Problem Statement
A streaming platform models student study habits where students alternate between focused study state ($S$) and distraction state ($D$). The monthly transition matrix is given by:
$$T = \begin{pmatrix} 0.85 & 0.10 \\ 0.15 & 0.90 \end{pmatrix}$$
where column 1 represents state $S$ and column 2 represents state $D$.
(a) If a student begins in state $S$ at month 0, find the probability they are in state $S$ after 3 months. [2 marks]
(b) Find the steady-state probability vector for the system. [4 marks]
Verified Solution & Marking Scheme
Part (a): Compute state vector at month 3
Initial state vector $\vec{s}_0 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$.
After 3 months: $\vec{s}_3 = T^3 \vec{s}_0$.
$$T^3 = \begin{pmatrix} 0.85 & 0.10 \\ 0.15 & 0.90 \end{pmatrix}^3 \approx \begin{pmatrix} 0.528 \\ 0.472 \end{pmatrix}$$
Hence the probability of being in state $S$ after 3 months is $0.528$ (to 3 s.f.).
Part (b): Set up steady-state matrix equation T s = s
Let steady-state vector be $\vec{s} = \begin{pmatrix} s_1 \\ s_2 \end{pmatrix}$ where $s_1 + s_2 = 1$.
$$\begin{pmatrix} 0.85 & 0.10 \\ 0.15 & 0.90 \end{pmatrix} \begin{pmatrix} s_1 \\ s_2 \end{pmatrix} = \begin{pmatrix} s_1 \\ s_2 \end{pmatrix}$$
From the first equation:
$$0.85s_1 + 0.10s_2 = s_1 \implies 0.10s_2 = 0.15s_1 \implies s_2 = 1.5s_1$$
Part (b): Solve with total probability s₁ + s₂ = 1
$$s_1 + 1.5s_1 = 1 \implies 2.5s_1 = 1 \implies s_1 = \frac{1}{2.5} = 0.40$$
$$s_2 = 1 - 0.40 = 0.60$$
Steady-state probability vector is $\begin{pmatrix} 0.40 \\ 0.60 \end{pmatrix}$.