I want $x$ to be Poisson distributed.
I will call occupation probability $p(x=n) =: p(n)$ and the transition probability $p(x=n \rightarrow x=n+1) =: p(n \rightarrow n+1)$
The value of $x$ is changed in a stochastic iterative process. Changes of $\pm1$ are possible. I need the transition probabilities $p(n \rightarrow n+1)$ and $p(n \rightarrow n-1)$ that lead to a poisson distributed $x$ after $m \to \infty$ iterations.
The detailed balance condition $$p(n) \cdot p(n \rightarrow n+1) = p(n+1) \cdot p(n +1 \rightarrow n)$$ leads to $$p(n+1)= p(n) \frac{ p(n \rightarrow n+1)}{ p(n +1 \rightarrow n)}$$
The occupation probability of a Poisson-distributed variable can be written as $$ p(n+1)=p(n) \frac{\lambda}{n+1}$$
Therefore we get $$\frac{ p(n \rightarrow n+1)}{ p(n +1 \rightarrow n)}= \frac{\lambda}{n+1}$$
I thought it'd be good in order to obtain the best efficiency to force a change in every iteration by setting
$$ p(n \rightarrow n+1) + p(n \rightarrow n-1) = 1 $$
So i get the recursion rules:
$$ p(n \rightarrow n+1) = 1- p(n \rightarrow n-1) $$ $$ p(n \rightarrow n-1) = \frac{n}{\lambda} p(n-1 \rightarrow n)$$
This defines the series but for a constant $p(0 \rightarrow 1) =: p_0$
However, i clearly messed something up, because if i calculate the explicit values for the transition probability i obtain something like e.g. $$ p(1 \rightarrow 2) =1- \frac {p_0 }{ \lambda} $$ which is obviously not a probability since $\lambda$ is an arbitrary parameter and if $\lambda \lt p_0$ this leads to negative results...