Probability distribution of binary time series

Question

I am unable to understand concepts related to the probability distribution of binary time series. This is from the book Binary time series by Benjamin Kedem, vol 52

Let $X_t$, t =0,1,... be a binary stationary Markov chain of $p^{th}$ order, such that $Pr(X_t = 1) = p, Pr(X_t=0) = q,$

$p+q =1$

We define, $Pr(X_t = x_t|X_{t-1} = x_{t-1},\ldots,X_{t-k} = x_{t-k}) = p_{x_t x_{t-1}\ldots x_{t-k}}$ Consider a stationary AR(1) process $Z_t = \phi Z_{t-1} + u_t$, where $|\phi| < 1$ and $u_t$ are independent $N(0,\sigma^2)$ variates. Then $Z_t$ is a zero mean stationary Gaussian process. Let $X_t$ be the clipped series at level zero. The joint distribution of the binary series of length (tuples) $n$ is given by $Pr(X_1 = x_1,\ldots,X_n = x_n) = \frac{1}{2} \prod_{i=2}^{n} \prod p_{y_iy_{i-1}\ldots y_1}^{I_i I_{i-1}\ldots I_1}$ Eq(1)

such that the second product is over all $2^i$ 0-1 tuples $(y_i,y_{i-1},\ldots,y_1)$ and $I_i = x_i$ if $y_i = 1$ otherwise $I_i = 1- x_i$ if $y_i = 0$

This is the joint distribution of any 0-1 time series $X_1,\ldots,X_n$ provided $Pr(X_1 = x_1) = 1/2$ and the products $I_tI_{t-1}\ldots I_1$ are the sufficient statistics for the chain.

My problem is that I have never seen any notation given in Eq(1) and I am unfamiliar with this kind of distribution where there is something on the power of probability. Can somebody please explain to me what is the meaning of the expression in Eq(1) and the notations $I$, power of probability? I remember reading about binomial distribution where the probability is raised to a power. But, this distribution and the meaning is complex to understand. Thank you for help.

Answer 1

It looks like Eq(1) is just a (very confusing) way of writing the chain rule of probability: $$ Pr(X_1 = x_1, ..., X_n = x_n) = Pr(X_1 = x_1) Pr(X_2 = x_2 | X_1 = x_1) Pr(X_3 = x_3 | X_1 = x_1, X_2 = x_2) \cdots \\ = \frac{1}{2} \prod_{i=2}^n Pr(X_i = x_i | X_{i-1} = x_{i-1}, ..., X_1 = x_1) $$ This is because $$ \prod_{y_i, ..., y_1} p_{y_i \cdots y_1}^{I_i \cdots I_1} = p_{x_i \cdots x_1} $$ according to the definition of $I_i$. Consider the case when all $y$'s are one. Then each $I_i = x_i$, so $$ p_{y_i \cdots y_1}^{I_i \cdots I_1} = p_{1 \cdots 1}^{x_i \cdots x_1} $$ If any $x_j$ in the exponent is zero, the whole exponent is zero. Since any probability raised to the zeroth power equals 1, this factor would contribute nothing to the product. It only contributes when all $x_i, \cdots, x_1$ equal 1. Applying the same argument for any pattern of $y$ values shows that the only factor that contributes to the product is the one where $y_i = x_i, ..., y_1 = x_1$. This is how you get the equation above.