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!
 A: $F_Y(y) := P(Y\le y)$ is the probability that the random variable $Y$ is less than or equal to a given real value $y$.  
This function is then known as the cumulative distribution function.
For a continuous random variable it is the integral of the probability density function up to $y$, while for a discrete random variable it is the partial sum up to $y$ of the probability mass function. 
For example if $Y$ is the sum from rolling two standard fair dice then $F_Y(4)=\frac1{36}+\frac2{36}+\frac3{36}=\frac16$.    
A: 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.
A: Not the 'entire' course, but a few ideas that may be helpful, when you put them together.
First, always pay attention to the support of a random variable.
For example, $X$ has support $(0, \infty),$ which implies $P(X > 0) = 1.$ This is the reason that $P(-\sqrt{y} < X < \sqrt{y})$ becomes 
$P(0 < X < \sqrt{y}).$  
Then when you move on to $Y = X^2,$ you must also have $P(Y > 0) = 1.$
One way to find the density function of $Y$ is to find its
cumulative distribution function (CDF) and then take the derivative to
get the density function. The CDF of $Y$ is:
$$F_Y(y) = P(Y \le y) = P(X^2 \le y) = \cdots = P(X \le \sqrt{y})\\
= 1 - P(X > \sqrt{y}) = 1 - e^{-\lambda(\sqrt{y})^2} = 1 - e^{-\lambda y},$$
for $y > 0.$
Just try to figure out each equal sign in this continued
equation. You have most of it in your Question. Then the density function of $Y$ is
$$f_Y(y) = F_Y^\prime(y) = \frac{d}{dy}(1 - e^{-\lambda y}) = ???,$$
for $y > 0.$
Note: You may be in a rush to get ready for your exam, but you won't save
time trying to start at the end of a string of ideas. Start at the
beginning with examples and problems, and move forward step-by-step
as efficiently as possible. 
A: You understand more than you think. 
Try plotting the function when $\lambda = 1$. Note we have a nice curve. This is a probability mass function (pmf). This tell us the probability that the value $x$ will occur. We express this by saying $P(X=x)$.
Consider the plot of our function, what if we want to know the probability that our random variable $X$ is greater than 1? Then we would, as you did for the general case, compute the definite integral from 1 to infinity. We describe this as $P(X > 1)$. Since we know (and have defined) the total probability of a random variable to be $1$, we could also describe this as $1 - P(X < 2)$.
When we define $Y = X^2$, the probability that $Y$ is less than $0$ or $P(Y<0)$ must be zero because $X^2$ can never be negative.
In the last equation when we substitute $y$ for $x^2$ we do so because we are talking in terms of $Y$, which we agreed earlier is $X^2$. Just a change of variable.
