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

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The discussion revolves around proving that the sequence \( T_n = \prod_{i=1}^n X_i^{1/n} \) converges in probability to \( e^{\mu} \), given that \( (X_n)_{n \in \mathbb{N}} \) are positive i.i.d. random variables with a finite positive expected logarithm \( E[\ln X_n] = \mu \). The key idea is to apply the law of large numbers, which implies that the average of the logarithms converges to the expected value. Consequently, the logarithm of \( T_n \) converges to \( \mu \), leading to the conclusion that \( T_n \) converges to \( e^{\mu} \). No solutions were provided by other participants, but a solution is offered by the original poster. This highlights the importance of understanding convergence in probability within the context of independent random variables.
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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.