Is the Inverse Calculation for a Piecewise CDF Correct?

[Mogarrr]

Homework Statement

Show that the given function is a cdf (cumulative distribution function) and find F_X^{-1}(y)
(c) F_X(x) = \frac {e^{x}}4, if x<0, and 1-(\frac {e^{-x}}4), if x \geq 0

Homework Equations

for a strictly increasing cdf, F_X^{-1}(y) = x \iff F_X(x) = y

and for a non-decreasing (a.k.a. difficult problem) cdf, F_X^{-1}(y) = inf \{ x: F_X(x) \geq y \}

The Attempt at a Solution

It's not so hard to show that F is a cdf. The lim_{x \to -\infty} F_X(x)= 0, the lim_{x \to \infty} F_X(x) = 1, the function is non-decreasing, and right-continuous.

I have the solution for the inverse, but it doesn't seem right to me. The given solution is

F_X^{-1}(y) = ln(4y) for 0<y< \frac 14 and -ln(4(1-y)) for \frac 14 \leq y<1

But this solution doesn't seem to agree with the definition of inverse F or the inverses I found.

so if y = e^{\frac {x}4}, then doesn't this imply x = 4lny? and doesn't y= 1 - (e^{\frac {-x}4 }), imply that x = -4ln(1-y)?
For example, using the inverses I have, the set of x's such that F_X(x) \geq \frac 14 would include 4ln( \frac 12) \approx -2.772 and -4ln(1- \frac 12) \approx 2.772, and the infimum of the set (greatest number that is less than all other numbers in the set, I think) is -2.772. So I should use the first inverse, F_X^{-1}(y) = 4lny, since this function would have the smallest x's.

Am I right here? Please help.

 

[Mogarrr]

so if y = e^{\frac {x}4}, then doesn't this imply x = 4lny? and doesn't y= 1 - (e^{\frac {-x}4 }), imply that x = -4ln(1-y)?

I see my mistake now.