2
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

For $A=0.8$ the distribution function looks something like this:

enter image description here

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.

$\endgroup$
  • $\begingroup$ Yeah, thank you, what you are saying makes sense. Setting the imaginary terms to zero means 20% of the lifetimes is 0. The 20% indicates the percentage of university CubeSats which don't function at deployment (aka, dead on arrival) $\endgroup$ – Rob Mar 24 '17 at 18:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.