2
$\begingroup$

Let's suppose we know that if $X \ge 0$ and X is integer-valued, then: $E(X) = \sum_{n \ge 1} Pr(X \ge n)$ For more info please visit: Proof of $\sum_{k=0}^n k \text{Pr}(X=k) = \sum^{n-1}_{k=0} \text{Pr}(X>k) -n \text{Pr}(X>n)$

My question is how to get $E(X^2)$.

Thanks,

  • 0
    Thanks @Jean-Sébastien. Do you have any hint on how to get E(X^2). I'll edit my question and I'll ask E(X^2) part only.2012-11-07

1 Answers 1

2

$X^2=\sum_{n\geqslant1}(2n-1)\cdot\mathbf 1_{X\geqslant n}\implies\mathbb E(X^2)=\sum_{n\geqslant1}(2n-1)\cdot\mathbb P(X\geqslant n)$

  • 0
    Nice answer...thanks.2012-11-07