skellam: Skellam Distribution Family Function

Description

Estimates the two parameters of a Skellam distribution by maximum likelihood estimation.

Usage

skellam(lmu1 = "loge", lmu2 = "loge", imu1 = NULL, imu2 = NULL,
        nsimEIM = 100, parallel = FALSE, zero = NULL)

Arguments

lmu1, lmu2

Link functions for the $\mu_1$ and $\mu_2$ parameters. See Links for more choices and for general information.

imu1, imu2

Optional initial values for the parameters. See CommonVGAMffArguments for more information. If convergence failure occurs (this VGAM family function seems to require good init

nsimEIM, parallel, zero

See CommonVGAMffArguments for more information. In particular, setting parallel=TRUE will constrain the two means to be equal.

Value

An object of class "vglmff" (see vglmff-class). The object is used by modelling functions such as vglm and vgam.

Warning

This VGAM family function seems fragile and very sensitive to the initial values. Use very cautiously!!

Details

The Skellam distribution models the difference between two independent Poisson distributions. 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.

References

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.

Examples

Run this code

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, sdata, trace = TRUE, crit = "c")
fit2 <- vglm(y ~ x2, skellam(parallel = TRUE), 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 = fit2@df.residual - fit1@df.residual, lower.tail = FALSE)

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