Here's another strategy for finding (x1,...,xn) with f(x1,...,xn) ? a.
(c) Set g?(x) to be the conditional expectation
91(x) = E[f(X1,..., Xn) | X1 = x].
Show that Ex? [91 (X1)] ? a and determine how many evaluations of g? you need in order to
find x1 with g(x1) ? a.
(d) Now let g2(x) be the conditional expectation
92(x) = E[f(X1,..., Xn) | X? = x1, X2 = x],
where x1 is from part (c). Argue that E[g2(X2)] ? a and determine how many evaluations of
g2 you need to find x2 with g(x2) ? a.
(e) Repeat the above process n times. On the i-th step, define the conditional expectation
gi(x) = E[f(X1,..., Xn) | X? = x1,..., Xi-1 = Xi-1, Xi = x],
where x; is such that gj(x) ? a for each j < i. Argue that x1,... In satisfy f(x1,...,xn) ? a.
How many total steps does it take to find x1, x2,..., In? Compare this to your answer in part
(b).
