Suppose the probability mass function $\{p_n\}_{n \geq 0}$ takes the form \begin{equation}\label{eq:p*n_example} p_n = \left\{ \begin{array}{ll} n\,p_0\,r^n, & \quad \text{if}~ n \geq 1,\\ p_0, &\quad \text{if}~ n = 0 \end{array} \right. \end{equation} for some $0 < r < 1$. May I know is there any "standard name" for such (discrete) probability distribution?

  • $\begingroup$ Geometric distribution. You can write it as $p_n = r(1 - r)^n $, for $n = 0, 1, 2, ...$ $\endgroup$
    – rumathe
    Mar 18 at 17:47
  • 1
    $\begingroup$ Hello, you clearly forgot the factor $n$ in front of the $p_n\,r^n$ term $\endgroup$ Mar 18 at 17:50
  • $\begingroup$ Sorry, long day xD So it looks like kind of modified geometric distribution anyway. $\endgroup$
    – rumathe
    Mar 18 at 17:51
  • $\begingroup$ Of course I know this distribution is some sort of modification of the famous geometric distribution, that's why I am asking if there is any "standard name" to which I can refer this distribution... $\endgroup$ Mar 18 at 17:54
  • $\begingroup$ Shifted geometric distribution? $\endgroup$
    – rumathe
    Mar 18 at 17:57

1 Answer 1


This distribution does not have a special name but it can be related back to known distributions.

Let $Z\sim\operatorname{Bernoulli}(p_1)$ with $$ p_1=\frac{\frac{r}{(1-r)^2}}{1+\frac{r}{(1-r)^2}} $$ and $X\sim\operatorname{NegativeBinomial}(2,1-r)$ be independent Bernoulli and negative binomial random variables. Now define $$ W=Z(X+1). $$ Now notice that $$ \mathsf P(W=0)=\mathsf P(Z=0)=1-p_1=\frac{1}{1+\frac{r}{(1-r)^2}}. $$ Likewise, for $n=1,2,\dots$ $$ \begin{aligned} \mathsf P(W=n) &=\mathsf P(Z=1\cap X=n-1)\\ &=\mathsf P(Z=1)\mathsf P(X+1=n)\\ &=\frac{\frac{r}{(1-r)^2}}{1+\frac{r}{(1-r)^2}}nr^n. \end{aligned} $$ Thus $$ \mathsf P(W=n)= \begin{cases} 1-p_1, &n=0\\ p_1nr^n, &n\geq 1, \end{cases} $$ which is the distribution in question.

  • $\begingroup$ Ok, so this distribution car arise from a product random variable.... But still, maybe there is no "standard name" for such a distribution... Also I believe your computation is problematic. For instance, for $n = 4$ the probability $\mathbb{P}(W = 4)$ should contain the term $\mathbb{P}(Z =2)\cdot \mathbb{P}(X = 1)$ as well. $\endgroup$ Mar 18 at 19:02
  • $\begingroup$ Not quite. $P(W=4)=P(Z=1)P(X+1=4)$ since $Z$ can only take on values of zero and one. $\endgroup$ Mar 18 at 19:10
  • $\begingroup$ @MathandYuGiOhlover In this case, the distributiondoesn't have it's own name but it can be related back to distributions with known names. $\endgroup$ Mar 18 at 19:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .