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%I #32 Sep 14 2022 10:23:53
%S 1,1,2,3,6,12,22,44,88,169,338,646,1292,2584,5055,10110,20220,39915,
%T 79830,157008,314016,628032,1244631,2489262,4978524,9899008,19798016,
%U 39596032,78879609,157759218,313777086,627554172,1255108344,2502016784,5004033568,10008067136,19971766007
%N Number of monotonic lattice paths of length n which do not pass above the line y = x/(log_2(3)-1).
%C A monotonic lattice path is one which starts at (0,0) consists entirely of steps of length 1, moving only rightwards (x,y) -> (x+1,y) or upwards (x,y) -> (x,y+1).
%C a(n+1) <= 2*a(n).
%C Consider a stock with an initial price of x. At every time step it either gains or loses 50% of its value with equal probability (x -> 0.5*x or x-> 1.5*x). An investor buys this stock and sells it if its current price exceeds its initial price and holds onto it otherwise. What is the probability the stock is not sold after n time steps?
%C The answer to this is a(n) / 2^n. Limit_{n->oo} a(n) / 2^n = c ~ 0.2863153965300.
%C Notice that the expected value of the stock remains constant after each time step ((0.5*x + 1.5*x)/2 = x), but the expected log of the stock price is constantly decreasing, at a rate of (1/2) * log(3/4), each time step.
%C Using the central limit theorem, there is a 100% likelihood that the stock price falls below any arbitrarily small positive value, p > 0 in the long run.
%C Since there is a probability 1-c that the investment yields a profit, a probability c that the stock is never sold, and the stock maintains a constant expected value, the expected profit if sold is cx/(1-c) ~ 0.401179169535*x.
%H Alois P. Heinz, <a href="/A355850/b355850.txt">Table of n, a(n) for n = 0..3323</a>
%e n, a(n), list of paths
%e n = 0: 1 {(0,0)}
%e n = 1: 1 {(0,0) -> (1,0)}
%e n = 2: 2 {(0,0) -> (1,0) -> (2,0), (0,0) -> (1,0) -> (1,1)}
%e n = 3: 3 {(0,0) -> (1,0) -> (2,0) -> (3,0), (0,0) -> (1,0) -> (2,0) -> (2,1) (0,0) -> (1,0) -> (1,1) -> (2,1)}
%Y Cf. A001405, A020857, A126042.
%K nonn
%O 0,3
%A _C. Krishna_, Jul 18 2022