pgenf and genfHazard are computed numerically, and may not be precise!dgenf(x, kappa = 5, eta = 1.5, gamma = .8, lambda = 1, forceExpectation = F)
pgenf(q, kappa = 5, eta = 1.5, gamma = .8, lambda = 1, forceExpectation = F)
genfHazard(x, kappa = 5, eta = 1.5, gamma = .8, lambda = 1, forceExpectation = F)TRUE, the expectation of the distribution is forced to be 1 by letting theta be a function of the other parameters.