# What would be the value of $X_1$ in case of a Limiting Distribution?

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, $$X_1 = \lambda$$.

• 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? – астон вілла олоф мэллбэрг Mar 21 at 4:26