# Tag Info

5

At the moment $T_{k-1}$ we are on a boundary of non-visited and visited points. Now what Lawler$^*$ says is that there are two possibilities (both having probability $1/2$): We step into the non-visited area. Then we have $T_{k} - T_{k-1} = 1$. We step back into the visited area. Then $$T_{k} - T_{k-1} = 1+ \text{time to reach the boundary again} \\+ ... 2 The notation is easier if we have n+1 cat: 0,1,2,\dots,n. Assume that p\neq 1/2 (the symmetric case is considered here; thanks to @MarcusM for giving a link) and denote q=1-p. Theorem The probability P_k that the ball finishes at cat k is proportional to (\frac{q}{p})^{n-k}, i.e. it is equal to$$ P_k = \frac{p^{k} q^{n-k}}{\sum_{k=1}^n ...

2

Elaborating on the comments, here is an answer to your question. Basically all the confusion lies in the difference between expected value and expected distance- at least it did for me. For a distribution centred at the origin, this difference is huge, and for large $N$ tends towards: $$\sqrt{\dfrac{2\sigma^2 N}{\pi}}$$ This is clearly a lot bigger than ...

1

Eigenvectors of circulant matrices are Fourier modes. $\mathbf K_1$ isn't quite a circulant matrix, but you can arrange the phase such that the missing entries don't contribute. The eigenvalue formula then follows by \begin{eqnarray} &&2\sin ax-\sin a(x+1)-\sin a(x-1)\\&=&2\sin ax-\sin ax\cos a-\cos ax\sin a-\sin ax\cos a+\cos ax\sin ...

1

In the second line, there's no conceptual leap, just two typos – if you correct the $P$ at the beginning to a $p$ and add a closing parenthesis at the end, this line becomes a mere arithmetic rearrangement of the previous one. In the first line, the idea is that if player $A$ has $i$ units and heads is flipped, then player $A$ has $i+1$ units, so the ...

1

In case @joriki's "mere arithmetic rearrangement" is still unclear, \begin{align*} \mathbf{P}[E_{i+1}]\cdot p-\mathbf{P}[E_i]&=-(1-p)\cdot\mathbf{P}[E_{i=1}]\\ \mathbf{P}[E_{i+1}]\cdot p+(p-1-p)\cdot\mathbf{P}[E_i]&=(p-1)\cdot\mathbf{P}[E_{i=1}] \end{align*} Basically we've just added and subtracted a $p$ from the coefficient of $\mathbf{P}[E_i]$ on ...

1

In view of symmetry, $S_{\tau_n}$ has a very simple distribution: it is $n$ or $-n$ with probabilities $1/2$, so $$E[e^{S_{\tau_n}}] = \frac{e^{nt} + e^{-nt}}2 = \cosh nt.$$ If you want to find the pgf of $\tau_n$, denote by $E_m$ the expectation under the condition that $S_0 = m$. Then, clearly, f_m(z):= E_m[z^{\tau_n}] = \frac ...

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