raneys: finish proof on unif rand perms.

I realize this is a small proof to a relatively easy fact. I may
consider moving it out into prose.

raneys_lemmas/raneys_lemmas.html

@@ -178,7 +178,10 @@ <h1 id="random-sequences"><span class="header-section-number">3</span> Random se
       <p>From here on, the term <span class="math">\(n-\)</span><em>permutation</em> refers to a bijective map <span class="math">\(\pi:\{1,\ldots,n\}\to\{1,\ldots,n\}\)</span>. A <em>swap of <span class="math">\(i\)</span> and <span class="math">\(j\)</span></em> is a permutation <span class="math">\(\rho\)</span> such that the distinct elements <span class="math">\(i, j\)</span> in its range have <span class="math">\(\rho(i) = j\)</span>, <span class="math">\(\rho(j) = i\)</span>, and for which <span class="math">\(\rho (k) = k\)</span> when <span class="math">\(k \ne i, j\)</span>.</p>
       <p><strong>Remark</strong>  <em>Suppose we have a probability space over the set of <span class="math">\(n-\)</span>permutations, and that, for any measurable set <span class="math">\(S\)</span> and swap <span class="math">\(\rho\)</span>, <span class="math">\(\text{Prob}(S) = \text{Prob}(\{\rho(s)|s\in S\})\)</span>. Intuitively, this means we can apply any swap to a set <span class="math">\(S\)</span>, and the result has the same probability as the original.</em></p>
       <p><strong>Proof</strong>  Let <span class="math">\(\rho_{ij}\)</span> denote the swap of <span class="math">\(i\)</span> and <span class="math">\(j\)</span>. Then the set of swaps <span class="math">\(\{\rho_{ij} | 1\le i &lt; j \le n\}\)</span>, when closed under composition, generates the set of all <span class="math">\(n-\)</span>permutations. One way to see this is by considering the Steinhaus-Johnson-Trotter algorithm, which enumerates all <span class="math">\(n-\)</span>permutations using a single swap between any two successive elements <span class="citation">(Wikipedia 2015)</span>.</p>
-      <p>TODO <span class="math">\(\Box\)</span></p>
+      <p>Since any pair of <span class="math">\(n-\)</span>permutations <span class="math">\(\pi_1\)</span> and <span class="math">\(\pi_2\)</span> are a finite pair of swaps apart, they must have the same probability. More formally, there must exist a finite sequence of swaps <span class="math">\(\rho_1, \ldots, \rho_k\)</span> so that <span class="math">\(\pi_1 = \rho_1 \circ \ldots \circ \rho_k \circ \pi_2\)</span>. Then <span class="math">\[\text{Prob}(\pi_2) = \text{Prob}(\rho_k \circ \pi_2)
+                           = \ldots
+                           = \text{Prob}(\rho_1 \circ \ldots \circ \rho_k \circ \pi_2)
+                           = \text{Prob}(\pi_1).\]</span> <span class="math">\(\Box\)</span></p>
       <hr />
       <p>(TODO stuff below here will be heavily edited or rewritten altogether.)</p>
       <p>In this section, we'll consider a finite sequence <span class="math">\(x\)</span> of length <span class="math">\(n\)</span> whose elements <span class="math">\(x_i\)</span> are independent random variables, each uniformly distributed in the interval <span class="math">\([-1, 1]\)</span>.</p>

raneys_lemmas/raneys_lemmas.md

@@ -449,8 +449,16 @@ Steinhaus-Johnson-Trotter algorithm, which enumerates all
 $n-$permutations using a single swap between any two successive
 elements [@s-j-t-alg].
 
-
-TODO
+Since any pair of $n-$permutations $\pi_1$ and $\pi_2$ are 
+a finite pair of swaps apart, they must have the same probability.
+More formally, there must exist a finite sequence of swaps
+$\rho_1, \ldots, \rho_k$ so that
+$\pi_1 = \rho_1 \circ \ldots \circ \rho_k \circ \pi_2$.
+Then
+$$\text{Prob}(\pi_2) = \text{Prob}(\rho_k \circ \pi_2)
+                     = \ldots
+                     = \text{Prob}(\rho_1 \circ \ldots \circ \rho_k \circ \pi_2)
+                     = \text{Prob}(\pi_1).$$
 $\Box$