How Do I Calculate the Variance of a Transformed Random Variable?

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SUMMARY

The discussion focuses on calculating the variance of a transformed random variable defined as \(\frac{1}{\log{X} + 2}\), where \(X\) is a random variable. Participants explore the relationship between the variance of \(X\) and the transformed variable, questioning whether it can be expressed as \(\frac{1}{\log{var(X)}}\). Additionally, the maximum likelihood estimator for a Pareto distribution is introduced, with the estimator \(\hat{\theta} = \frac{n}{\sum{ln(X_i)}_{i=1}^n - n \ln(k)}\) and its expected value \(E(\hat{\theta}) = \frac{\theta}{\theta-1}\) discussed.

PREREQUISITES
  • Understanding of random variables and their transformations
  • Familiarity with variance and expectation in probability theory
  • Knowledge of maximum likelihood estimation (MLE)
  • Basic concepts of Pareto distribution
NEXT STEPS
  • Study the properties of transformed random variables in probability theory
  • Learn about variance calculation techniques for non-linear transformations
  • Explore maximum likelihood estimation methods for different probability distributions
  • Investigate the characteristics and applications of the Pareto distribution
USEFUL FOR

Students in statistics, data scientists, and anyone involved in probability theory and statistical modeling, particularly those working with random variables and maximum likelihood estimators.

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Homework Statement



How do I calculate the variance of
[tex]\frac{1}{\log{X} + 2}[/tex]

where X is a random variable?


The Attempt at a Solution



Is it:
[tex]\frac{1}{\log{var(X)}}[/tex]




Homework Equations





The Attempt at a Solution

 
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What is the expectation value of the probability distribution?
 
The forst post was just something I made up. The estimator I am trying to calculate the variance of:
[tex]\hat{\theta} = \frac{n}{\sum{ln(X_i)}_{i=1}^n - n \ln(k)}[/tex]

Which is the maximum likelihood to a pareto distribution
When I calculated the expected value I got:

[tex]E(\hat{\theta}) = \frac{\theta}{\theta-1}[/tex]
[tex]E(X) = \frac{\theta k}{\theta - 1}[/tex]
 
Last edited:

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