What does this definition mean: $F_Y(y) =P(YI am doing calculations on $F_Y(y) := P(Y<y)$, but I am clueless as to what $P(Y<y)$ means. 
For instance the following question:
Given function: $f_X(x)= 2\lambda x e^{-\lambda x^2}$ when $x \geq 0$ (parameter $\lambda>0$)
Show that for $x>0$ $P(X>x)=e^{-\lambda x^2}$
I did the calculation (integration) and that's fine. I just don't know what it is I am doing. What does $P(X>x)$ mean?
Because the following question I'm not sure how to solve:
Compute the probability mass function of $Y=X^2$ 
So the teacher says it should be solved as follows, but again I don't know what it means:
For $y<0,\ F_Y(y)=P(Y<y)=0$ (Why does this equal zero?) 
For $y>0,\ F_Y(y)=P(X^2<y)$ (substitute for the definition, but why?)
$=P(-\sqrt{y} < X < \sqrt{y})$ (okay)
$=P(0<X<\sqrt{y})$ (why is it zero?)
$=1-P(x>\sqrt{y})=1-e^{-\lambda y}$ (why is this so? the integral is $e^{\lambda x^2}$, how come I am allowed to substitute the y for $x^2$ and how does all of this have anything to do with $Y=X^2$?) 
And then you have to differentiate, because you've got Fy, but you want $f_y$, which I get.
Would appreciate the help a lot! Got an exam on the 30th!

P(Y < y) means the probability that Y is less than y.
Similarly P(X > x) is the probability that X is greater than x.
I hope I answered your basic question.
