Why is $1 = \sum_{k \geq 1} \mathbb{1}_{ \{N=k\} }$

Question

N is a random variable taking integer values, why is it true that $1 = \sum_{k \geq 1} \mathbb{1}_{ \{N=k\} }$ ?

$\mathbb{1}_{ \{N=k\} } = 1 , N=k $

$\mathbb{1}_{ \{N=k\} } = 0 , N \neq k $

taking the sum of indicators for $k \geq 1 $ gives :

$\sum_{k \geq 1} \mathbb{1}_{ \{N=k\} } = \mathbb{1}_{ \{N=1\} } + \mathbb{1}_{ \{N=2\} } + \dots \ $

I don't really get this, this is supposed to be a sum of functions where N is a random variable taking integer values, so for k from 1 to $\infty$, why is it equal to 1 ?

Answer 1

Because $$ \sum_{k\geq 1}1_{\{N=k\}}=1_{\bigcup_{k\geq 1} \{N=k\}}=1_\Omega = 1 $$ if we assume that $N$ can only take on values in $\{1,2,\ldots\}$.

Answer 2

$\sum_{k\geq 1}\mathbb 1_{N=k}=\mathbb 1_{{\mathbb N^*}}$ not $1$.

Answer 3

Consider a simpler case when $N$ takes values in $\{1,2\}$. Then $\{N=1\}\cup \{N=2\}=\Omega$, where $\Omega$ is the underlying sample space. In other words, for any $\omega\in\Omega$, $$ 1\{N(\omega)=1\}+1\{N(\omega)=2\}=1\{N(\omega)\in\Omega\}=1. $$ Similarly, when $N$ takes values in $\{1,2,\ldots,K\}$, $$ 1\{N(\omega)=1\}+1\{N(\omega)=2\}+\cdots+1\{N(\omega)=K\}=1. $$ Now generalize this to your case.