What does X-2 mean given continous probability distribution X? I have the continous probability distribution X: $f(x) = 2x e^{-x^2} \, x \geq 0$ and zero everywhere else. One of my homework problems is to find the probability distribution of X-2, -2X, and X^2 but intuitively it doesnt make much sense to me.
For example if i consider X-2:
$f(x) = 2xe^{-x^2} - 2 \, x \geq 0$ and $-2$ everywhere else. This doesnt make sense and isnt a probability distribution. Neither is:
$f(x) = 2xe^{-x^2} - 2 \, x \geq 0$ and $0$ everywhere else.
A little bit of input would be highly appreciated.

If function $F_{X}$ is prescribed by $x\mapsto1-e^{-x^{2}}$ for
$x\geq0$ and $x\mapsto0$ otherwise then it has the characteristics
of a CDF and the function $f_{X}:=f$ as mentioned in the question
is the PDF that belongs to it. 
Then $F_{X-2}\left(x\right)=P\left(X-2\leq x\right)=P\left(X\leq x+2\right)=F_{X}\left(x+2\right)$
and $f_{X-2}$ can be found by taking the derivative of $F_{X-2}$
leading to $f_{X-2}\left(x\right)=f_{X}\left(x+2\right)=f\left(x+2\right)$.
Likewise the CDF and PDF of $-2X$ and $X^2$ can be found.
