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.

The probability density function is not the same thing as the value of the random variable.  What the density tells us is the likelihood of observing a given outcome of a random variable.  If $X$ has density $f(x)$, then $X-2$ has the density $f(x+2)$.
Let's look at a simple example.  Suppose we have a discrete random variable $Y$ which has probability mass function $$\Pr[Y = 0] = 1/3, \quad \Pr[Y = 3] = 2/3,$$ and any other value of $Y$ does not occur (so either $Y = 0$ or $Y = 3$).  Now what is the support of $Y - 2$?  That is to say, $Y - 2$ can equal only $-2$ or $1$, because $Y$ itself can be only $0$ or $3$.  Then $$\Pr[Y-2 = -2] = \Pr[Y = 0] = 1/3,$$ and $$\Pr[Y - 2 = 1] = \Pr[Y = 3] = 2/3,$$ just as before.  All we have done is shifted the probability mass function.  The same reasoning applies to your $X$.
In general, for a one-to-one transformation $Y = g(X)$ of a random variable $X$, the probability density of $Y$ as a function of the probability density of $X$ is given by $$f_Y(y) = f_X(g^{-1}(y)) \left| \frac{dg^{-1}}{dy} \right|.$$  For the function $y = g(x) = x-2$, $g^{-1}(y) = y+2$, hence $dg^{-1}/dy = 1$ and we get $f_Y(y) = f_X(y+2)$, as claimed.  For $g(x) = 2x$, $g^{-1}(y) = y/2$, and $dg^{-1}/dy = 1/2$.
The last case requires a bit of subtlety, because in general, $x^2$ is not a one-to-one function.  But it is one-to-one on the support of $X$, since $X > 0$.  So we can still use this transformation rule:  $g(x) = x^2$, $g^{-1}(y) = \sqrt{y}$, and $dg^{-1}/dy = 1/(2\sqrt{y})$, and the rest is straightforward.
