Why is an unbiased random walk non-ergodic? Wikipedia says "An unbiased random walk is non-ergodic."
Let's look at a simple random walk. It's defined as: take independent random variables $Z_{1},Z_{2}$, where each variable is either $1$ or $−1,$ with a 50% probability for either value, and set $S_{0}=0\,\!$ and $S_{n}=\sum _{j=1}^{n}Z_{j}$.
If we calculate (let's say) the mean mean for an ensemble of size $N$ it will be $(\sum _{j=1}^{n}Z_{j})/N$ and the mean for a single realisation of length $N$ will be exactly the same $(\sum _{j=1}^{n}Z_{j})/N.$
So, why is it non-ergodic?
 A: That Wikipedia article writes,

The process $X(t)$ is said to be mean-ergodic or mean-square ergodic in the first moment if the time average estimate $${\hat {\mu }}_{X}={\frac {1}{T}}\int _{0}^{T}X(t)\,\mathrm{d}t$$ converges in squared mean to the ensemble average $\mu _{X}$ as $T\rightarrow \infty.$

The problem is that $\hat\mu$ becomes more and more variable as $T$ increases.  This becomes apparent when $X(t)$ is the discrete Binomial random walk described in the question, because the time average is
$$\hat\mu(X) = \frac{1}{T} \sum_{i=1}^T X(t) = \frac{1}{T} \sum_{i=1}^T \sum_{j=1}^i Z(i) = Z(1) + \frac{T-1}{T}Z(2) + \cdots + \frac{1}{T}Z(T).$$
Notice how the early terms persist: $Z(1)$ appears with coefficient $1$ and the coefficients of the subsequent $Z(i)$ converge to $1$ as $T$ grows.  Their contributions to the time average therefore do not get averaged out and consequently the time average cannot converge to a constant.

In the context and notation of the Wikipedia article, let's prove this result  by finding the mean and variance of the time average.
The expectation of $\hat{\mu}_X$ is
$$\mathbb{E}(\hat{\mu}_X) = {\frac {1}{T}}\int _{0}^{T}\mathbb{E}(X(t))\,\mathrm{d}t = \frac{1}{T}\int_0^T 0\, \mathrm{d}t = 0.$$
Therefore its variance is the expectation of its square,
$$\eqalign{
\operatorname{Var}(\hat{\mu}_X) &= \mathbb{E}\left(\hat{\mu}_X^2\right)\\
&= \mathbb{E}\left({\frac {1}{T}}\int _{0}^{T}\mathbb{E}(X(t))\,\mathrm{d}t \ {\frac {1}{T}}\int _{0}^{T}\mathbb{E}(X(s))\,\mathrm{d}s \right) \\
&= \left(\frac {1}{T}\right)^2 \int_0^T \int_0^T \mathbb{E}(X(t)X(s))\,\mathrm{d}t \mathrm{d}s  \\
&= \left(\frac {1}{T}\right)^2 \int_0^T \int_0^T \min(s,t)\,\mathrm{d}t \mathrm{d}s \\
&= \left(\frac {1}{T}\right)^2 \int_0^T \left(\int_0^s t\,\mathrm{d}t + \int_s^T s\,\mathrm{d}t\right)\mathrm{d}s \\
&= \left(\frac {1}{T}\right)^2 \int_0^T \left(\frac{s^2}{2} + (T-s)s\right)\mathrm{d}s \\
&= \left(\frac {1}{T}\right)^2 \frac{T^3}{3} \\
&= \frac{T}{3}.
}$$
Because this grows ever larger as $T$ grows, $\hat\mu_X$ cannot possibly converge to a constant as required by the definition of ergodicity--even though it has a constant average of zero.  Whence Wikipedia writes (to quote the passage fully),

An unbiased random walk is non-ergodic. Its expectation value is zero at all times, whereas its time average is a random variable with divergent variance.

