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If I'm not missing something: conditioned on $W_u$, the random variable $W_s - W_u$ is independent with distribution $ \mathcal{N}(0, s - u)$. Therefore you can compute the expectation by rewriting $2W_s = 2(W_s - W_u) + 2W_u$: $$ \mathbf{E}[2 W_s + W_u | W_u=2] = \mathbf{E}[2(W_s - W_u) + 3W_u |W_u=2] = \mathbf{E}[2(W_s - W_u) | W_u] + 6 =6. $$ Edit: if $...


3

Assume WLOG $A=0$, $B=1$, so that we are inside the interval $[0,1]$, then $$x_{n+1}=\begin{cases} \frac{x_{n}}{2} & p=\frac12\\ \frac12 +\frac{x_{n}}{2} & p=\frac12\\ \end{cases}$$ with $x_0=1/2$. Consider the fractional part of the binary representation of $x_n$. The above transition rule corresponds to shifting the fractional part one place to ...


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In a Markov chain with limiting distribution $\vec{\pi}$, it takes $\frac1{\pi_i}$ expected time to return to state $i$ starting from that state. So it's enough to compute the limiting distribution of this Markov chain. Rather than have reflective barriers at $-\frac n2$ and $\frac n2$, it is equivalent to make the Markov chain periodic modulo $n$, so that ...


1

One approach - I don't know if it's the intended one - is to replace the simple random walk by another irreducible and recurrent Markov chain. For this problem, let's take the state space to be $\{0,1,2,\dots\}$ where states $\{1,2,\dots\}$ act like the simple random walk on $\mathbb Z$, but state $0$ transitions to state $k$ with probability $1$. We can ...


1

About the inequality, $||\mu_n-\mu_n^\prime||_{TV}\leq 2\mathbb{\hat{P}}(T>n)$, it's a bit more involved since by the given definition $T$ is not the coupling time of $S$ and $S^\prime$ (coupling time $\tau=\min\{n\geq 0:S_m=S_m^\prime\text{ for all } m\geq n\}$) . As you have correctly said $||\mu_n-\mu_n^\prime||_{TV}\leq 2\mathbb{\hat{P}}(S_n\neq S_n^\...


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We don't need to do anything as exotic as Ising models here. Just let $G$ consist of two complete graphs $K_t$ with some bridges between them: In this example with $t=5$, $H$ is the graph on vertex set $\{a,b,c,d,e,f\}$ we get by forgetting the indices on the vertices of $G$; we do the same thing for general $t$. Then $h(G) = \frac2{t+1}$ is achieved by ...


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