Iterating this argument, we conclude that $\mathbbE[X_n] = \mathbbE[X_1]$ for all $n \geq 1$.
Actually, Williams’ own famous example: ( M_n = \prod_i=1^n (1 + X_i) ) where ( X_i ) are independent with mean 0 but ( \mathbbE[X_i^2] ) small? No — that explodes. The clean one: ( M_n = ) number of female births in branching process? Not quite. david williams probability with martingales solutions best
The “best” solution in his sense is the one that justifies each step with a theorem from earlier in the book, no hand-waving. Iterating this argument, we conclude that $\mathbbE[X_n] =
Ironically, the very act of hunting for the best solutions teaches you something: Williams’ exercises often have multiple valid solution paths. Comparing solutions from Wood, MathStackExchange, and GitHub reveals the creativity hidden in the problems. Iterating this argument