An additive property of weak records from geometric distributions

Abstract

Let $\{W_m\}{_{m\ge 1}}$ be the sequence of weak records from a discrete parent random variable, $X$, supported on the non-negative integers. We obtain a new characterization of geometric distributions based on an additive property of weak records: $X$ follows a geometric distribution if and only if for certain integers, $n,\, s\ge 1, W_{n+s}\stackrel{d}{=}W_n+W^{\prime }_s$, with $W^{\prime }_s$ independent of $W_n$ and $W^{\prime }_s\stackrel{d}{=} W_s$.

Appendix: Some results about combinatorial numbers

In the proof of our main result we used some properties of the combinatorial numbers. We collect these properties in this section.

Let $a$ and $b$ be non-negative integers. If $a\ge b, \left(\begin{array}{l} a\\ b \end{array}\right)=a!/(b!(a-b)!)$. If $a<b, \left(\begin{array}{l} a\\ b \end{array}\right)=0$.

Lemma 1

Let $a, b, c$ and $k$ be non-negative integers. Then

• (i) If $b\ge c$,

  $$\begin{aligned} \sum _{j=0}^c \left(\begin{array}{c} a+j\\ j \end{array}\right) \left(\begin{array}{l} b-j\\ c-j \end{array}\right) = \left(\begin{array}{c} a+b+1\\ c \end{array}\right). \end{aligned}$$

  (36)

• (ii) If $a\ge 1$

  $$\begin{aligned} \left(\begin{array}{l} k+a\\ k+1 \end{array}\right) =\sum _{j=0}^{a-1} \left(\begin{array}{c} k+a-j-1\\ k \end{array}\right). \end{aligned}$$

  (37)

• (iii) For all $k\ge 0$,

  $$\begin{aligned} \left(\begin{array}{c} k+a+b-j\\ k+1 \end{array}\right) -\left(\begin{array}{c} k+a-j\\ k+1 \end{array}\right) -\left(\begin{array}{c} k+b-j\\ k+1 \end{array}\right) \ge 0,\quad j\ge 0. \end{aligned}$$

  (38)

 

Proof

 

• (i) For any $\alpha \ge 0, (1-z)^{-(\alpha +1)}=\sum _{j=0}^\infty \left(\begin{array}{c} \alpha +j\\ j \end{array}\right) z^j, |z|<1$. Consider the identity

  $$\begin{aligned} (1-z)^{-(a+1)}(1-z)^{-(b-c+1)}=(1-z)^{-(a+b-c+2)}, \quad |z|<1, \end{aligned}$$

  (39)

  expanding both sides of (39) and equating the coefficients of powers of degree $k\ge 0$ in $z,$ we obtain

  $$\begin{aligned} \sum _{j=0}^k \left(\begin{array}{c} a+j\\ j \end{array}\right) \left(\begin{array}{c} b-c+k-j\\ k-j \end{array}\right) = \left(\begin{array}{c} a+b-c+k+1\\ k \end{array}\right), \end{aligned}$$

  (40)

  and (36) follows from (40) with $k=c$.

• (ii) It is not difficult to check by induction on $a$ that

  $$\begin{aligned} \left(\begin{array}{l} k+a\\ k+1 \end{array}\right) =\sum _{i=0}^{a-1} \left(\begin{array}{c} k+i\\ k \end{array}\right) ,\quad k\ge 0, \end{aligned}$$

  (41)

  and (37) follows from (41) with the change of indexes $i=a-1-j$.

• (iii) By induction on $k$. If $k=0$ the result is straightforward. Suppose that the inequality is valid for certain $k\ge 0$ and note that for any non-negative integers, $\alpha , \beta ,$

  $$\begin{aligned} \left(\begin{array}{l} \alpha +1\\ \beta +1 \end{array}\right) =\frac{\alpha +1}{\beta +1} \left(\begin{array}{l} \alpha \\ \beta \end{array}\right) \end{aligned}$$

  (42)

  (this formula is trivially true if $\alpha <\beta$). Then, for any $j\ge 0,$ using (42) and the hypothesis of induction,

  $$\begin{aligned} \left(\begin{array}{c} k+1+a+b-j\\ k+2 \end{array}\right)&= \frac{k+1+a+b-j}{k+2} \left(\begin{array}{c} k+a+b-j\\ k+1 \end{array}\right)\\&\ge \frac{k+1+a+b-j}{k+2}\left\{ \left(\begin{array}{c} k+a-j\\ k+1 \end{array}\right) + \left(\begin{array}{c} k+b-j\\ k+1 \end{array}\right) \right\} \\&\ge \frac{k+1+a-j}{k+2} \left(\begin{array}{c} k+a-j\\ k+1 \end{array}\right)\\&+\frac{k+1+b-j}{k+2} \left(\begin{array}{c} k+b-j\\ k+1 \end{array}\right)\\&\ge \left(\begin{array}{c} k+1+a-j\\ k+2 \end{array}\right) + \left(\begin{array}{c} k+1+b-j\\ k+2 \end{array}\right). \end{aligned}$$

  $\square$