Law of large numbers for inverse of sample mean

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SUMMARY

The discussion centers on the convergence of the inverse of the sample mean, specifically whether \(\frac{1}{\overline{X}_n}\) converges to \(\frac{1}{\mu}\) when \(\overline{X}_n\) converges to \(\mu\). It is established that \(E\left(\frac{1}{\overline{X}_n}\right) \neq \frac{1}{\mu}\) unless \(E\left(\frac{1}{X_1 + \cdots + X_n}\right) = \frac{1}{n \mu}\). The continuity of the function \(x \to \frac{1}{x}\) is noted, emphasizing the need for careful application of expectations in this context.

PREREQUISITES
  • Understanding of the Weak Law of Large Numbers
  • Familiarity with expectations in probability theory
  • Knowledge of continuous functions and their properties
  • Basic concepts of convergence in probability
NEXT STEPS
  • Study the implications of the Weak Law of Large Numbers on sample means
  • Explore properties of expectations for functions of random variables
  • Investigate the continuity of functions and its effects on convergence
  • Learn about the behavior of inverse functions in probability contexts
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Students of probability theory, statisticians, and mathematicians interested in the properties of sample means and their inverses.

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Homework Statement


If [tex]\overline{X}_n[/tex] converges to [tex]\mu[/tex], does [tex]\frac{1}{\overline{X}_n}[/tex] converge to [tex]\frac{1}{\mu}[/tex]?

Homework Equations


http://mathworld.wolfram.com/WeakLawofLargeNumbers.html

The Attempt at a Solution


[tex]\frac{1}{\overline{X}_n} = \frac{n}{X_1 + \cdots + X_n}[/tex]

[tex]E\left(\frac{1}{\overline{X}_n}\right) = E\left(\frac{n}{X_1 + \cdots + X_n}\right)[/tex]

[tex]E\left(\frac{1}{\overline{X}_n}\right) = n E\left(\frac{1}{X_1 + \cdots + X_n}\right)[/tex]

The RHS does not equal [tex]1/{\mu}[/tex] unless [tex]E\left(\frac{1}{X_1 + \cdots + X_n}\right) = \frac{1}{n \mu}[/tex] but how can I show that?

I know [tex]E\left( \frac{1}{\overline{X}_n} \right) \neq \frac{1}{\mu}[/tex] but I'm not sure how to apply that here.
 
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Note that as long as [tex]x \ne 0[/tex] the function [tex]x \to \frac 1 x[/tex] is continuous.
 

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