Denote by p(i,j) the TP from (i,j) to (i+1,j), and let's set q(i,j) = 1-p(i,j) as the TP from (i,j) to (i,j+1). I seem to have showed that, given all the target probabilities T(i,j) > 0, then all the transition probabilities (TP's) are uniquely determined (UD). This holds with the conditions that I) T(i,j) > 0 for all i,j >= 0. II) For all k, T(0,k) + T(1,k-1) + . . . + T(k,0) = 1 (implicit in the statement of the problem). Proof: For k >= 0, let L_k denote all points of form (j,k) or (k,j), for j >= k. All points in the first quadrant are the disjoint union of all the (L-shaped) sets L_k. Then: * Clearly all the TP's on L_0 are UD. Hence the TP's along each edge that's emanating from and perpendicular to L_0 are also UD. * Inductively assume all the TP's that are below and/or to the left of L_k have been determined. * Then all the TP's that lie along L_k are UD. Then, so are the TP's emanating from and perpendicular to L_k are also UD. Q.E.D. by induction. ((( Example: Suppose all TP's along L_0 have been UD, and (by subtracting from 1) also the emanating perpendicular TP's as well. Now to find, say, the TP p(1,1) (from (1,1) to (1,2)) we just note that (since there are only 2 points from which a random walk can go to (1,2) in 1 step) we have: T(1,2) = T(0,2)*q(0,2) + T(1,1)*p(1,1). And so p(1,1) = (T(1,2) - T(0,2)*q(0,2)) / T(1,1) as long as T(1,1) > 0 (condition I). This method will lead to all TP's along L_1 being UD. Now suppose we want to find a TP emanating from and perpendicular to L_1, like q(1,3). We already know the *other* TP emanating from (1,3), namely p(1,3), and so q(1,3) is UD via q(1,3) = 1 - p(1,3). ))) The TP's are clearly *not* UD if the Q(i,j) have enough 0's among them. But I think in that case all the TP's going to point (i,j) with Q(i,j) > 0 *are* UD, so the TP's that are non-UD would seem to be irrelevant. --Dan ________________________________________________________________________________________ It goes without saying that .