# What would be the value of $P$ in this example? (2)

A Markov Chain $$(X_n)_n$$ has the following transition matrix:
$$P = \begin{bmatrix} 0.1 & 0.3 & 0.6\\ 0 & 0.4 & 0.6\\ 0.3&0.2&0.5 \end{bmatrix}$$ with initial distribution $$\alpha = (0.2, 0.3, 0.5)$$.

Find $$P(X_0 = 3|X_1 = 1)$$ .

My solution:

$$P(X_0 = 3|X_1 = 1) = \frac{P(X_0 = 3, X_1 = 1)}{P(X_1 = 1)}$$
$$\Rightarrow P(X_0 = 3|X_1 = 1) = \frac{P(X_1 = 1, X_0 = 3)}{P(X_1 = 1)}$$
$$\Rightarrow P(X_0 = 3|X_1 = 1) = \frac{P(X_1 = 1| X_0 = 3)\cdot P(X_0=3)}{P(X_1 = 1)}$$

Now,

• from $$P$$, we have $$P(X_1=1|X_0=3) = 0.30$$
• from $$\alpha$$, we have $$P(X_0=3) = 0.50$$, and
• from $$\alpha P$$, we have $$P(X_1=1) = 0.17$$

So,

$$P(X_0 = 3|X_1 = 1) = \frac{0.30 \cdot 0.50}{0.17} \approx 0.88$$.

I have two questions regarding this:

1. although, I wrote, "from $$P$$, we have $$P(X_1=1|X_0=3) = 0.30$$", I have a simple confusion here. When the system is at state $$X_0$$, it has an initial distribution of $$\alpha$$. So, theoretically, shouldn't it change its state according to $$\alpha$$ rather than $$P$$? This is actually practically impossible as $$\alpha$$ is a $$(1 \times 3)$$ matrix and hence there would be no $$\alpha_{(3,1)}$$. But, the question remains.

2. is there any better/easier method than what I used here?

1. No, what you did is correct. Given that the chain is at $$X_0 = 3$$ you are interested in the transition probability to the state $$1$$. $$\alpha(i)$$ defines $$\mathbf{P}(X_0 = i)$$. It does not specify how the chain changes states but rather how it spontaneously starts.
• Is $P(X_1 = 1) =0.17$ or $0.20$? – user366312 Mar 12 at 18:03
• I get $\alpha P = (0.17, 0.28, 0.55)$ so $P(X_1 = 1) = 0.17$. – ippiki-ookami Mar 12 at 18:06