| ZTNegativeBinomial {distributions3} | R Documentation |
Create a zero-truncated negative binomial distribution
Description
Zero-truncated negative binomial distributions are frequently used to model counts where zero observations cannot occur or have been excluded.
Usage
ZTNegativeBinomial(mu, theta)
Arguments
mu |
Location parameter of the negative binomial component of the distribution. Can be any positive number. |
theta |
Overdispersion parameter of the negative binomial component of the distribution. Can be any positive number. |
Details
We recommend reading this documentation on https://alexpghayes.github.io/distributions3/, where the math will render with additional detail.
In the following, let X be a zero-truncated negative binomial random variable with parameter
mu = \mu.
Support: \{1, 2, 3, ...\}
Mean:
\mu \cdot \frac{1}{1 - F(0; \mu, \theta)}
where F(k; \mu, \theta) is the c.d.f. of the NegativeBinomial distribution.
Variance: m \cdot (\mu + 1 - m), where m is the mean above.
Probability mass function (p.m.f.):
P(X = k) = \frac{f(k; \mu, \theta)}{1 - F(0; \mu, \theta)}
where f(k; \mu, \theta) is the p.m.f. of the NegativeBinomial
distribution.
Cumulative distribution function (c.d.f.):
P(X = k) = \frac{F(k; \mu, \theta)}{1 - F(0; \mu, \theta)}
Moment generating function (m.g.f.):
Omitted for now.
Value
A ZTNegativeBinomial object.
See Also
Other discrete distributions:
Bernoulli(),
Binomial(),
Categorical(),
Geometric(),
HurdleNegativeBinomial(),
HurdlePoisson(),
HyperGeometric(),
Multinomial(),
NegativeBinomial(),
Poisson(),
ZINegativeBinomial(),
ZIPoisson(),
ZTPoisson()
Examples
## set up a zero-truncated negative binomial distribution
X <- ZTNegativeBinomial(mu = 2.5, theta = 1)
X
## standard functions
pdf(X, 0:8)
cdf(X, 0:8)
quantile(X, seq(0, 1, by = 0.25))
## cdf() and quantile() are inverses for each other
quantile(X, cdf(X, 3))
## density visualization
plot(0:8, pdf(X, 0:8), type = "h", lwd = 2)
## corresponding sample with histogram of empirical frequencies
set.seed(0)
x <- random(X, 500)
hist(x, breaks = -1:max(x) + 0.5)