Proving that if $X$ is a non-negative random variable, $E[X]=\sum_{j\ge 1}{P(X\ge j)}$. Prove that if $X$ is a non-negative random variable, $E[X]=\sum_{j\ge 1}{P(X\ge j)}$. An explanation I came by is: $\sum_{j\ge 1}{P(X\ge j)}=\sum_{i\ge 1}i{P(X=i)}=E[X]$ because "when we write $P(X ≥j) =\sum_{i≥j}P(X=i)$ we deduce that on the for any i = 1, 2, . . . the term P(X = i)
is counted precisely i times on the left hand side.". 
Firstly, I sense that there is a grammatical mishap here, and since I am not native, it is twice as hard for me to understand as any natural English speaker. Secondly, I still don't understand what way the aforementioned probability repetitions can be counted in. I would really appreciate any clarifying on this.  
 A: You want $X$ to be an integer-valued random variable in addition to being nonnegative. The result that you want to show is best explained by
writing out the right hand side as
\begin{aligned}
\sum_{j=1}^\infty P(X\geq j) &=\\
&\quad \begin{matrix}P(X=1) &+& P(X=2) &+& P(X=3) &+& P(X=4) &+ \cdots\\
& + & P(X=2) &+& P(X=3)&+& P(X=4) &+ \cdots\\
& &  &+& P(X=3)&+& P(X=4) &+ \cdots\\
& &  & & &+& P(X=4) &+ \cdots\\
& &  & & & &  &+ \ddots\\
\end{matrix}
\end{aligned}
where the sums of the rows are $P(X\geq 1), P(X\geq 2), P(X\geq 3), \ldots$ etc.
The sums of the columns are $P(X=1),~ 2P(X=2), 3P(X=3), \ldots$ etc.
I will leave the details for you to work out.
A: Note that
$$
P(x\ge k)=\sum_{j=k}^\infty P(x=j)\tag{1}
$$
Therefore,
$$
\begin{align}
\sum_{k=1}^\infty P(x\ge k)
&=\sum_{k=1}^\infty\sum_{j=k}^\infty P(x=j)\tag{2}\\
&=\sum_{j=1}^\infty\sum_{k=1}^j P(x=j)\tag{3}\\
&=\sum_{j=1}^\infty j\,P(x=j)\tag{4}\\
\end{align}
$$
Explanation:
$(2)$: apply $(1)$
$(3)$: change order of summation (summing over $1\le k\le j$)
$(4)$: sum in $k$
A: If $p_{i,j}\geq0$ for each $i,j\in\left\{ 1,2,\dots\right\} $ then:
$$\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}p_{i,j}=\sum_{j=1}^{\infty}\sum_{i=1}^{\infty}p_{i,j}$$
Let $X$ be a random variable taking values in $\mathbb Z$.
Filling in $p_{i,j}=P\left\{ X=i\right\} $ if $i\geq j$ and $p_{i,j}=0$
otherwise this results in:$$\sum_{i=1}^{\infty}iP\left\{ X=i\right\} =\sum_{j=1}^{\infty}P\left\{ X\geq j\right\}$$
If $X$ is nonnegative then the LHS of this equation equals $\mathbb EX$.
