Squaring a Pareto random variable: Is there a proper name for this transformation? Let $Y= X^2$ be a function of a random variable where $X$ has a Pareto Type II (Lomax) distribution with parameters $\alpha = 8$, $\theta = 4000$, $x > 0$. Find the distribution of $Y$.
I basically went through the ropes as typical for this kind of question. I started with $F_Y(y) = P(Y < y) = P(-\sqrt{y} < x < \sqrt{y})$. 
Of course, the Lomax distribution is defined for $x > 0$ and so we take $x > 0$ as the lower bound in the integral rather than -$\sqrt{y}$.
This gave me, using the Fundamental Theorem of Calculus,
$[(8*(4000)^8) / ((y^{1/2} + 4000)^9)] * (1/2) * (y^{-1/2})$, 
which is just the density of $x$ evaluated at $y^{1/2}$ and using the chain rule where the derivative of $y^{1/2}$ is of course $(1/2)* (y^{-1/2})$.
I am fairly certain this is correct, but I have been asked to name this distribution, in which I am stuck. I can't recognize the distribution of Y, and I've tried manipulating the expression to make it look like something I recognize, with limited success. Are there any slick ways of doing this? Or does this transformation even produce a distribution that is "well known"?
EDIT: For anyone who cares, a Pareto IV where u = 0 is a Burr distribution.

The Pareto II (Lomax) CDF is given by $$F_X(x) = 1 - \left(1 + \frac{x}{\sigma}\right)^{-\alpha}, \quad x > 0.$$  Your parametrization is slightly different, but the calculations are comparable.
The Pareto IV CDF is given by $$F_Y(y) = 1 - \left(1 + \left(\frac{y-\mu}{\sigma}\right)^{1/\gamma} \right)^{-\alpha},$$ so it seems natural that if $Y = \mu + (\sigma X)^\gamma$, then we immediately find $$F_Y(y) = \Pr[Y \le y] = \Pr\left[X^\gamma \le \frac{y - \mu}{\sigma} \right] = F_X(((y-\mu)/\sigma)^{1/\gamma}),$$ hence we see that your transformation $Y = X^2$ turns a Pareto II (Lomax) into a Pareto IV with $\alpha$, $\mu = 0$ unchanged, but $\gamma = 2$ and the Pareto IV $\sigma$ parameter is the square of the Pareto II value of $\sigma$.

Okay, so I think you are using the same parametrization, so $\sigma = \theta$, in which case you have the CDF $$F_X(x) = 1 - \left(1 + \frac{x}{\theta}\right)^{-\alpha} = 1 - \frac{\theta^\alpha}{(\theta+x)^\alpha}.$$  With the transformation $Y = X^2$, we obtain $$F_Y(y) = F_X(\sqrt{y}) = \color{red}{\boxed{1 - \left(1 + \frac{y^{1/2}}{\theta}\right)^{-\alpha}}} = 1 - \frac{\theta^\alpha}{(\theta+ y^{1/2})^\alpha},$$ and its derivative is $$f_Y(y) = \alpha \left(1 + \frac{y^{1/2}}{\theta}\right)^{-(\alpha+1)} \frac{1}{2\theta y^{1/2}}.$$  But since we are comparing CDFs, all you need to do is observe that the expression in red corresponds to the Pareto IV distribution I wrote above, but with values $\mu = 0$, $\gamma = 2$, $\theta = \sigma^{1/\gamma}$.
Incidentally, I am using the definitions given by Wikipedia.  These may not match the nomenclature or parametrization used by SOA/CAS for the exams.
