- #1

Mogarrr

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## Homework Statement

Show that the given function is a cdf (cumulative distribution function) and find [itex]F_X^{-1}(y)[/itex]

(c) [itex]F_X(x) = \frac {e^{x}}4 [/itex], if [itex]x<0[/itex], and [itex]1-(\frac {e^{-x}}4) [/itex], if [itex]x \geq 0 [/itex]

## Homework Equations

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

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

## The Attempt at a Solution

It's not so hard to show that F is a cdf. The [itex] lim_{x \to -\infty} F_X(x)= 0[/itex], the [itex]lim_{x \to \infty} F_X(x) = 1 [/itex], 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

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

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

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

For example, using the inverses I have, the set of x's such that [itex]F_X(x) \geq \frac 14 [/itex] would include [itex] 4ln( \frac 12) \approx -2.772 [/itex] and [itex] -4ln(1- \frac 12) \approx 2.772 [/itex], 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, [itex] F_X^{-1}(y) = 4lny [/itex], since this function would have the smallest x's.

Am I right here? Please help.

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