Density of probability/function of random variables question

[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(x_k)=y , Fy(y)=sumFx(x_k)/|g`(x_k)|
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 don't 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(x<sub>k</sub>)/|g`(x_k)|=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.

 

[Stephen Tashi]

. to find the intervals i use the formula [g(-1);g(1)] but i don't know if its right and in this case g(-1)=g(1)=2;
What i am doing wrong?

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.

 

[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]

 

[Stephen Tashi]

The minimum valuea is 1 and the maximum value is 2 for x=[-1;1]

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.)

 

[blue_raver22]

Hi there,

It looks like you are working on finding the density of probability or function of random variables for a given function g(x). In order to do this, you need to find the values of x that satisfy the equation g(x) = y. In the first exercise, you correctly identified the two possible solutions for x, which are x = +sqrt(y-1) and x = -sqrt(y-1). However, you are using the wrong interval to calculate Fy. Since g(x) is bijective on [-1;1], the correct interval to use is [-1;1]. This means that Fy will be defined on the interval [g(-1);g(1)] = [2;2], which is a single point. This means that Fy(y) = 1 for y = 2 and Fy(y) = 0 for all other values of y.

In the second exercise, you have correctly identified that x = -sqrt(y-3) is not a possible solution since it falls outside of the interval [0;2]. However, your calculation for Fy is incorrect. Remember that Fy is the sum of the probabilities of all possible values of x that satisfy g(x) = y, divided by the absolute value of the derivative of g(x). In this case, the absolute value of the derivative is 2x, so Fy(y) = sumFx(xk)/|g`(xk)| = 1/2*1/2x = 1/4x. This means that Fy(y) = 1/4x for y = [g(0)=3;g(2)=7], which is the interval [3;7]. For all other values of y, Fy(y) = 0.

I hope this helps clarify your understanding of finding the density of probability/function of random variables. Keep up the good work on your homework!