# Probability - Find PDF

1. Mar 5, 2013

### dkotschessaa

1. The problem statement, all variables and given/known data

Let the random variable X have the continuous probability density function f(x). find the probability density function of Y= F(x)

2. Relevant equations

I *think* we are meant to use something like $h(y) = f(g^{-1}(y)) * ((dg^{-1}(y)) /dy)$

(That is as good as I know how to format right now - I hope that's readable).

But I'm given just f(x) and I'm being asked for F(x). And yet it's not asking for a CDF, it's asking for a PDF. So I'm really confused on what's being asked for here.

(Unless this is another one of my book typos) Does the question even make sense?

-Dave K

2. Mar 5, 2013

### Ray Vickson

Find the cdf of Y as $F_Y(y) = P\{ Y \leq y \} = P\{ F(X) \leq y \},$ then find the density of Y (the pdf) by differentiation. This method applies even when the formula you use fails (for example, because of intervals or points where f(x) = 0 but inside the range of X). Doing a bit of on-line research, or looking into a book on simulation, may be very helpful.

3. Mar 7, 2013

### dkotschessaa

So am I to integrate first for the CDF, and then differentiate back down again?

I think it's perhaps so trivial that I'm getting confused. I even have the "back of book" answer but I don't know the reasoning.

4. Mar 7, 2013

### Ray Vickson

That is why I suggested you do a bit of on-line research or look into a book on simulation. Specifically, look at 'methods of generating random samples' or 'random variable generation' or something similar.

5. Mar 7, 2013

### dkotschessaa

I've looked everywhere online. But I'll try that.

6. Mar 7, 2013

### vela

Staff Emeritus
Is Y=F(x) exactly how it was written in your book? If so, it looks like a typo or two. You may want to ask your professor for clarification.

Perhaps you're supposed to find the cdf F(x) and then define the new random variable Y=F(X) (a function of X, not x) and find its density function.

7. Mar 8, 2013

### dkotschessaa

Yes, and the book is full of typos. And the professor gets lost in the middle of class. and..and.. you get the idea.

I'm not sure. I'm really confused and frustrated. I have something written down and I'm handing it in.

Thanks for the help everyone.

-Dave K

8. Mar 8, 2013

### ArcanaNoir

Dang David, next time try running it by me. I already had that class so I should be able to work it out if I can see the problem in the book. :)

9. Mar 8, 2013

### haruspex

No, I think Ray's interpretation is right. Yes, it should be Y=F(X), not F(x), but I doubt there's any relationship between f and F.

10. Mar 8, 2013

### vela

Staff Emeritus
But it doesn't sound like F(X) was specified, which makes it impossible to find fY(y).

11. Mar 8, 2013

### Ray Vickson

No. Finding $f_Y(y)$ is "easy" (once you see it!); this is one of the important standard results that are used all the time in Monte-Carlo simulation.

12. Mar 8, 2013

### vela

Staff Emeritus
I don't see it. You're given random variable X with probability density fX(x). Then you're asked to find fY(y) where Y=F(X). Haruspex has suggested F(X) bears no relation to fX(x), so it's completely arbitrary. You have virtually no information about Y other than it's a function of X.

13. Mar 8, 2013

### haruspex

The answer will involve F'.

14. Mar 8, 2013

### Ray Vickson

I will be happy to elucidate as soon as I am convinced the OP has already handed in his solution.

15. Mar 8, 2013

### Ray Vickson

In my previous responses I have been assuming all along that F is the cdf of X (just because of the notation you have used). If that is not what you meant then all my previous responses are irrelevant.

16. Mar 9, 2013

### ArcanaNoir

I can attest to the fact that the assignment has been handed in. OP is on vacation now so you probably won't hear from him for a week.

17. Mar 10, 2013

### Ray Vickson

OK, so if X has pdf f(x) and cdf F(x), then Y = F(X) is uniformly distributed on (0,1). This is true even if X does not have a density (such as for discrete random variables, or mixed random variables that are partly continuous and partly discrete); the only thing that matters is that F(x) is a well-defined function (defined so as to be right-continuous). See, eg.,
http://en.wikipedia.org/wiki/Inverse_transform_sampling
for illustration and proof.

This method is of fundamental importance in Monte-Carlo simulation.

Note: the OP did not actually *say* that F was the cdf of X, but I just assumed he meant that. If not, then F is a more-or-less arbitrary function and there will be no simple, general result.

18. Mar 15, 2013

### dkotschessaa

Oh, the whole assignment was a mess, as you know. I'd be bugging you for every question.

I am dropping the class anyway. My first drop, ever. I took some time on spring break to think about it. (I'm still there, just at a hotel in the middle of nowhere, PA). My final act was to get together with the class (we are down to about 8 people) and we agreed on some solutions to the assignment.

Thanks everyone.

-Dave K

19. Mar 15, 2013

### HallsofIvy

Staff Emeritus
In post 17, Ray Vickson is assuming that F(x) is the integral of f(x). Is that true? My first thought was that they were unrelated functions.

In that case, $$P(y< Y)= P(F(x)< Y= \int_{-\infty}^Y F(t)f(t)dt$$ and the PDF is the derivative of that, with respect to Y.

20. Mar 15, 2013

### dkotschessaa

The question was word for word as found in the book. There was no further information and I don't know what we were supposed to assume.