I would like to find the estimates of the parameters in a truncated (at zero) negative binomial distribution. Suppose $Z$ has this distribution with parameters ($\alpha,\beta$). (The parametrization used is $p = \dfrac{\beta}{\beta +1}$).The probability distr.
$\Big(P(Z =z) = \frac{\Gamma(\alpha+z)}{\Gamma(\alpha)z!} \big( \frac{\beta}{\beta+1} \big)^{\alpha}\big( \frac{1}{\beta+1} \big)^{z}\Big)$ .
$P(Z = z \mid Z \geq 1) = \dfrac{P(Z =z, Z \geq 1)}{P(Z \geq1)} = \dfrac{ \frac{\Gamma(\alpha+z)}{\Gamma(\alpha)z! } \big( \frac{\beta}{\beta+1} \big)^{\alpha}\big( \frac{1}{\beta+1} \big)^{z} } {1-(\frac{\beta}{\beta+1})^{\alpha} } $
Finding the expectation and variance gives me two equations to estimate the parameters
$$E[Z \mid Z\geq 1] = \dfrac{\alpha}{\beta\big(1-(\frac{\beta}{\beta+1})^{\alpha}\big)} = \bar{x}$$
$Var[Z \mid Z\geq1 ] = \dfrac{\alpha(\beta + 1)}{\beta^2\big(1-(\frac{\beta}{\beta+1})^{\alpha}\big)^2} = s^2$ , ($s^2$ sample variance)
However i can't even find an expression for "atleast" one of the variables in terms of the other, hence then I could solve one equation numerically. (But if I there is a way to solve for both thats just great)
Best Answer
Are you thinking that somehow the inequality (≥) factors in to the calculation? It does not. That inequality is just telling you you are talking about a truncated distribution.
In each case, mean and variance, you have three items:
X=Y=Z
Lop off the "X=", and use the remaining items. You'll need to solve for the two unknowns, alpha and beta, which you do using Newton's method.
I actually need to do that very thing, so I'll let you know how it goes.