Is the Inverse Calculation for a Piecewise CDF Correct?

In summary: Since F_X(x) is defined differently for x<0 and x>=0, the inverse function must also be defined differently for y<1/4 and y>=1/4. For y<1/4, we have x=4lny, and for y>=1/4, we have x=-4ln(1-y). Therefore, the correct solution for F_X^{-1}(y) is:F_X^{-1}(y) = 4lny for 0<y<1/4 and -4ln(1-y) for 1/4<=y<1In summary, to show that the given function is a cdf, we can observe that it satisfies the necessary conditions such as being
  • #1
Mogarrr
120
6

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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  • #2
Mogarrr said:
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]?

I see my mistake now.
 

FAQ: Is the Inverse Calculation for a Piecewise CDF Correct?

What is the inverse of a CDF?

The inverse of a CDF (Cumulative Distribution Function) is the function that maps probabilities to their corresponding values in a probability distribution. It allows us to determine the value that corresponds to a given probability in a distribution.

What is the purpose of finding the inverse of a CDF?

The inverse of a CDF is useful in statistical analysis as it allows us to determine the probability of a certain value occurring in a distribution. It is also used in hypothesis testing and calculating confidence intervals.

How is the inverse of a CDF calculated?

The inverse of a CDF is typically calculated using numerical methods such as interpolation or root finding algorithms. These methods involve using a computer program to iteratively approximate the inverse function.

What is the relationship between the CDF and the inverse of a CDF?

The CDF and its inverse are mathematical inverses of each other. This means that applying the CDF to a value and then applying the inverse of the CDF to that result will yield the original value. They are essentially two sides of the same coin.

How is the inverse of a CDF used in practice?

The inverse of a CDF is commonly used in statistical modeling and data analysis. It can help us determine critical values for hypothesis testing, calculate probabilities for confidence intervals, and generate random values from a distribution with a specified probability.

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