# What does this definition mean: $F_Y(y) =P(Y<y)$?

I 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!

• I edited your question a bit, maybe you could take a look how the edit was done, and continue the editing on your own :-) Jun 26, 2015 at 19:55
• @GSassatelli: Thanks for some further editing. The first attempt by another "editor" verged on vandalism. Jun 26, 2015 at 20:00
• I am sorry for you, but this site is not supposed to teach you the entire Probability course from scratch in $3$ days.
– user228113
Jun 26, 2015 at 20:00
• @BruceTrumbo, sorry for that, I actually was that first (actually second) one. I'll be more careful next time.
– user228113
Jun 26, 2015 at 20:08
• @Zhanxiong: When editing, please make sure that your edit displays correctly. Otherwise, you end up doing more harm than helping, as it happened here (I shall roll things back). Jun 26, 2015 at 20:32

$F_Y(y) := P(Y\le y)$ 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 $F_Y(4)=\frac1{36}+\frac2{36}+\frac3{36}=\frac16$.

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.

• I will look tomorrow. Post a Comment if there is part of this you can't figure out. Jun 26, 2015 at 20:53

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.

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.