If $E(X^2)=3, E(X)=1$, find $\sum\limits_{i=1}^{\infty} i P(X \geq i)$
Hint given:
$\sum\limits_{i=1}^{\infty} i P(X \geq i)=\frac12\left[E(X^2)-E(X)\right]$
My attempt:
From Formula similar to $EX=\sum\limits_{i=1}^{\infty}P\left(X\geq i\right)$ to compute $E(X^n)$?, I know that $E(X^n)=\sum\limits_{i=1}^{\infty} \{i^n-(i-1)^n\} P(X \geq i)$
Now, $E(X^2)=\sum\limits_{i=1}^{\infty} \{i^2-(i-1)^2\} P(X \geq i)$
$=\sum\limits_{i=1}^{\infty} (2i-1) P(X \geq i)$
Similarly, $E(X)=\sum\limits_{i=1}^{\infty} \{i-(i-1)\} P(X \geq i) =\sum\limits_{i=1}^{\infty} P(X \geq i)$
This gives me $\frac12[E(X^2)+E(X)]=\sum\limits_{i=1}^{\infty} i P(X \geq i)$
which would give my result as $\frac{3+1}{2}=2$, instead of the given answer $1$.
Am I wrong?
Let $P(X=0)=P(X=1)=1/2$. Then $E[X^2]-E[X]=0$, but $1\cdot P(X\geq 1)=1/2$, so the hint is wrong, and you're right.