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It is known that if $(X_n)_{n \in \mathbb{N}} \subset L^2$ are uncorrelated and identically distributed random variables, then $$\frac{1}{n} \sum_{k=1}^n X_k \to \mathbb{E}[X_1]\text{ almost surely.}$$ Is it true to claim that $$ \frac{1}{n}\sum_{k=1}^n\tanh (X_k)\to \mathbb{E}\left[\tanh (X_1)\right]\text{ almost surely?} $$

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Try to construct examples of a sequence $X_n$ in which the $X_n$ are uncorrelated but not independent. There's a good chance you'll find a sequence that violates the stated claim.

For example, let $C$ (stands for "coin") be a random variable with $\mathbb{P}(C=H) = \mathbb{P}(C=T) = \frac{1}{2}$. Let $X_n$ be a sequence such that \begin{eqnarray} \mathbb{P}(X_n=1 | C=H) &=& 1 \\ \mathbb{P}(X_n=0 | C=T) &=& \mathbb{P}(X_n=2 | C=T) = \frac{1}{2} \end{eqnarray} and such that the $X_n$ are mutually independent conditioned on $C=T$. Note that the $X_n$ are indeed uncorrelated. But if you compute the almost sure limit of $\frac{1}{n}\sum_{k=1}^n \tanh(X_k)$, the values will differ depending on the value of $C$, so the claim is false.

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