This is Exercise 14.4.1 in Rosenthal's A First Look at Rigorous Probability Theory, Second Edition, page 173.
Let $\{Z_i\}$ be i.i.d. with $P(Z_i = 1) = P(Z_i = -1) = 1/2$.
Let $X_0 = 0, X_1 = Z_i$, and for $n > 2$, $X_n = X_{n-1} + (1 + Z_1 +
\cdots
+ Z_{n-1})(2Z_n -
1)$. (Intuitively, this corresponds to wagering, at each time
n,
one
dollar more than the number of previous victories.)
Prove that $X_n$ is a martingale.
Here's my attempt, very standard:
Letting $F_n$ be the $\sigma$-algebra generated by $X_0,...,X_n$
$$\Bbb E (X_{n+1} \mid F_n) = \Bbb E (X_n \mid F_n) + \Bbb E (2Z_{n+1}-1 \mid F_n) + \sum_{i=1}^n \Bbb E ((2Z_{n+1}-1)Z_i\mid F_n)$$ by linearity of the conditional expectation. Now the first summand is $X_n$, since $X_n$ is $F_n$-measurable, the second is $-1$, since $Z_{n+1}$ is independent from $F_n$, and in the third I can factor out the $Z_i$ in each expectation since they're also $F_n$-measurable. This leaves us with
$$\Bbb E (X_{n+1} \mid F_n) = X_n - 1 + \Bbb E (2Z_{n+1}-1 \mid F_n) \sum_{i=1}^n Z_i$$
once again $\Bbb E (2Z_{n+1}-1 \mid F_n) = \Bbb E (2Z_{n+1}-1) = -1$, and we're left with
$$\Bbb E (X_{n+1}\mid F_n) = X_n - 1 - \sum_{i=1}^n Z_i$$
which is far from the desired result of $X_n$. Did I make some mistake or is there an error in the text?