"Ka Lee" <kaloklee@gmail.com> wrote in message <glgbf1$lim$1@fred.mathworks.com>...
> Hi, do you mean this is all I need to do?
>
> x = exprnd(lambda);
> y = x+tao;
>
> Is y the desired lefttruncated exponential distribution? Thanks!
Yes. If you have doubts on that score, Ka Lee, consider this. An exponential distribution that stops on the left at x = tao must have conditional density of the form
f(x) = exp(lambda*x)/K
with K the probability that x >= tao:
K = int(exp(lambda*x),tao,inf) = 1/lambda*exp(lambda*tao).
Hence
f(x) = exp(lambda*x)*lambda/exp(lambda*tao)
= lambda*exp(lambda*(xtao))
This means that your distribution has the same density as the usual one for an x which is less by tao  that is, adding tao to the x of the usual distribution will give you the density of your distribution. Therefore
tao+exprnd(lambda)
should give you the distribution you want.
By the way, when you write 'tao' do you mean the Greek letter 'tau'?
Roger Stafford
