Comments (1)
Hello,
I found the solution thanks to a colleague: using the logdcf function.
d = InverseGaussian(1.65,590)
exp(logcdf(d, 2.))
gives the correct answer, that is to say 1 in this case
For most distributions, logcdf function simply takes the logarithm of cdf. But for some distributions that may have numerical problems for certain parameter values (such as the inverse Gaussian distribution), the logcdf function implements the calculation in a more subtle way and avoids numerical problems by scaling to log.
Best ragards
Laurent
from distributions.jl.
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