How do I convert between cdf and pdf?

[scot72001]

hi
i'm looking for help when going from a cdf function:
F(x) = { 1- e^-αx^β x ≥ 0, α>0, β>0
{ 0 x < 0

to getting the corresponding pdf

also i am looking to do the opposite(pdf to cdf)
for:
f(x) = { (1 + α)/2 for -1 ≤ x ≤ 1, -1 ≤ α ≤ 1
{ 0 otherwise

i'm unsure as to how to integrate and differentiate these parts.
can you help please

thanks
michael

 

[HallsofIvy]

hi
i'm looking for help when going from a cdf function:
F(x) = { 1- e^-αx^β x ≥ 0, α>0, β>0
{ 0 x < 0

Is that $1- e^{-\alpha x^\beta}$ rather than $1- e^{-\alpha}x^\beta$? If so, let $u= \alpha x^\beta$ and use the chain rule: df/dx= (df/du)(du/dx).

to getting the corresponding pdf

also i am looking to do the opposite(pdf to cdf)
for:
f(x) = { (1 + α)/2 for -1 ≤ x ≤ 1, -1 ≤ α ≤ 1
{ 0 otherwise

Then integrate: $F(x)= \int_{1}^x (1+\alpha)/2 dt$ for $-1\le x\le 1$. That should be easy. Of course, F(1) must be 1. That will require that $\alpha$ have a specific value. In fact, since f(x) is a constant, this is a uniform probability and you should be able to do it without integrating.

i'm unsure as to how to integrate and differentiate these parts.
can you help please

thanks
michael