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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 ...
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 $$\pi_2-\pi_{n}=\sum_{j=2}^{n-1}\... 1 Hint: X_n is a Markov chain and the current state is either the same or increases by 1, up to k. That means \min(2, X_n) = X_n as long as X is in state 1. The only possible transition (except 1 \to 1) is 1 \to 2 (why?) What happens with Y_n then? Can you draw it as a markov chain? 1 In fact, there is a deeper relation between the Laplacian and Brownian motion. Let (M, g=\langle\cdot, \cdot\rangle) be a smooth Riemannian manifold without boundary. The Laplace-Beltrami operator is defined as the contraction of the covariant derivative of the differential of any smooth function on M$$\forall f \in C^\infty(M): \Delta_M f := \mathrm{...