Mean reversion in AR(1) process Mean reverting level in following AR(1) process is $b/(1-a)$. 
$$x(t) = a + bx(t-1).$$ I understand this.
I read that the mean reverting level for AR(1) process given below with finite differencing is $-(b/a)$:
$$x(t)-x(t-1) = a+ bx(t-1).$$ Can someone explain how is this true? Can you please help me with the derivation? 
 A: I have the impression that your question contains typos: the "mean reverting level" is $a/(1-b)$ and $-a/b$ respectively. Think that if what you wrote was correct, then for the first process, if it so happened that the constant term was equal to $a=1$, the process wouldn't have a mean-reverting level. And for the second process, if it so happened that $a=0$, it too would not have a mean-reverting level. Does these consequences appear reasonable? Why would the existence or not of a mean-reverting level depend on whether a stand-alone constant term, a shifter, acquired a specific value or not?
In a deterministic difference-equation context, the "mean-reverting level" is the fixed point of the process.
So for the first process,
$$x(t) = x(t-1) = x \Rightarrow (1-b)x = a \Rightarrow x = a/(1-b)$$
and analogously for the second process.
But since you are calling these processes $AR(1)$ etc, it appears that you work in 
a statistical framework, where the "mean reverting level" is the expected value of a stationary process ($b<1$),the value at which the process revolves, and to which it "tends" if it "runs long enough". And in order to calculate that, you need to express it in terms of the innovation term, typically a white noise (when we run a regression we do not "eliminate" the innovation -we just unavoidably ignore it since it is unknown).  
We have, solving recursively
$$x_t = a + bx_{t-1} +u_t = a + b\Big(a + bx_{t-2} +u_{t-1}\Big) +u_t $$
$$ = a(1+b) + b^2\Big(a + bx_{t-3} +u_{t-2}\Big) +bu_{t-1}+u_t=...$$
$$... = a(1+b+b^2+...) + \sum_{i=0}^{\infty} b^{i}u_{t-i}$$
$$= \frac {a}{1-b} + \sum_{i=0}^{\infty} b^{i}u_{t-i}$$
So 
$$E(x_t) = \frac {a}{1-b}  + E\left(\sum_{i=0}^{\infty} b^{i}u_{t-i}\right) = \frac {a}{1-b}$$
the last equality because the innovations are independent and each has a zero-expected value. 
You can use the same approach for the second process.
A: If you are willing to make some stationarity assumptions, namely $|b|<1$ in the first case and $|1+m|<1$ in the second, here's a quick way:
Let
$$X_t=a+bX_{t-1}+\varepsilon_t,\ \varepsilon_t \overset{iid}{\sim}(0,\sigma²)$$
Then
$$\begin{align}
\mathbb{E}X_t&=a+b\mathbb{E}X_{t-1}\\
&=a+b\mathbb{E}X_t\\
&=\frac{a}{1-b}
\end{align}$$
Now consider the first difference data generating process with the same assumptions on $\varepsilon_t$, viz. $$\begin{align}
X_t-X_{t-1}&=n+mX_{t-1}+\varepsilon_t\\
&\iff \\
X_t&=n+(1+m)X_{t-1}+\varepsilon_t\\
&\implies\\
\mathbb{E}X_t&=\frac{n}{1-(1+m)}=-n/m
\end{align}$$
