned out to be the issue in the second problem I described. The output now matches derivations.

[summersyu Yu]

Dear Bruno,

Thank you for your answer. Following is my understanding:

 [X meaneffective sigmaeffective] = TruncatedGaussian(...)
If meaneffective is 5, but I want the mean to be 15.
Then shift=10
X = shift + TruncatedGaussian(sigma, range-shift, ...)

This is how we generate random variables from truncated normal distribution with mean 15, variance sigmaeffective and range [a,b].

But what is the CDF of a random variable from truncated normal distribution?

Let y be the cdf of a truncated normal variable x, then is the following equation right? I am not sure of the mean and variance in normcdf

y = (normcdf(x) - normcdf(a)) ./ (normcdf(b) - normcdf(a))

 

[Bree]

Hi Bruno,

Will you generalize this code into the multivariate one?
Trying to generate this kind of numbers, but in high dimensionality. Say something n = [128, 1e6] (rather than n=[1, 1e6].

[Bruno Luong]

Hi Bree, not in my intention.

[Ling Xue]

what does "meaneffective sigmaeffective "mean? We need to need the mean and variance before generating Gaussian random variable, right? It seems that meaneffective is changing even the range and sigma remains the same.

[Manu]

thanks!

[milad babaei]

Hi Bruno,is that possible to change the bound in a code and generate random numbers??for example in this zone n=[1, 1e4].

[Bruno Luong]

Midal, the RANGE second argument of TruncatedGaussian can be used for bound.

[Zack]

Hi bruno
Can you please tell me how to use your function for a truncated gaussian with mean mu non zero a standard deviation sigma and a range [-1 1]?

is it possible to extend this function for a [a b] with a<>b and maybe a or b equal to infinity?

Thanks

[Bruno Luong]

Zack,
first question: see my comment #2 above.
second question: yes the function accept infinity as range.

[Ricardo Giglio]

Hi Bruno, I'd like to generate two series with same mean but different variances as described below, but I always get different means... Could you help me understanding why? Thanks

A_bar = 1.5; A_var = 4.5;
B_bar = 1.5; B_var = 6.5;

A = A_bar + TruncatedGaussian(sqrt(A_var), [1 inf], 10);
B = B_bar + TruncatedGaussian(sqrt(B_var), [1 inf], 10);
mean(A(:))
mean(B(:))

ans =

    4.9641

ans =

    5.6081

[Bruno Luong]

The effective mean of the truncated Gaussian changes with the range. To get the effective mean of the date, call in two outputs form:

>> [X meaneffective] = TruncatedGaussian(sqrt(4.5), [1 inf], [1,1e6]);
>> meaneffective

meaneffective =

    3.7034

>> mean(X)

ans =

    3.7016

>> [X meaneffective] = TruncatedGaussian(sqrt(6.5), [1 inf], [1,1e6]);
>> meaneffective

meaneffective =

    4.2673

>> mean(X)

ans =

    4.2684

HINT: to get the mean you aim for, use FZERO in this function:

function deltamean = meanTG(xbar, meantarget, stdtarget, range)

[~, meaneffective] = TruncatedGaussian(stdtarget, range-xbar, 0);
deltamean = meantarget - (meaneffective + xbar);

end % meanTG

% Call fzero
meantarget = 6;
stdtarget = 3;
range = [1 Inf];

xbarstart = 10;
xbar = fzero(@(xbar) meanTG(xbar, meantarget, stdtarget, range), xbarstart)

% Create a shift TrunccatedGaussian
X = xbar + TruncatedGaussian(stdtarget, range-xbar, [1 1e6]);

% Check
fprintf('mean(X) = %f\n', mean(X));
fprintf('std(X) = %f\n', std(X));
fprintf('min(X) = %f\n', min(X));
fprintf('max(X) = %f\n', mean(X));

[milad babaei]

tnx bruno,what does "meaneffective sigmaeffective "mean?FOR example: what's the definition of this:

mean(C)=623.066, estimated=624.239
sigma=147.65, effective=91.8246, estimated=90.8613

[Bruno Luong]

Mean effective, sigma effective are the real mean and standard deviation of the random variable after truncation. The word "effective" is to make the distinction with the mean and standard deviation of the full Gaussian variable BEFORE the truncation.

[Bruno Luong]

Please note the typo in the last line in my comment from August 26th 2011

fprintf('max(X) = %f\n', max(X));

[Raymond]

Dear Bruno,

How do a generate from the normal distribution with mean zero and standard deviation 1 a truncated normal on the range [2 500]? I calculate the this random variable has a standard deviation of 0.3381. When I try [X meaneff sigmaeff] = TruncatedGaussian(0.3381, [2 500], 1) I get an error message.

[Bruno Luong]

Raymond, there exist no random variable in [2 500] with mean = 0. The mean must be in [2, 5000] regardless the distribution. Thus what you ask is mathematically impossible.

[Raymond]

Dear Bruno,

I'm very sorry not to have been more precise. I wanted to generate a truncated standard normal random variable, truncated to the interval [2,500]. Is this possible?

Raymond

[Bruno Luong]

@Raymond, you still do not elaborate what is "truncated standard normal random variable". I can only guess:

x = TruncatedGaussian(-1,[2 500],[1 1e6]);

Beside that you need to read the help of TruncatedGaussian

[mortain]

Ciao Bruno,

thanks for the code you've written. I have a similar doubt as Ricardo. I'd assume the code would divide the domain of the "range" in a number of points equal to the number of points specified in the command "n", TruncatedGaussian(sigma, [range], [n]);

However, the values in X are defined randomly, hence the mean estimated will always be different, even though the range and sigma are the same
>> range=[-3 3];
>> sigma=1;
>> [X meaneff sigmaeff] = TruncatedGaussian(sigma, range, [1 1e3]);
>> fprintf('mean(X)=%g, estimated=%g\n',meaneff, mean(X))
mean(X)=0, estimated=0.0555218
>> fprintf('sigma=%g, effective=%g, estimated=%g\n', sigma, sigmaeff, stdX)
sigma=1, effective=1, estimated=0.992506
>> [X meaneff sigmaeff] = TruncatedGaussian(sigma, range, [1 1e3]);
>> fprintf('mean(X)=%g, estimated=%g\n',meaneff, mean(X))
mean(X)=0, estimated=0.00445842
>> fprintf('sigma=%g, effective=%g, estimated=%g\n', sigma, sigmaeff, stdX)
sigma=1, effective=1, estimated=0.992506

This can be seen as well from hist, using same number of divisions.
Van I ask you if I correctly got your coding, and how to obtain an uniform division of the domain.

Thanks