# How do I determine the appropriate likelihood function for a custom CDF

I'm doing some work describing how things dissolve in solution, and I've determined that a particular parameterization of the 3P Weibull provides a good fit. Now what I want to do is specify target amounts dissolved at a particular time and use MLE to find the values for the parameters. Since there isn't just one 3-parameter Weibull distribution, I can't just turn to R or Python to get their pre-defined distributions. Instead, I need to figure out how to determine the likelihood function (negative log likelihood?) that I should pass into a minimizing function.

The CDF I'm using is $$a * (1-e^{-(x/b) ^ c})$$ where a, b, and c are the parameters and x represents an amount of elapsed time.

Is there any direction on how I go from this CDF (I don't know the PDF) to a likelihood function? Do I even need to assume this is a distributional problem? Is there a better way to do what I'm trying to do?

Before you get to the likelihood function, the first issue here is to confirm that this is a valid cumulative distribution function. If you examine the requirements for this, you will see that you must have $$a=1$$ for validity of the form, so your distribution will reduce to the two-parameter Weibull distribution.

Once you have narrowed down the CDF, you then obtain the PDF using differentiation and use the latter to form the likelihood function for an arbitrary sample.

• So really I just have a non-linear function that to my untrained eye resembled a CDF. Thank you.
– Mark
Commented Sep 10, 2021 at 12:04

You can get easily get the PDF.

1. If your values are in the range $$0$$ to $$+\infty$$ then you should have $$a=1$$; this because CDF goes to $$1$$ as $$x$$ goes to infinity.

2. The PDF is simply the derivative of the CDF, it is $$f(x)=\frac{a*c}b*\left({\frac{x}b}\right)^\left(c-1\right)*e^{-\left(\frac{x}b\right)^c}$$