3
$\begingroup$

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)

$\endgroup$
5
  • 3
    $\begingroup$ 1. Are you required to use method-of-moments estimators or are you looking for any good estimator? 2. It seems to me that $\bar x^2 / s^2 =\alpha$ would be a good start at solving these equations (assuming they are correct, which I have not checked). BTW, what is "$m$"? $\endgroup$ – whuber Feb 3 '15 at 21:27
  • $\begingroup$ yes i need to use method-of-moments @whuber $\endgroup$ – Danny Feb 3 '15 at 21:29
  • $\begingroup$ ok @whuber but that is $\frac{\alpha}{\beta + 1} = \bar{x}^2/s^2$ anyhow it will help me enough to solve this..thx $\endgroup$ – Danny Feb 3 '15 at 21:56
  • $\begingroup$ Right--I overlooked a power of $\beta+1$. That's going to be messier, but at least it only leads to quadratic equations. $\endgroup$ – whuber Feb 3 '15 at 22:08
  • $\begingroup$ are you required to solve this analytically or estimate it? You can get arbitrarily close to the true solution by any root finding algorithm, that's what most GMM estimation looks like in practice. $\endgroup$ – jayk Feb 5 '15 at 1:26
-1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.