11. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=0$, and for any $n \in \mathbf{Z}_{+}, a_{n+1}$ takes the value $a_{n}-1$ or $a_{n}+1$ with probability $\frac{1}{2}$ each.
(1) Let the value of $a_{2 n+1}$ be the random variable $X$. Find the probability distribution of $X$;
(2) Find the expected value of the absolute value of $X$, $E|X|$.
11. (1) Let $d_{n}=a_{n+1}-a_{n}$. Then for any positive integer $n$, the probability that $d_{n}$ takes 1 or -1 is $\frac{1}{2}$, and
$$
a_{2 n+1}=a_{1}+\sum_{i=1}^{2 n}\left(a_{i+1}-a_{i}\right)=\sum_{i=1}^{2 n} d_{i} \text {. }
$$

Let $a_{2 n+1}=k$. Clearly, $|k| \leqslant 2 n$, and suppose at this point $d_{1}, d_{2}$, $\cdots, d_{2 n}$ have $x$ 1's and $2 n-x$ -1's. Then
$$
x-(2 n-x)=k \text {. }
$$

Therefore, $k=2(x-n)$ can only take even values in the range $[-2 n, 2 n]$.

For even numbers $2 m(m=0, \pm 1, \cdots, \pm n)$, the event $\{X=2 m\}$ is equivalent to having $n+m$ 1's and $n-m$ -1's in the $2 n$ numbers $d_{1}, d_{2}, \cdots, d_{2 n}$. Thus, the probability distribution of $X$ can be expressed as
$$
P(X=2 m)=\frac{\mathrm{C}_{2 n}^{n+m}}{2^{2 n}}(m=0, \pm 1, \cdots, \pm n) .
$$
(2) For any $1 \leqslant i \leqslant n$, it is easy to see that
$$
P(X=-2 m)=P(X=2 m) \text {. }
$$

Thus, $P(|X|=2 m)=\frac{\mathrm{C}_{2 n}^{n+m}}{2^{2 n-1}}(m=1,2, \cdots, n)$.
Therefore, by the properties of binomial coefficients and the binomial theorem, we have
$$
\begin{array}{l}
E|X|=\sum_{m=1}^{n}\left(2 m \cdot \frac{\mathrm{C}_{2 n}^{n+m}}{2^{2 n-1}}\right)=\frac{1}{2^{2 n-2}} \sum_{m=1}^{n} m \mathrm{C}_{2 n}^{n+m} \\
=\frac{1}{2^{2 n-2}} \sum_{m=1}^{n}\left[(n+m) \mathrm{C}_{2 n}^{n+m}-n \mathrm{C}_{2 n}^{n+m}\right] \\
=\frac{1}{2^{2 n-2}} \sum_{m=1}^{n}\left(2 n \mathrm{C}_{2 n-1}^{n+m-1}-n \mathrm{C}_{2 n}^{n+m}\right) \\
=\frac{1}{2^{2 n-2}}\left(2 n \sum_{m=1}^{n} \mathrm{C}_{2 n-1}^{n+m-1}-n \sum_{m=1}^{n} \mathrm{C}_{2 n}^{n+m}\right) \\
=\frac{1}{2^{2 n-2}}\left[2 n \times \frac{1}{2} \times 2^{2 n-1}-\frac{1}{2}\left(2^{2 n}-\mathrm{C}_{2 n}^{n}\right) n\right] \\
=\frac{n \mathrm{C}_{2 n}^{n}}{2^{2 n-1}} .
\end{array}
$$