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.
 A: 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. 
A: 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.
A: 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.
A: To find the distribution of $X-2$ lets introduce a variable $y = x-2$.  Then replacing $x$ with $y+2$, we find
$$
f(y)= 2(y+2)e^{-(y+2)^2}
$$
for $y+2 \geq 0$ (or $y \geq -2$) and zero elsewhere.
This can be simplified with a bit of algebra, but the concept is the important bit.  But while this is right, we kind of lucked out.
When we naively do this for $2X$, however, we will get
$$f_\mbox{wrong} (y) = y e^{\frac{y^2}{4}} \mbox{ for } y \geq 0
$$
which has the wrong integral (2 instead of 1).  The reason is that we need to include the fact that a small change in $x$ is half that of the change in $y$; that is, we need to divide by $\frac{dy}{dx}$, giving the correct answer of 
$$f (y) =\frac{1}{2} y e^{\frac{y^2}{4}} \mbox{ for } y \geq 0
$$
Now do you see why $X-2$ came out right?
Similarly, the distribution of $X^2$ starts with $y = x^2$ or $x = \sqrt{y}$ and becomes (the derivative is $\frac{dy}{dx} = 2x = 2\sqrt{y}$) 
$$
\frac{1}{2 \sqrt{y}} 2 \sqrt{y} e^{-y} = e^{-y} \mbox{ for } y \geq 0
$$
