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.

We approach the problem through the cumulative distribution functions, even though it is more inefficient than the method of transformations. 
1) Let $Y=X-2$. We want the density function of $Y$. First we find an expression for the cumulative distribution function of $Y$, that is, $\Pr(Y\le y)$.
We have
$$F_Y(y)=\Pr(Y\le y)=\Pr(X-2\le y)=\Pr(X\le 2+y).$$
For $y\le -2$, we have $\Pr(X\le 2+y)=0$, so $F_Y(y)=0$, and therefore the density function $f_Y(y)$ is $0$. For $y\gt -2$, we have
$$F_Y(y)=\int_0^{2+y}2xe^{-x^2}\,dx.$$
Now we have two options: (i) Calculate $F_Y(y)$, and differentiate to find $f_Y(y)$ or (ii) Use the Fundamental Theorem of Calculus to differentiate the above integral. That is easier, and gives
$$f_Y(y)=2(2+y)e^{-(2+y)^2}$$
for $y\gt -2$. 
2) Let $Y=-2X$. We have $F_Y(y)=\Pr(Y\le y)=\Pr(-2X\le y)=\Pr(X\ge -\frac{y}{2}$.
Now work much as in 1).
3) Let $Y=X^2$. For $Y\le 0$, we have $F_Y(y)=0$, so the density function is $0$. For $y\gt 0$, we have
$$F_Y(y)=\Pr(X^2\le y)=\Pr(X\le \sqrt{y})=\int_0^{\sqrt{y}} 2xe^{-x^2}\,dx.$$
Now calculate the integral, and differentiate, or differentiate under the integral sign. 
Remark: In the title, you ask what $X-2$ means. It is easier to explain with a different function. Imagine an experiment in which we take a person at random, and measure her height. Let the random variable be the person's height in metres. Let $Y=100X$. Then $Y$ is a random variable, and measures the person's height in cm. 
Suppose the measurement was made when the person was wearing shoes with $2$ cm thick soles. Let $Z=100X-2$. Then the random variable $Z$ is the person's bare foot height in cm. 
