# Stats: figuring out ranges of transformed variables

1. Aug 23, 2011

### bennyska

1. The problem statement, all variables and given/known data
so in general, i am having difficulty in figuring out the range of a transformed variable. sometimes it's easy, i just plug in values and get the other values out, but sometimes i'm not seeing it. for example:
f(x)=4x3, for 0<x<1
and let U = (X-.5)2. Finding the pdf of U was not difficult, but finding the range is. the book says 0<u<.25.

2. Relevant equations

3. The attempt at a solution
i can see that if i plug in 0 for x, i get .25, and if i plug in 1 for x, i get .25. i guess, in this range, i can see that the max is .25 and the min is 0. in general, is that how i find it? just take the min and max? that doesn't seem right to me.

another example would be with f(x) = 1, and let Y = X(1-X). finding the pdf wasn't too difficult, but again i'm having problem with the range. actually, the book gives the same answer, and the graph is pretty similar, just rotated. so, really, i don't know where to get the ranges.

2. Aug 23, 2011

### Staff: Mentor

This is actually pretty straightforward. Just work through the inequalities.

You are given that 0 < x < 1,
so -.5 < x - .5 < .5,
If you square x - .5 (to get u), you will always get something that is >= 0, so you have
0 <= (x - .5)2 < (.5)2

or 0 <= u < .25

3. Aug 23, 2011

### Ray Vickson

Look at the graph of y = (x - .5)2. What range of y-values is obtained from the x-interval 0 <= x <= 1?

RGV

4. Aug 24, 2011

### bennyska

thanks, mark, that's totally obvious now that you put it out there.
i'm still having a little trouble with the next one, though.
0<x<1
0<x(1-x)<(1-x)
0<u<1-x... solving for x in terms u doesn't really seem to work. in short, i don't see how i can rid of that x and get .25 on that side...

ray: so is that all i have to do? if i look at the graph of the function i'm using to transform, and it's continuous, all i have to do is take the max and min? that seems wrong to me, since it's not one to one, but i really don't know.

5. Aug 24, 2011

### Ray Vickson

Yes, to get the *range* that's all you have to do. It is true that (x-.5)^2 does not have an inverse, but that is a separate issue (which you must address when computing the _distribution_ of Y, but not when getting the range of Y).

RGV

Last edited: Aug 24, 2011
6. Aug 24, 2011

### Staff: Mentor

For the 2nd problem, if u = x(1 - x), the graph of this equation is a parabola that opens down. The two x-intercepts are pretty easy to find, and the vertex is also easy to find. Once you know the coordinates of the vertex, finding the range of values of u is straightforward.

7. Aug 24, 2011

### bennyska

so i guess it makes sense that the min and max of the function i'm transforming is the range. it just seems weird to me because it's not one to one. but if you have to incorporate that fact elsewhere, then i'm more okay with it. it just seemed like i was leaving information out. thanks.