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))