Having a little trouble with functions of random variables

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SUMMARY

The discussion focuses on finding the probability density function (PDF) of the random variable Y, defined as Y = 1 - e^(-X) where X follows a uniform distribution X ~ UNIF(0,1). The solution involves calculating the cumulative distribution function (CDF) Fy = Pr(Y < y) and transforming it through various steps leading to the integral F(y) = ∫_0^{-ln(1-y)} dx = -ln(1-y). Participants clarify the evaluation process and address common mistakes encountered during the derivation.

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  • Understanding of uniform distributions, specifically X ~ UNIF(0,1)
  • Knowledge of probability density functions (PDF) and cumulative distribution functions (CDF)
  • Familiarity with exponential functions and their properties
  • Basic calculus, particularly integration techniques
NEXT STEPS
  • Study the derivation of PDFs from CDFs in continuous random variables
  • Learn about transformations of random variables, particularly using the exponential function
  • Explore the properties of the uniform distribution and its applications
  • Practice solving problems involving the integration of functions in probability theory
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Students studying probability theory, statisticians, and anyone involved in mathematical modeling of random variables.

bennyska
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Homework Statement


Let X ~UNIF(0,1), and Y=1-e-x. Find the PDF of Y


Homework Equations





The Attempt at a Solution


So i have Fy=Pr(Y<y)
=Pr(1-e-x<y)
=Pr(-e-x<y-1)
=Pr(e-x>1-y)
=Pr(-x>ln(1-y)
=pr(x<-ln(1-y)
=Fx(-ln(1-y))
(here is where I'm having a breakdown)
=x|0 to -ln(1-y) which doesn't really make sense... (that's my notation for evaluate from 0 to...)
i'm having a similar problem, and i think if i can get this figured out, i'll be able to do that one as well.
 
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OK, so you got

[tex]F(y) = \int_0^{-\log (1-y)} dx = -\log (1-y)[/tex]

Why do you think it doesn't make sense?
 
i was making a few dumb mistakes. got it taken care of. thanks.
 

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