How to get the distribution is explained in http://www.dartmouth.edu/~chance/tea...k/Chapter7.pdf
Are random variables for a Weibull distribution additive?
Let's say there are random variables and with different Weibull distributions. The Weibulls are very similar, but not one and the same as both parameters vary slightly.
How would the distribution of look like? Would it be Weibull, and if so, what are the parameters?
How to get the distribution is explained in http://www.dartmouth.edu/~chance/tea...k/Chapter7.pdf
That is more for discrete distributions. How about I lay down my problem here for everyone.
I actually have four random variables based on slightly different three-parameter (beta shift) Weibull distributions.
Notice that is the same for all PDFs. I want to find the PDF, and more importantly, the CDF of the random variable .
It seems complicated, but treating z with the same parameters as one of the x's would lose a lot of possible outcomes. In my calculations, forcing z to follow an x would lose 44% of possible outcomes.
Therefore, I either need a whole new distribution that is dependent upon the four Weibulls OR I need to form a new Weibull with new parameters dependent upon the parameters from each of the four Weibulls.
Thanks for anyone willing to look into this!