# Why are there different forms of the negative binomial distribution?

I have found the following two forms of the negative binomial distribution:

1) Let random variable $X$ be the number of failures before $r$ successes are obtained. Then the pmf of $X$ is given by

$$P(X = x) = {x+r-1\choose r-1}p^r(1-p)^x$$

2) Let random variable $X$ be the number of trials performed before r successes are obtained. Then the pmf of $X$ is given by

$$P(X = x) = {x-1\choose r-1}p^r(1-p)^{x-r}$$

In either case, the random variable $X$ is defined differently. Obviously, the expected value of $X$ in either case then, $\mu$ is different. In the first case, $$\mu = \frac{r(1-p)}{p}$$ and in the second case,$$\mu = \frac{r}{p}$$

Incidentally or otherwise, however, the variance $Var (X)$ is the same in either case $$Var(X) = \frac{r(1-p)}{p^2}$$

a) What is the motivation behind having alternative forms of the same probability distribution (or are they different) ?

b) Is there some intuition behind their having the same variance?

• The two random variables in case 1 and case 2 are related by a simple equation. Try to think of that. – StubbornAtom Feb 18 '18 at 10:00

Let $X$ be the number of failures before $r$ successes are obtained. Let $Y$ be the number of trials before $r$ successes are obtained. Suppose further that each trial is iid Bernoulli distributed.
We have that $Y=X+r$. Hence, $E(Y) = E(X)+r$, and $Var(X)=Var(Y)$.
• Shouldn't it be $Y = X + r$ ? – Eulerian Feb 18 '18 at 10:44
• @NihalJain youâ€™re right. For some reason I had in my mind that $r=1$. Thanks for pointing this out. – Theoretical Economist Feb 18 '18 at 11:21