E x1 m where the last conclusion follows from the

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: = ab ba ba ba To give a rigorous proof now, we use Theorems 5.13 and 5.11 to conclude 2 0 = E0 (S⌧ ^n ⌧ ^ n) = a2 P (S⌧ = a, ⌧ n) + b2 P (S⌧ = b, T n) 2 + E (Sn ; ⌧ > n) E0 (⌧ ^ n) 2 P (⌧ < 1) = 1 and on {⌧ > n} we have S⌧ ^n max{a2 , b2 } so the third term tends to 0. To handle the fourth term we note that by (1.6) E0 (⌧ ^ n) = n X P (⌧ m=0 Putting it all together, we have m) " 1 X P (⌧ m) = E 0 ⌧ . (5.13) m=0 0 = a2 P0 (S⌧ = a) + b2 P0 (S⌧ = b) E0 ⌧ and we have proved the result. Consider now a random walk Sn = S0 + X1 + · · · + Xn where X1 , X2 , . . . are i.i.d. with mean µ. From Example 5.2, Mn = Sn nµ is a martingale with respect to Xn . 170 CHAPTER 5. MARTINGALES Theorem 5.15. Wald’s equation. If T is a stopping time with ET < 1, then E (ST S0 ) = µET Recalling Example 5.9, which has µ = 0 and S0 = 1, but ST = 1 shows that for symmetric simple random walk E1 V0 = 1. Why is this true? Theorems 5.13 and 5.11 give ES0 = E (ST ^n ) µE (T ^ n) As n " 1, E0 (T ^ n) " E0 T by (5.13). To pass to the limit in the other term, we note that ! T X E |ST ST ^n | E |Xm |; T >...
View Full Document

Ask a homework question - tutors are online