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A Markov chain with continuous state space has a transition exponential density function: $$p(x_{t},x_{t+1})=\frac{1}{x_{t}}exp(-\frac{x_{t+1}}{x_{t}})$$ i.e. the realized value in period t is the mean value of the next period exponential distribution. By standard procedure, if I want to get the stationary distribution of this Markov chain, I should solve the equation: $$\phi(y)=\int_{0}^{\infty}p(x,y)\phi(x)dx$$ where p(x,y) is the transition density function defined as above and $\phi()$ is the objective stationary distribution. But is there explicit solution for this integration equation?

Furthermore,in the markov chain with continuous state space,what transition density has corresponding explicit stationary distribution in long run?

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As $\phi(x)=1/x$ is a solution, which is not integrable on $(0,\infty)$, you should probably not expect this Markov chain to have a stationary probability, except $\delta_0$ ...

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  • $\begingroup$ I think your point that $\phi(x) = 1/x$ solves the stationarity condition, plus some uniqueness argument should show that there is no normalizable density which is stationary under this transition law. $\endgroup$
    – josh
    Commented Jul 10, 2015 at 13:35
  • $\begingroup$ yes, that's my point and I want to know which transition density will lead to an analytical stationary distribution. $\endgroup$
    – Galor
    Commented Jul 10, 2015 at 14:21
  • $\begingroup$ At the same time, $1/x$ is one possible solution of this equation and other feasible solution may still exist.So "probably expect" is precise.I try to use Fredholm theorem to verify the existence and it seems to be valid.So I feel a little confused. $\endgroup$
    – Galor
    Commented Jul 10, 2015 at 14:32
  • $\begingroup$ Here's something else to consider: the form of the transition kernel is such that if $\phi(x)$ is a solution then so is $\phi_\lambda(x) \equiv \phi(\lambda x)$. In other words, you have a scale invariance in your solutions. Contrast this with, say, the heat kernel of the Laplacian on the real line. $\endgroup$
    – josh
    Commented Jul 10, 2015 at 14:41
  • $\begingroup$ I see.Thank you~I will try other function. $\endgroup$
    – Galor
    Commented Jul 10, 2015 at 14:53

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