# Non integer successes in negative binomial distribution.

How do we calculate the probability for negative binomial distribution when the number of successes are non-integer? It's easy to calculate when the failures are non-integer by using gamma relation, but I wanted to know the method to determine the probability if both successes and failures are non-integer.

For e.g., probability that there were 2.3 successes when 5.8 failures took place with probability of success being p. Also, the probability that there were 0.3 successes when 0.8 failures took place with probability of success being p.

• There can't be a noninteger number of successes, or of failures. Maybe one could ask e.g. probability of at least 2.3 successes, but that's the same as probability of 3 or more. – coffeemath Jan 12 '18 at 1:58
• Yeah, at least 2.3 successes. The latter part of your response solves my question. Thanks a lot. – RSA Jan 12 '18 at 2:01

https://stats.stackexchange.com/questions/310676/continuous-generalization-of-the-negative-binomial-distribution If I understand you correctly, you are just looking to generalize the discrete negative binomial distribution to a continuous domain. If not, please correct me.

Arguably one could compute a value corresponding to a non-integer number of successes by substituting $$\Gamma(x+1)$$ for $$x!$$ in the standard formula for the probability, i.e. instead of

$$\frac{\Gamma(x+n)}{\Gamma(n) x!} p^n (1-p)^x$$

use

$$\frac{\Gamma(x+n)}{\Gamma(n) \Gamma(x+1)} p^n (1-p)^x$$

You'd have to decide for yourself whether the numbers computed this way actually correspond to the outcome of some sensible stochastic process in the real world, but the formula does interpolate sensibly between the values of the "real" negative binomial probabilities. R code:

dnb2 <- function(x,prob,size,log=FALSE) {
lg <- lgamma(x+size)-lgamma(size)-lgamma(x+1)+size*log(prob)+x*log(1-prob)
if (log) lg else exp(lg)
}

png("cont_NB.png")
x <- 0:30
par(las=1,bty="l")
plot(x,dnbinom(x,prob=0.2,size=3),type="p",
ylim=c(0,0.065),ylab="probability")