Let $(X_n)_{n \in \Bbb N}$ be a sequence of positive i.i.d. random variables

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SUMMARY

The discussion focuses on the convergence in probability of the sequence \( (T_n)_{n \in \mathbb{N}} \), defined as \( T_n := \prod_{i=1}^n X_i^{1/n} \), where \( (X_n)_{n \in \mathbb{N}} \) are positive i.i.d. random variables with a constant finite positive expectation \( E[\ln X_n] = \mu \). It is established that \( T_n \) converges in probability to \( e^{\mu} \) as \( n \) approaches infinity. This result is significant in the context of the law of large numbers and the behavior of products of random variables.

PREREQUISITES
  • Understanding of independent and identically distributed (i.i.d.) random variables
  • Knowledge of the law of large numbers
  • Familiarity with the properties of logarithms and expectations
  • Basic concepts of convergence in probability
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  • Study the law of large numbers in detail
  • Explore the concept of convergence in probability in stochastic processes
  • Learn about the properties of logarithmic transformations in probability theory
  • Investigate applications of products of random variables in statistical modeling
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Mathematicians, statisticians, and students studying probability theory, particularly those interested in the behavior of sequences of random variables and their convergence properties.

Euge
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Here is this week's POTW:

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Let $(X_n)_{n \in \Bbb N}$ be a sequence of positive i.i.d. random variables such that $E[\ln X_n]$ is a constant finite positive number $\mu$. Show that if $$T_n := \prod_{i = 1}^n X_i^{1/n}\quad (n = 1,2,3,...)$$ then $(T_n)_{n\in \Bbb N}$ converges in probability to $e^{\mu}$.
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No one answered this week's problem. You can read my solution below.

Note $$T_n = \exp\left(\frac{1}{n}\sum_{i = 1}^n \ln X_n\right)$$ and by the weak law of large numbers, $$\frac{1}{n}\sum_{i = 1}^n\ln X_i \rightarrow \mu\quad \text{in probability}$$ Since, in addition, the map $x\mapsto e^x$ is continuous, then $T_n \to e^\mu$ in probability.
 

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