How to Convert Between CDF and PDF Functions in Mathematical Equations?

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Homework Help Overview

The discussion revolves around converting cumulative distribution functions (CDF) to probability density functions (PDF) and vice versa, specifically focusing on two mathematical functions. The original poster seeks assistance with the differentiation and integration processes involved in these conversions.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive the PDF from a given CDF and vice versa, expressing uncertainty about the integration and differentiation processes. Some participants suggest using the chain rule for differentiation and question the treatment of constants in the integration process.

Discussion Status

Participants are exploring different approaches to the problem, with some offering guidance on differentiation techniques and integration of constants. There is no explicit consensus on the correctness of the proposed functions, and multiple interpretations of the integration process are being discussed.

Contextual Notes

The original poster has acknowledged posting the question in the wrong section initially and is navigating the constraints of homework guidelines while seeking help.

scot72001
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cdf to pdf and vise versa
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
 
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You don't need to post the same question in multiple places.
 
sorry i just noticed it said put homework in the homework section after i had posted it in the first location
 
As for differentiating, you need to use the chain rule. Integrating that function should be simple since (1 + a)/2 is a constant with respect to X.
 
so would the following look ok for the first part?
f(x) = {e^-αx^β } {αβ [x^(β-1) } x ≥ 0, α>0, β>0

and for the second part it should look like
f(x) = { (1 + αx)/2 for -1 ≤ x ≤ 1, -1 ≤ α ≤ 1 to start off. apologies for missing the x as it is crucial. do you think i can still use 1+α as a constant
then the answer would look like
((1 +α)x^2)/4
 

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