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Consider the transition matrix on $\{0,1\}$ given by

$$ P(x,y)=\frac 12\text{ for all }x,y\in\{0,1\} $$

corresponding to a sequence of i.i.d. fair bits. Let $(Y_{t}){t\geq0}$ be a Markov chain with transition matrix $P$ started with a fair coin toss, and set $X_0=0$ and $X{t+1}=Y_t$ for $t\geq0.$ Both $(X_{t})$ and $(Y_{t})$ are Markov chains with transition matrix $P$, so $\{(X_t,Y_t)\}$ is a coupling. Moreover, the sequence $\{(X_t,Y_t)\}_{t\geq0}$ is itself a Markov chain, but it is not a Markovian coupling.

From page 61, Markov Chains and Mixing Times. 2nd ed. by David A. Levin, Yuval Peres, and Elizabeth L. Wilmer.

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If $Y \ge 0$ and $p > 0$, then

$$ \mathbb E[Y^p]=\int_0^\infty py^{p-1}\mathbb{P}[Y>y]\mathrm{d}y. $$

This equation has been frequently used in the proofs in probability theory.

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Let $X_1, X_2, . . .$ be IID uniform on $(−1, 1)$. Let $Y_i = X^2_i$ and note that $\mathbb E[Y_i ] = 1/3, \text{Var}[Y_i ] \le\mathbb E[Y^2_i ] \le 1$, and $\mathbb E[Y^4_i ] \le 1 < +\infty$. Then

$$ \frac{X_1^2+\cdots+X_n^2}n\to\frac13, $$

both in probability and almost surely. In particular, this implies for $ε > 0$

$$ \mathbb{P}\left[(1-\varepsilon)\sqrt{\frac n3}<\|X^{(n)}\|_2<(1+\varepsilon)\sqrt{\frac n3}\right]\to1, $$

where $X^{(n)} = (X_1, . . . , X_n).$ I.e., most of the cube is close to the boundary of a ball of radius $\sqrt{n/3}.$

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