Proving the variance of pareto random variable equals $(a\lambda)/((a-1)^2(a-2))$ So my PDF for the Pareto distribution is:
$$\dfrac{a\lambda^a}{x^{a+1}},\quad x\ge\lambda$$
To find the variance, you need to find the integral of $x^2\dfrac{a\lambda^a}{x^{a+1}}$ and subtract it from $[E(x)]^2$.
So, I worked out the integral and got $\dfrac{a\lambda^2}{a-2}$ and found that $[E(x)]^2=\dfrac{\lambda^2a^2}{(a-1)^2}$.
The problem is that I keep getting the variance equal to $\dfrac{a\lambda^2}{(a-1)^2(a-2)}$ instead of the supposed answer of $\dfrac{a\lambda}{(a-1)^2(a-2)}$.
What am I doing wrong?
 A: Note that if $X \sim \operatorname{Pareto}(a,\lambda)$, with density $$f_X(x) = \frac{a \lambda^a}{x^{a+1}}, \quad x \ge \lambda,$$ we have $$\int_{x=\lambda}^\infty f_X(x) \, dx = a \lambda^a \int_{x=\lambda}^\infty x^{-(a+1)} \, dx = a\lambda^a \left[-\frac{1}{a x^a} \right]_{x = \lambda}^\infty = 1,$$ so the density does indeed integrate to $1$ for $a > 0$.  So consider the $k^{\rm th}$ raw moment, which is given by $$\begin{align*} \operatorname{E}[X^k] &= \int_{x=\lambda}^\infty x^k \cdot \frac{a \lambda^a}{x^{a+1}} \, dx \\ &= a \lambda^a \int_{x=\lambda}^\infty \frac{1}{x^{a+1-k}} \, dx \\ &= \frac{a \lambda^a}{(a-k)\lambda^{a-k}} \int_{x=\lambda}^\infty \frac{(a-k) \lambda^{a-k}}{x^{a-k+1}} \, dx \\ &= \frac{a\lambda^k}{a-k}. \end{align*}$$  This is because the last integral is the integral of a $\operatorname{Pareto}(a-k, \lambda)$ density, so it equals $1$ whenever $a-k > 0$, for the same reason that we showed that a $\operatorname{Pareto}(a, \lambda)$ density integrates to $1$ for $a > 0$.  So now in particular we have $$\operatorname{E}[X] = \frac{a\lambda}{a-1}, \quad \operatorname{E}[X^2] = \frac{a\lambda^2}{a-2},$$ corresponding to the choices $k = 1$ and $k = 2$, respectively.  Note the first moment is defined for $a > 1$, and the second for $a > 2$.  The variance is then $$\begin{align*} \operatorname{Var}[X] &= \operatorname{E}[X^2] - \operatorname{E}[X]^2 \\ &= \frac{a\lambda^2}{a-2} - \frac{a^2 \lambda^2}{(a-1)^2} \\ &= \frac{((a-1)^2 - a(a-2))a\lambda^2}{(a-1)^2(a-2)} \\ &= \frac{a\lambda^2}{(a-1)^2(a-2)}.\end{align*}$$
