Distribution of Log of Random Variable

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SUMMARY

The discussion focuses on determining the distribution of the random variable Y, defined as Y = r ln X, where X follows a normal distribution X ~ N(μ, σ²). The transformation involves changing variables from x to y, leading to the probability density function for Y being expressed as an integral. The final result confirms that the distribution of Y is valid only for X values greater than zero, emphasizing the importance of the Gaussian integrand's properties in the derivation.

PREREQUISITES
  • Understanding of normal distribution, specifically N(μ, σ²).
  • Knowledge of transformation techniques in probability theory.
  • Familiarity with integration and probability density functions.
  • Basic concepts of random variables and their properties.
NEXT STEPS
  • Study the properties of the normal distribution and its applications.
  • Learn about variable transformations in probability and statistics.
  • Explore the derivation of probability density functions from transformations.
  • Investigate the implications of logarithmic transformations on data distributions.
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Statisticians, data scientists, and researchers interested in probability theory and the behavior of transformed random variables.

andrewcheong
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Let X and Y be random variables.

X ~ N(u,s^2)
Y = r ln X, where r is a constant.

What is the distribution of Y?

(This is not a homework problem. It's just related to something I was curious about, and I can't figure out how to solve this, if it is solvable...)
 
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You know that

1 = \int_{-\infty}^{\infty}dx~\frac{1}{\sqrt{2\pi \sigma^2}} \exp\left[\left(\frac{x-\mu}{\sigma}\right)^2\right] = \int_{0}^{\infty}dx~\frac{2}{\sqrt{2\pi \sigma^2}} \exp\left[\left(\frac{x-\mu}{\sigma}\right)^2\right]

So, make a change of variables y = r \ln x. The lower limit x = 0 becomes y = -\infty and the upper limit remains infinity. dy = r dx/x = r dx e^{-y/r}

Hence,

1 = \int_{-\infty}^{\infty}dy~\frac{2e^{y/r}}{r\sqrt{2\pi \sigma^2}} \exp\left[\left(\frac{e^{y/r}-\mu}{\sigma}\right)^2\right]

The integrand is thus the probability density function for y. Note that the distribution is only valid for values of x zero or greater, as y is not defined for x < 0. This is why in the first line I used the evenness of the gaussian integrand to write it in terms of x > 0 only.
 

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