# Density of probability/function of random variables question

1. Apr 20, 2013

### Drao92

Hi everyone,
I have the following exercise.
Fx(x)=0, x<-1 or x>1
Fx(x)=1/2, x=[-1;1]
g(x)=x^2+1 --- this is the function of random variable
I must calculate Fy which is the sum of solutions of g(xk)=y , Fy(y)=sumFx(xk)/|g(xk)|
g(x) is bijective on [-1;1]
y=x^2+1=> x=+sqrt(y-1) or x=-sqrt(y-1), since x=[-1;1] both are posible solutions.
And my question is on what interval is Fy defined... to find the intervals i use the formula [g(-1);g(1)] but i dont know if its right and in this case g(-1)=g(1)=2;
What i am doing wrong???

On a similar exercise i had
Fx(x)=1/2, x=[0;2]
Fx=0, out of the interval
g(x)=x^2+3
g(x)=2x
x=sqrt(y-3) and x=-sqrt(y-3), since x=[0;2], x=-sqrt(y-3) is not a posible solution.
Fy(y)=sumFx(xk)/|g`(xk)|=1/4*1/sqrt(y-3))
So
Fy(y)=1/4*1/sqrt(y-3) for y=[g(0)=3;g(2)=7];
Fy(y)=0 for y =[-infinite;3]U[7;+infinite]
On seminars we did only with g(x)=a*x+b which was easy and these are for homework.

Last edited: Apr 20, 2013
2. Apr 20, 2013

### Stephen Tashi

The values of Y that that have non-zero probability will range from the minimum to the maxium value of the function X^2 + 1 for X in the interval [-1, 1]. You should be able to determine these values "by inspection". If not, you could work it as max-min problem using calculus. You method doesn't work because g(-1) is not the minumum value that g(x) attains.

3. Apr 21, 2013

### Drao92

So there are more methods???
For the second exercise the solution is good??? The condition for density gives me 1. http://www.wolframalpha.com/input/?i=integral+from+3+to+7+from+1/4*1/sqrt(y-3)dy
Thanks for tip, ill look on seminars, i think we did somthing similar and ill post my solution.

Later edit.
This is the solution:
http://www.wolframalpha.com/input/?i=integral+from+1+to+2+from+1/2*1/sqrt(y-1)dy
It seems good :). Thanks very much. Its more logical like this because in theory y range on oy axis of graph x^2+1 for x=[-1;1].
Its everything corect, right????
The minimum valuea is 1 and the maximum value is 2 for x=[-1;1]

Last edited: Apr 21, 2013
4. Apr 21, 2013

### Stephen Tashi

Yes. I'm not going to check all your work. If you want that much help. you should post your questions in the sections on homework help. (You phrased the question well and showed your work, so your post would be OK for that section.)