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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!

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    $\begingroup$ I edited your question a bit, maybe you could take a look how the edit was done, and continue the editing on your own :-) $\endgroup$
    – Math-fun
    Jun 26, 2015 at 19:55
  • $\begingroup$ @GSassatelli: Thanks for some further editing. The first attempt by another "editor" verged on vandalism. $\endgroup$
    – BruceET
    Jun 26, 2015 at 20:00
  • $\begingroup$ I am sorry for you, but this site is not supposed to teach you the entire Probability course from scratch in $3$ days. $\endgroup$
    – user228113
    Jun 26, 2015 at 20:00
  • $\begingroup$ @BruceTrumbo, sorry for that, I actually was that first (actually second) one. I'll be more careful next time. $\endgroup$
    – user228113
    Jun 26, 2015 at 20:08
  • $\begingroup$ @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). $\endgroup$
    – Alex M.
    Jun 26, 2015 at 20:32

4 Answers 4

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

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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.

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

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  • $\begingroup$ Don't you mean the probability that X is greater than x? $\endgroup$
    – Chris
    Jun 27, 2015 at 2:12
  • $\begingroup$ @Chris Yes, I fixed it. $\endgroup$
    – Amy B
    Jun 28, 2015 at 1:28
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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.

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