Showing that ${\rm E}[X]=\sum_{k=0}^\infty P(X>k)$ for a discrete random variable Let $X$ be a discrete random variable whose range is $0,1,2,3,\ldots$. Prove that
$$
{\rm E}[X]=\sum_{k=0}^\infty P(X>k).
$$
How to prove this? I tried a bit but unable to post due to formatting issue. Can anyone help?
 A: Hint: Write
$$\begin{matrix}
P(X>0) &= &P(X=1) &+ &P(X=2) &+ &P(X=3) &+ &P(X=4) &+ &\cdots\\
P(X>1) &= & & & P(X=2) &+& P(X=3) &+  &P(X=4) &+ &\cdots\\
P(X>2) &= & & &  && P(X=3) &+  &P(X=4) &+ &\cdots\\
P(X>3) &= & & &  & &  &  &P(X=4) &+ &\cdots\\
\vdots &\vdots
\end{matrix}$$
Then, sum each column to get 
$$\displaystyle \sum_{k=0}^\infty P(X>k) = P(X=1)+2P(X=2)+3P(X=3) + 4P(X=4)+\cdots$$
A: The following is probably formal enough for your purposes. For brevity let $p_i=\Pr(X=i)$. Then
$$E(X)=p_1+2p_2+3p_3+4p_4+5p_5+6p_6+\cdots.\tag{1}$$
Let is subtract $p_1+p_2+p_3+p_4+\cdots$ from the right side of  (1). So we have subtracted $\Pr(X\gt 0)$. We conclude that
$$E(X)=\Pr(X\gt 0)+p_2+2p_3+3p_4+4p_5+\cdots.\tag{2}$$
Subtract $p_2+p_3+p_4+p_5+\cdots$ from the right side of (2). So we are subtracting $\Pr(X\gt 1)$, and we conclude that 
$$E(X)=\Pr(X\gt 0)+\Pr(X\gt 1)+p_3+2p_4+3p_5 +\cdots. \tag{3}$$
Subtract $p_3+p_4+p_5+\cdots$ from the right side of (3). So we are subtracting 
$\Pr(X\gt 2)$ and conclude that 
$$E(X)=\Pr(X\gt 0)+\Pr(X=1)+\Pr(X=2)+p_4+2p_5+\cdots. \tag{4}$$
Continue forever.
There is a more persuasive argument, that involves addition rather than subtraction.
On a line write $p_1+p_2+p_3+p_4+p_5+\cdots$.
On the line below, write down $p_2+p_3 +p_4+p_5+\cdots$, aligning the $p_2$ with the $p_2$ in the line above, the $p_3$ with the $p_3$, and so on.
Now on the next line write $p_3+p_4+p_5+\cdots$, aligning the $p_3$ with the ones above, the $p_4$ with the $p_4$'s, and so on.
And so on.
The column sums give $E(X)$.
The row sums give $\Pr(X\gt 0)$, $\Pr(X\gt 1)$, $\Pr(X\gt 2)$, and so on. 
A: \begin{align*}\mathbb E[X]=&\,\sum_{k=0}^{\infty}k\cdot\mathbb P(X=k)\\
=&\,\mathbb P(X=1)\\
+&\,\mathbb P(X=2)+\mathbb P(X=2)\\
+&\,\mathbb P(X=3)+\mathbb P(X=3)+\mathbb P(X=3)\\
+&\,\mathbb P(X=4)+\mathbb P(X=4)+\mathbb P(X=4)+\mathbb P(X=4)\\
+&\,\cdots
\end{align*}
Now, try summing vertically instead of horizontally. The sum of the first column will be $$\sum_{k=1}^{\infty}\mathbb P(X=k)=\mathbb P(X>0).$$ The sum of the second column will be $$\sum_{k=2}^{\infty}\mathbb P(X=k)=\mathbb P(X>1).$$ The sum of the third column will be $$\sum_{k=3}^{\infty}\mathbb P(X=k)=\mathbb P(X>2)$$ and so forth. Hence, $$\mathbb E[X]=\sum_{k=0}^{\infty}\mathbb P(X>k)$$ as claimed.
A: Under suitable conditions to exchange two series, we have
\begin{align}
\sum_{k=0}^\infty\mathbb{P}(X>k)
=
\sum_{k=0}^\infty\sum_{x=0}^\infty\mathbb{1}_{\{x>k\}}\mathbb{P}(X=x)
=
\sum_{x=0}^\infty\sum_{k=0}^\infty\mathbb{1}_{\{x>k\}}\mathbb{P}(X=x)
=
\sum_{x=0}^\infty\sum_{k=0}^x\mathbb{P}(X=x)
=
\mathbb{E}(X).
\end{align}
The proof is complete.
