Let there be a distribution $F_{X}(x)=1-x^{-\alpha}, x \geq 1$, find $\operatorname{E}(X) \;\text{and} \operatorname{Var}(X)$ and for what values of $\alpha$ they are defined.
My solution is:
\begin{align} \operatorname{E}(X) &= \int^{\infty}_{1}(1-F_{X}(x))dx=\int^{\infty}_{1}x^{-\alpha}dx \\ &= \lim_{x\to \infty} \frac{x^{-\alpha+1}}{-\alpha+1} - \frac{1}{-\alpha+1}\\&= \frac{1}{\alpha-1}\\ \\\operatorname{E}(X^{2})&= \int^{\infty}_{1}x^{2}\alpha x^{-\alpha-1}dx =\lim_{x\to \infty}\frac{ \alpha x^{2-\alpha}}{2-\alpha} - \left. \frac{ \alpha x^{2-\alpha}}{2-\alpha}\right\rvert_{x=1}= -\frac{\alpha}{2-\alpha} \\\operatorname{Var}(x)&= -\frac{\alpha}{2-\alpha} - \left( \frac{1}{\alpha-1} \right)^{2} \end{align} As long as $\alpha>2$ we have both moments defined and Var(X) > 0.
I'd like to know if this is correct and I'd like an explanation about the following step when computing the 2nd moment through the survival function:
\begin{align} E(X^2) &= \int_0^{\infty} x^2 f(x) \, dx \\ &= \int_0^{\infty} \int_0^x 2t f(x) \, dt \, dx \\ &= \int_0^{\infty} \int_t^{\infty} 2t f(x) \, dx \, dt \\ &= \int_0^{\infty} 2t G(t) \, dt, \end{align} G(t) being the survival function. I understand the change in the upper bound, but why the lower bound turns into t?
Note that if $a > 2$, your calculation of $\operatorname{E}[X] = 1/(a-1)$ would suggest that $\operatorname{E}[X] < 1$; yet we know the support of $X$ is on $[1, \infty)$. Thus your expectation cannot be correct.
To understand why this happens, you should compute the expectation the usual way, via $$\operatorname{E}[X] = \int_{x = 1}^\infty x f_X(x) \, dx,$$ where $$f_X(x) = F'_X(x) = \frac{d}{dx}\left[1 - x^{-a}\right] = ax^{-a-1}, \quad x \ge 1, \quad a > 0.$$ Consequently $$\operatorname{E}[X] = \int_{x=1}^\infty ax^{-a} \, dx = \left[\frac{a}{1-a} x^{1-a}\right]_{x=1}^\infty = \frac{a}{1-a} \left(\lim_{x \to \infty} x^{1-a} - 1 \right) = \frac{a}{a-1}, \quad a > 1. $$ So, why did the "shortcut" $$\operatorname{E}[X] = \int_{x = 0}^\infty S_X(x) \, dx$$ not work? The reason has to do with the lower limit of integration. If $F_X(x) = 0$ for $x \le c$, for some constant $c$ that represents where the lower endpoint of the support of $X$, then when you integrate $S_X(x)$ from $0$ to $c$, this part contributes a nonzero amount to the value of the integral because $F_X = 0$ implies $S_X = 1$. You cannot simply ignore it just because the density is zero. So to use your shortcut, you must write instead $$\operatorname{E}[X] = \int_{x=0}^\infty S_X(x) \, dx = \int_{x=0}^1 S_X(x) \, dx + \int_{x=1}^\infty S_X(x) \, dx = 1 + \frac{1}{a-1} = \frac{a}{a-1}.$$ To give you a more geometric understanding of this phenomenon, suppose we have a continuous uniform distribution on $[0,1]$. Its expectation is clearly $1/2$. The survival function is obviously $S_X(x) = 1-x$ on this interval. If we perform a location transformation by shifting it over $1$ unit, its expectation would also shift by this amount, giving $3/2$. But if you were to just integrate $2-x$ over $[1,2]$ you would get the wrong answer because the survival is now $1$ for $[0,1]$; specifically, $$S_X(x) = \begin{cases} 1, & 0 \le x < 1 \\ 2-x, & 1 \le x \le 2 \\ 0, & 2 < x. \end{cases}$$ Now in light of the above, consider how to proceed with the variance calculation. To check your work, calculate the second moment via the density; namely $$\operatorname{E}[X^2] = \int_{x=1}^\infty x^2 f_X(x) \, dx.$$