uniform and independent X1,X2,.∈{0，l} a pattern:x∈{0，l}k substring:X[i,j]=XiX:+1Xj stopping time T:first time encountering x as a substring 七 1 x[1,]=x[k-j+1, otherwise k Theorem: E[T]=>x2 j=1 ∀t,i 0 i<t let 2i-t+1-1 t<i<t+k,X t,i]=x[1,i-t 2k-1 i≥t+k,Xt,t+k-1]=x -1 otherwise x=0110 i let Zi= X=101001000110101110.. t=1 ￥⊙：-1-1-1-13.1000000000000
Y (t) i = 8 >>>< >>>: 0 i<t 2it+1 1 t i<t + k,X[t, i] = x[1, i t] 2k 1 i t + k,X[t, t + k 1] = x 1 otherwise a pattern: x ∈ {0,1}k stopping time T: first time encountering x as a substring substring: X[i, j] = XiXi+1 ··· Xj j = ( 1 x[1, j] = x[k j + 1, k] 0 otherwise E[T] = X k j=1 j2 Theorem: j let 8t, i Zi = X 1 t=1 Y (t) i let X = 1 0 1 0 0 1 0 0 0 1 1 0 1 0 1 1 1 0 ... x = 0 1 1 0 i Y (t) i : -1 -1 -1 -1 3 -1 0 0 0 0 0 0 0 0 0 0 0 0 ... uniform and independent X1, X2, ... ∈ {0,1}
uniform and independent X1,X2,.∈{0，l} a pattern:x∈{0，l}k substring:X[i,j]=XiX:+1Xj stopping time T:first time encountering x as a substring 七 1 x[1,]=x[k-j+1,] otherwise k Theorem: ET网=X2 j=1 ∀t,i 0 i<t Y= let 2i-t+1-1 t<i<t+k,X t,i]=x[1,i-t 2k-1 i≥t+k,Xt,t+k-1]=x -1 otherwise x=0110 i let Zi= X=101001000110101110.. t=1 y0:-1-1-11-1-11-17-10000000
a pattern: x ∈ {0,1}k stopping time T: first time encountering x as a substring substring: X[i, j] = XiXi+1 ··· Xj j = ( 1 x[1, j] = x[k j + 1, k] 0 otherwise E[T] = X k j=1 j2 Theorem: j let 8t, i Zi = X 1 t=1 Y (t) i let X = 1 0 1 0 0 1 0 0 0 1 1 0 1 0 1 1 1 0 ... x = 0 1 1 0 i Y (t) i : -1 -1 -1 -1 -1 -1 -1 -1 7 -1 -1 0 0 0 0 0 0 0 ... Y (t) i = 8 >>>< >>>: 0 i<t 2it+1 1 t i<t + k,X[t, i] = x[1, i t] 2k 1 i t + k,X[t, t + k 1] = x 1 otherwise uniform and independent X1, X2, ... ∈ {0,1}