Proof that $E[X^2]$ = $\sum_{n=1}^\infty (2n-1) P(X\ge n)$

Question

X is a random variable with values from $\Bbb N\setminus{0}$

I am trying to show that $E[X^2]$ = $\sum_{n=1}^\infty (2n-1) P(X\ge n)$ iff $E[X^2]$ < $\infty$.

I rewrote $P(X \ge n)$:

$E[X^2]$ = $\sum_{n=1}^\infty (2n-1)\sum_{x=1}^\infty 1_{x \ge n}P(X=x)$

Now I tried to rearrange the sums:

$E[X^2]$ = $\sum_{x=1}^\infty \sum_{n=1}^x (2n-1)P(X=x)$

But I think that I made a mistake. Could you give me some hints?

Answer 1

Just as you did it: \begin{align} \sum^\infty_{n=1}(2n-1)P(X\geq n) &=\sum^\infty_{n=1}(2n-1)\Big(\sum^\infty_{j=n}P(X=j)\Big)\\ &=\sum^\infty_{j=1}\sum^j_{n=1}P(X=j)(2n-1)\\ &=\sum^\infty_{j=1}P(X=j)\Big(\sum^j_{n=1}(2n-1)\Big)\\ &=\sum^\infty_{j=1}P(X=j)j^2 \end{align}

The last line follows from $\sum^j_{n=1}(2n-1)=2\frac{j(j+1)}{2}-j$

Answer 2

Using a bit more machinery.

First note that $$ X^2=\sum_{n=1}^\infty(2n-1)I(X\geq n)\tag{0} $$ with probability one where $I$ is the indicator function. To see that this identity is true note that when $X^2=k^2$ (so $X=k$) for an integer $k\geq 1$ the RHS is $$ \sum_{n=1}^k(2n-1)=\sum_{n=1}^k [n^2-(n-1)^2]=k^2 $$ as desired.

Take expectations of $(0)$ (use the Monotone Convergence Theorem to justify the interchanging of sums and expectations) to yield that
$$ EX^2=\sum_{n=1}^\infty(2n-1)EI(X\geq n)=\sum_{n=1}^\infty (2n-1)P(X\geq n) $$ as desired.

A similar result can be realized from the identity $$ X=\sum_{n=1}^\infty I(X\geq n)\tag{1} $$ whence by the MCT $$ EX=\sum_{n=1}^\infty P(X\geq n) $$ In general for $m\geq 1$ an integer we have that $$ X^m=\sum_{n=1}^\infty [n^m-(n-1)^m]I(X\geq n)\tag{2} $$ whence $$ EX^m=\sum_{n=1}^\infty [n^m-(n-1)^m]P(X\geq n). $$ The identities (1) and (2) hold with probability one.