Toss a coin $X_0 = 5$ times. Let $X_1 = \text{No. of heads}$. Toss the coin $X_1$ times. Let $X_2 = \text{No. of heads}$, and so on. Find $P(X_2 = 2 \mid X_4=1)$. (Using Markov chain)
If I define the states as the number of heads in each toss, I can define the transition matrix as follows from 0 heads to 5 heads:
\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0\\ \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 & 0\\ \frac{1}{4} & \frac{1}{2} & \frac{1}{4} & 0 & 0 & 0\\ \frac{1}{8} & \frac{3}{8} & \frac{3}{8} & \frac{1}{8} & 0 & 0\\ \frac{1}{16} & \frac{4}{16} & \frac{6}{16} & \frac{4}{16} & \frac{1}{16} & 0\\ \frac{1}{32} & \frac{5}{32} & \frac{10}{32} & \frac{10}{32} & \frac{5}{32} & \frac{1}{32} \end{bmatrix}
Is the question asking for $P_{21}^{(2)}$ which will be $\frac{3}{8}$, or it's asking for $f_{21}^{(2)}$? Also, This way we're not considering the probability of number of heads in our $X_0$ which we should? right? Also, here, given the future state = 1, it's asking for the probability of past state to be 2. Is it different from normal Markov chains?
Any idea?
Thanks.
