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add definitions here
\begin{align*} T(a,b;c,d) &= \max_{\pi \in \Pi_{(a,b)\to(c,d)}} \sum_{(i,j)\in \pi} X_{i,j} \quad \textrm{(last passage percolation time from }(a,b)\textrm{ to }(c,d) \textrm{)} \\ \mathbf{V}^n &= (V^n_1,\ldots,V^n_{n-1}) \quad \qquad \qquad \qquad \qquad \qquad \qquad \textrm{(block finishing times)} \\ \mathbf{U}^n &= (T(1,1; n,1), T(1,1; n-1,2), \ldots, T(1,1; 1,n)) \qquad \textrm{(point to line last passage percolation)} \\ \mathbf{W}^n &= (T(1,1; n,1), T(2,1; n,2), \ldots, T(n,1; n,n)) \qquad \textrm{(line to line last passage percolation)} \end{align*}
Theorem. $ \mathbf{U}^n \stackrel{D}{=} \mathbf{W}^n $
Proof. add this
Conjecture. $ \mathbf{V}^n \stackrel{D}{=} \mathbf{W}^n $
Proposition. Let $f_n(x_1,\ldots,x_n)$ denote the joint density function of the random vector $\mathbf{U}^n$. Then we have the recurrence relation $$ f_{n+1}(x_1,\ldots,x_{n+1}) = \exp\left( - \sum_{k=1}^{n+1} x_k \right) \int_0^{x_1 \wedge x_2} \int_0^{x_2 \wedge x_3} \ldots \int_0^{x_n \wedge x_{n+1}} f_n(y_1,\ldots,y_n) \exp\left( \sum_{k=1}^{n+1} y_{k-1} \vee y_k \right) dy_n \, dy_{n-1} \ldots dy_2 \, dy_1 $$
add this
add an explanation of the algorithm for computing the joint density of the finishing times
explain how the joint distribution of a pair $V_n(k), V_n(k+1)$ can be expressed in terms of an array of last passage percolation times to which a jeu de taquin operation has been applied.