Suppose we have the following stochastic process:
$$x_t=(1+\gamma)x_t, \ \ \ with \ \ 1/2 \ \ probability$$ $$x_t=(1-\gamma)x_t, \ \ \ with \ \ 1/2 \ \ probability$$
where $\gamma$ is some constant from $(0,1)$. In other words, the above stochastic process is a multiplicative random walk.
From the definition, we have:
$$\Delta x_t=\gamma x_t, \ \ \ with \ \ 1/2 \ \ probability$$ $$\Delta x_t=-\gamma x_t, \ \ \ with \ \ 1/2 \ \ probability$$
Therefore, the expected change of $x_t$ process for $[t, t+1]$ time interval is given by:
$$E [\Delta x_t]=1/2 \times [+\gamma x_t- \gamma x_t]=0.$$
Now, let's transform the multiplicative random walk into an additive random walk by taking $log$ of both sides:
$$log(x_t)=log(1+\gamma)+log(x_t), \ \ \ with \ \ 1/2 \ \ probability$$ $$log(x_t)=log(1-\gamma)+log(x_t), \ \ \ with \ \ 1/2 \ \ probability$$
From here the expected change in $log(x_t)$ is given by
$$E[\Delta log (x_t)]=1/2 \times [log(1+\gamma)+log(1-\gamma)]= 1/2 \times log (1-\gamma^2) \neq0$$
Question: How to explain inconsistence between the results, i.e. $E \Delta x_t=0$, but $E[\Delta log (x_t)] \neq 0$?