Verifying Solution for PDF to CDF and Inverse CDF Calculations

[zzmanzz]

Homework Statement

I was hoping someone could just verify this solution is accurate.

p(x) =
0 , x < 0
4x, x < .5
-4x + 4 , .5 <= x < 1

Find CDF and Inverse of the CDF.

Homework Equations

The Attempt at a Solution

CDF =
0 , x < 0
2x^2 , 0 <= x < .5
-2x^2 + 4x - 1 , .5 <= x <= 1
1, x > 1

Inverse of the CDF

0 , x < 0
sqrt( x / 2) , 0 <= x < .5
1 + sqrt ( 1 - x) / sqrt ( 2 ) , .5 <= x <= 1
1, x > 1

Thanks[/B]

 

[Ray Vickson]

Homework Statement

I was hoping someone could just verify this solution is accurate.

p(x) =
0 , x < 0
4x, x < .5
-4x + 4 , .5 <= x < 1

Find CDF and Inverse of the CDF.

Homework Equations

The Attempt at a Solution

CDF =
0 , x < 0
2x^2 , 0 <= x < .5
-2x^2 + 4x - 1 , .5 <= x <= 1
1, x > 1

Inverse of the CDF

0 , x < 0
sqrt( x / 2) , 0 <= x < .5
1 + sqrt ( 1 - x) / sqrt ( 2 ) , .5 <= x <= 1
1, x > 1

Thanks[/B]

You chose the wrong root for $.5 \leq x \leq 1$. Can you see why?

 

[zzmanzz]

You chose the wrong root for $.5 \leq x \leq 1$. Can you see why?

Thanks for pointing that out. The value for that should also be between .5 and 1, and with that root, it can be greater than 1?

 

[Ray Vickson]

Thanks for pointing that out. The value for that should also be between .5 and 1, and with that root, it can be greater than 1?

Yes, just look at it: you have $1 + \text{something positive}$.

 

[zzmanzz]

Yes, just look at it: you have $1 + \text{something positive}$.

Thank you!