Markov chain (DTMC) is described by transition matrix: $$\textbf{P} = \begin{pmatrix}0 & 1\\ \frac{1}{2} & \frac{1}{2} \end{pmatrix}.$$ Initial distribution $X_1 = \left(\frac{1}{4}, \frac{3}{4}\right)$. Calculate conditional entropy $H_2\!\left(X_2|X_1\right)$.
I know formula $H_b\!\left(X|Y\right) = -\sum_{i,j} p\left(x_i, y_j\right)\cdot\log_b\!\left(p\left(x_i|y_j\right)\right) = -\sum_{i,j} p\left(x_i, y_j\right)\cdot\log_b\!\left(\frac{p\left(x_i, y_j\right)}{p\left(y_j\right)}\right)$.
But I'm confused... How and where to use $X_1$? Thanks for step-by-step solution.
We have - being more explicit with the $p(x_i, y_j)$ terms, which are shorthand for $\def\Pr{\mathbb P}\Pr(X=x_i, Y=y_j)$, that $$ \def\p#1#2{\Pr(X_2 = #1, X_1 = #2) = \Pr(X_2 = #1 \mid X_1 = #2) \Pr(X_1 = #2)}\p ab $$ The first term can be read off from the matrix, the second one from $X_1$'s distribution. Let's start - we call the two states $0$ and $1$ \begin{align*} p(0,0) = \p 00 &= 0 \cdot \frac 14 = 0\\ \end{align*} Now calculate the other $p$'s and use them in $$ H_2(X_2 \mid X_1) = \sum_{i,j=0}^1 p(i,j)\log_2 \Pr(X_2 = i \mid X_1 = j) $$