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A Skellam Distribution describes the difference between two variables that have Poisson distributions. Is there a similar distribution that describes the difference between variables that follow negative binomial distributions?

My data is produced by a Poisson process, but includes a fair amount of noise, leading to overdispersion in the distribution. Thus, modeling the data with a negative binomial (NB) distribution works well. If I want to model the difference between two of these NB data sets, what are my options? If it helps, assume similar means and variance for the two sets.

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  • $\begingroup$ There are many distributions that are easy to describe which do not have standard names. $\endgroup$
    – Glen_b
    Jan 10, 2014 at 0:54

2 Answers 2

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I don't know the name of this distribution but you can just derive it from the law of total probability. Suppose $X, Y$ each have negative binomial distributions with parameters $(r_{1}, p_{1})$ and $(r_{2}, p_{2})$, respectively. I'm using the parameterization where $X,Y$ represent the number of successes before the $r_{1}$'th, and $r_{2}$'th failures, respectively. Then,

$$ P(X - Y = k) = E_{Y} \Big( P(X-Y = k) \Big) = E_{Y} \Big( P(X = k+Y) \Big) = \sum_{y=0}^{\infty} P(Y=y)P(X = k+y) $$

We know

$$ P(X = k + y) = {k+y+r_{1}-1 \choose k+y} (1-p_{1})^{r_{1}} p_{1}^{k+y} $$

and

$$ P(Y = y) = {y+r_{2}-1 \choose y} (1-p_{2})^{r_{2}} p_{2}^{y} $$

so

$$ P(X-Y=k) = \sum_{y=0}^{\infty} {y+r_{2}-1 \choose y} (1-p_{2})^{r_{2}} p_{2}^{y} \cdot {k+y+r_{1}-1 \choose k+y} (1-p_{1})^{r_{1}} p_{1}^{k+y} $$

That's not pretty (yikes!). The only simplification I see right off is

$$ p_{1}^{k} (1-p_{1})^{r_{1}} (1-p_{2})^{r_{2}} \sum_{y=0}^{\infty} (p_{1}p_{2})^{y} {y+r_{2}-1 \choose y} {k+y+r_{1}-1 \choose k+y} $$

which is still pretty ugly. I'm not sure if this is helpful but this can also be re-written as

$$ \frac{ p_{1}^{k} (1-p_{1})^{r_{1}} (1-p_{2})^{r_{2}} }{ (r_{1}-1)! (r_{2}-1)! } \sum_{y=0}^{\infty} (p_{1}p_{2})^{y} \frac{ (y+r_{2}-1)! (k+y+r_{1}-1)! }{y! (k+y)! } $$

I'm not sure if there is a simplified expression for this sum but it could be approximated numerically if you only need it to calculate $p$-values

I verified with simulation that the above calculation is correct. Here is a crude R function to calculate this mass function and carry out a few simulations

  f = function(k,r1,r2,p1,p2,UB)  
  {

  S=0
  const = (p1^k) * ((1-p1)^r1) * ((1-p2)^r2)
  const = const/( factorial(r1-1) * factorial(r2-1) ) 

  for(y in 0:UB)
  {
     iy = ((p1*p2)^y) * factorial(y+r2-1)*factorial(k+y+r1-1)
     iy = iy/( factorial(y)*factorial(y+k) )
     S = S + iy
  }

  return(S*const)
  }

 ### Sims
 r1 = 6; r2 = 4; 
 p1 = .7; p2 = .53; 
 X = rnbinom(1e5,r1,p1)
 Y = rnbinom(1e5,r2,p2)
 mean( (X-Y) == 2 ) 
 [1] 0.08508
 f(2,r1,r2,1-p1,1-p2,20)
 [1] 0.08509068
 mean( (X-Y) == 1 ) 
 [1] 0.11581
 f(1,r1,r2,1-p1,1-p2,20)
 [1] 0.1162279
 mean( (X-Y) == 0 ) 
 [1] 0.13888
 f(0,r1,r2,1-p1,1-p2,20)
 [1] 0.1363209

I've found the sum converges very quickly for all of the values I tried, so setting UB higher than 10 or so is not necessary. Note that R's built in rnbinom function parameterizes the negative binomial in terms of the number of failures before the $r$'th success, in which case you'd need to replace all of the $p_{1}, p_{2}$'s in the above formulas with $1-p_{1}, 1-p_{2}$ for compatibility.

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  • $\begingroup$ Thanks. I'll need some time to digest this, but your help is much appreciated. $\endgroup$ Jul 25, 2011 at 20:33
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Yes. skewed generalized discrete Laplace distribution is the difference of two negative binomial distributed random variables. For more clarifications refer the online available article "skewed generalized discrete Laplace distribution" by seetha Lekshmi.V. and simi sebastian

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    $\begingroup$ Can you provide a complete citation & a summary of the information in the paper so future readers can decide if it's something they want to pursue? $\endgroup$ Nov 16, 2016 at 2:41
  • $\begingroup$ The article mentioned by @simi-sebastian (the author?) is ijmsi.org/Papers/Volume.2.Issue.3/K0230950102.pdf. However, unless I'm mistaken, it only addresses the case of the Negative Binomial variables $X$ and $Y$ both having the same dispersion parameter, rather than the more general case described by the original poster. $\endgroup$ Jun 7, 2017 at 18:14

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