# Issue with Probability Integral Transform

I currently have a survivability function of the form: $S = A*e^{-(t/\theta)^\beta}$ and I would like to sample its PDF using PIT. Due to the definition of the survival function, I know that that the CDF of the PDF is: $F = 1-A*e^{-(t/\theta)^\beta}$. If I then invert the CDF I get that $t = \theta*[-ln((1-F)/A)]^{1/\beta}$. With a uniform distribution for $1-F$, I can plug that into my inverted CDF to get a sampling of my original PDF.

However, for the given survival function, $A=0.8$ and $\beta=0.5$, so for values of the uniform distribution that are larger than $A$, the $ln$ will be positive which ultimately leads to the inverted CDF producing an imaginary number. How am I supposed to sample the PDF if some of my outputs are imaginary?

For $A=0.8$ the distribution function looks something like this: You are complaining that for a random number less than $0.2$ the inverse spits out a complex number. You, unfortunately, handle the method mechanically. If your random number is between $0$ and $0.2$ then you have to ignore the inverse of the cdf. The PIT spits out a zero. The cdf above tells you that the probability that the result is zero is $0.2$
This cdf tells that $20\%$ of the life times is $0$??? Yes, $20\%$ of the newborn ones are stillborn ones.