<s>[INST] Finding density function of random variable $Y=X^2$ 
Suppose $X$ is continuous r.v with c.d.f 
$$ F(x) = \begin{cases} 1 - \left( \frac{2}{x} \right)^2 \; \; \; \;
 \; \; x>2 \\ 0, \; \; \; \; \text{otherwise} \end{cases} $$
Put $Y=X^2$. Please find density function of $Y$.

Attempt
By definition $f_Y(y) = \frac{d}{dy} F_Y(y) $. We know 
$$ F_y(y) = P(Y \leq y ) = P(X^2 \leq y ) = P(X \leq \sqrt{y} )$$
And hence 
$$ F_y(y) = \int\limits_2^{\sqrt{y}} 1 - \left( \frac{2}{x} \right)^2 = \sqrt{y} + \frac{4}{\sqrt{y}}- 4  $$
It follows that 
$$ f_Y(y) = \frac{1}{2 \sqrt{y}} - \frac{4}{2 y^{2/3} } $$
But,
my answer key say $f_Y(y) = \frac{4}{y^2}$.
What am I doing wrong here?
 [/INST] You were already given the CDF of $X$.  Integrating again is not necessary.  We simply write $$F_Y(y) = F_X(\sqrt{y}) = 1 - \left(\frac{2}{\sqrt{y}}\right)^2 = 1 - \frac{4}{y}, \quad y > 4,$$ thus $$f_Y(y) = \frac{4}{y^2}$$ as claimed.
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