Expected value of a random variable at a stopping time.

Question

Let $x_1,x_2 \dots$ be adapted to the filtration $\mathcal{F}_1, \mathcal{F}_2, \dots$.
Let $\tau$ be a stopping time that is also adapted to the filtration.
Say that
$\mathbb{E}[x_i \mid \mathcal{F}_{i-1}] = 0$.
Is it true that
$$\mathbb{E}[x_{\tau}] = 0?$$
One idea I had was to write $x_{n} = x_0 + (x_1 – x_0) + \dots + (x_{n}- x_{n-1}) = x_0 + \sum\limits_{i=1}^{n}z_i$ where $z_i = x_{i} – x_{i-1}$.
I thought perhaps the partial sums $S_n = \sum\limits_{i=1}^{n}z_i$ could be a martingale, and I could use the optional sampling theorem, but it doesn't seem to be the case.

Answer 1

Now I see this couldn't be the case: Suppose each $x_i$ is random variable that is $-1$ with probability $1/2$ and $+1$ with probability $1/2$. Then define $\tau = \inf\{ n \geq 1 \mid x_n =1\}$. Let $\mathcal{F}_n = \sigma(x_1,\dots, x_n)$. Then the assumptions hold but $\mathbb{E}[x_{\tau}] = 1$.