skellam.RdEstimates the two parameters of a Skellam distribution by maximum likelihood estimation.
skellam(lmu1 = "loglink", lmu2 = "loglink", imu1 = NULL, imu2 = NULL,
nsimEIM = 100, parallel = FALSE, zero = NULL)Link functions for the \(\mu_1\) and \(\mu_2\) parameters.
See Links for more choices and for general information.
Optional initial values for the parameters.
See CommonVGAMffArguments for more information.
If convergence failure occurs (this VGAM family function seems
to require good initial values) try using these arguments.
See CommonVGAMffArguments for information.
In particular, setting parallel=TRUE will constrain the
two means to be equal.
The Skellam distribution models the difference between two independent Poisson distributions (with means \(\mu_{j}\), say). It has density function $$f(y;\mu_1,\mu_2) = \left( \frac{ \mu_1 }{\mu_2} \right)^{y/2} \, \exp(-\mu_1-\mu_2 ) \, I_{|y|}( 2 \sqrt{ \mu_1 \mu_2}) $$ where \(y\) is an integer, \(\mu_1 > 0\), \(\mu_2 > 0\). Here, \(I_v\) is the modified Bessel function of the first kind with order \(v\).
The mean is \(\mu_1 - \mu_2\) (returned as the fitted values), and the variance is \(\mu_1 + \mu_2\). Simulated Fisher scoring is implemented.
An object of class "vglmff" (see vglmff-class).
The object is used by modelling functions such as vglm
and vgam.
This VGAM family function seems fragile and very sensitive to the initial values. Use very cautiously!!
Skellam, J. G. (1946). The frequency distribution of the difference between two Poisson variates belonging to different populations. Journal of the Royal Statistical Society, Series A, 109, 296.
Numerical problems may occur for data if \(\mu_1\) and/or \(\mu_2\) are large.
if (FALSE) { # \dontrun{
sdata <- data.frame(x2 = runif(nn <- 1000))
sdata <- transform(sdata, mu1 = exp(1 + x2), mu2 = exp(1 + x2))
sdata <- transform(sdata, y = rskellam(nn, mu1, mu2))
fit1 <- vglm(y ~ x2, skellam, data = sdata, trace = TRUE, crit = "coef")
fit2 <- vglm(y ~ x2, skellam(parallel = TRUE), data = sdata, trace = TRUE)
coef(fit1, matrix = TRUE)
coef(fit2, matrix = TRUE)
summary(fit1)
# Likelihood ratio test for equal means:
pchisq(2 * (logLik(fit1) - logLik(fit2)),
df = df.residual(fit2) - df.residual(fit1), lower.tail = FALSE)
lrtest(fit1, fit2) # Alternative
} # }