Random Walk Process in Time Series Is it true that the mean of a random walk process does not depend on time and the sequence can be considered mean stationary?  
 A: Yes, it is mean stationary (if the mean exists at all)
A simple symmetric random walk $\{ X_t | t \in \mathbb{Z} \}$ is a Markov chain on the integers with fixed transition probabilities given by:
$$\mathbb{P}(X_{t} = x+k | X_{t-1} = x) = \begin{cases}
\ \tfrac{1}{2} & & & \text{for } k = -1, \\[6pt]
\ \tfrac{1}{2} & & & \text{for } k = 1, \\[6pt]
\ \ 0 & & & \text{otherwise}. \\[6pt]
\end{cases}$$
This means that the expected value of the increments is:
$$\begin{align}
\mathbb{E}(X_t-X_{t-1}) 
&= \sum_{k \in \mathbb{Z}} k \cdot \mathbb{P}(X_{t} = x+k | X_{t-1} = x) \\[6pt]
&= (-1) \cdot \frac{1}{2} + (1) \cdot \frac{1}{2} = 0. \\[6pt]
\end{align}$$
Every random walk of this kind is fully defined by an "anchoring" distribution on $X_0$ (or any other anchoring point in the process).  Suppose that this distribution has a mean value $\mu \in \mathbb{R}$.  Then at every time point $n$ in the process we have:
$$\begin{align}
\mathbb{E}(X_n) 
&= \mathbb{E} \Bigg( X_0 + \sum_{t=1}^n (X_t-X_{t-1}) \Bigg) \\[6pt]
&= \mathbb{E}(X_0) + \sum_{t=1}^n \mathbb{E}(X_t-X_{t-1}) \\[6pt]
&= \mu + \sum_{t=1}^n 0 = \mu. \\[6pt]
\end{align}$$
This confirms mean stationarity of the process.  (Note that we assume here that the mean of the anchoring distribution exists.)  Mean stationarity is a consequence of the fact that the expected value of the increments of the random walk is zero.
A: TLDR: Yes
Let us consider a random walk in discrete time:
At every timestep $t = n\Delta t$ with $n \in \mathbb{N}$ the process goes one step forward or one step back with equal probability.
Leading: 
$P(\Delta x_i = 1) = 1/2$ 
$P(\Delta x_i = -1) = 1/2$ 
(where $i$ denotes the timesteps)
The process is at $t = n\Delta t$ is $X_t = \sum_{i=0}^{n}{\Delta x_i}$.
Taking the average $\mathbb{E}X_t = \mathbb{E}(\sum_{i=0}^{n}{\Delta x_i}) = \sum_{i=0}^{n}{\mathbb{E}(\Delta x_i)} = \sum_{i=0}^{n}{(\frac{1}{2}(1)+\frac{1}{2}(-1))} = 0$. This is clearly not dependent on time.
