For the two-state chain with

$$P = \begin{bmatrix} 1-p & p\\ q & 1-q \end{bmatrix}$$

the limiting distribution can be shown to be

$$\lambda = (\frac{q}{p+q}, \frac{p}{p+q})$$

Find the distribution of $X_1$.

My solution:

Since, for every Limiting Distribution $\lambda P =\lambda$, so, $X_1 = \lambda$.

Is that a correct answer?

If not, why?

  • 1
    $\begingroup$ Doesn't the distribution of $X_1$ depend on the distribution of $X_0$, which has to be specified for the problem to be solved? Also what would the limiting distribution have to do with this? $\endgroup$ – астон вілла олоф мэллбэрг Mar 21 at 4:26

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