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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 Jul 10 '15 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 Jul 10 '15 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 Jul 10 '15 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 Jul 10 '15 at 14:41
  • $\begingroup$ I see.Thank you~I will try other function. $\endgroup$ – Galor Jul 10 '15 at 14:53

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