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As I said, the WLLN is: under independence conditions as stated, for any fixed $\epsilon>0$ $$\lim_{n\rightarrow \infty}P\left(\left|\frac{X_1+\cdots+X_N}{n}-\mu\right|>\epsilon\right)=0.$$ Your forumla says that for any $k$: $$P\left(\left|\frac{X_1+\cdots+X_N}{n}-\mu\right|>k\frac{\sigma}{\sqrt{n}}\right)\leq\frac{1}{k^2}.$$ So let $\epsilon$ be ...


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Don't begin by writing everything in terms of $\pi_0$. The first few terms are complicated, but for $n\geq 3$ we have the regular pattern $p\pi_{n-1}+q\pi_{n+1}=\pi_n$ which can be rewritten $p(\pi_{n-1}-\pi_n)=q(\pi_{n}-\pi_{n+1})$. By induction, $$\pi_n-\pi_{n+1}=\left({p\over q}\right)^{n-2}(\pi_2-\pi_3),$$ and therefore $$\pi_2-\pi_{n}=\sum_{j=2}^{n-1}\...


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Hint: $X_n$ is a Markov chain and the current state is either the same or increases by $1$, up to $k$. That means $\min(2, X_n) = X_n$ as long as $X$ is in state $1$. The only possible transition (except $1 \to 1$) is $1 \to 2$ (why?) What happens with $Y_n$ then? Can you draw it as a markov chain?


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In fact, there is a deeper relation between the Laplacian and Brownian motion. Let $(M, g=\langle\cdot, \cdot\rangle)$ be a smooth Riemannian manifold without boundary. The Laplace-Beltrami operator is defined as the contraction of the covariant derivative of the differential of any smooth function on $M$ $$\forall f \in C^\infty(M): \Delta_M f := \mathrm{...



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