Exam MAS-I · practice
Exam MAS-I — Markov Chains Practice Flashcards
Thirty exam-realistic multiple-choice problems on CAS Exam MAS-I discrete-time Markov chains — reading transition matrices, $n$-step probabilities via Chapman–Kolmogorov and matrix powers, distributions after $n$ steps from an initial vector, state classification (recurrence, transience, period), stationary distributions solving $\pi=\pi P$, and absorbing-chain analysis through the fundamental matrix $N=(I-Q)^{-1}$ for expected time to absorption and absorption probabilities including gambler's-ruin — each with a fully worked solution.
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- Chapman-KolmogorovA Markov chain on states $\{1,2\}$ has one-step transition matrix $P=\begin{pmatrix} 0.6 & 0.4 \\ 0.3 & 0.7 \end{pmatrix}$. Given $X_0=1$, calculate $\Pr(X_2=2)$. (A) $0.40$ (B) $0.46$ (C) $0.52$ (D) $0.58$ (E) $0.70$**Answer: (C).** By Chapman–Kolmogorov, $\Pr(X_2=2\mid X_0=1)=(P^2)_{12}$. Condition on the intermediate state $X_1$: $(P^2)_{12}=P_{11}P_{12}+P_{12}P_{22}=0.6(0.4)+0.4(0.7)=0.24+0.28=0.52$. The two paths are $1\to 1\to 2$ (prob $0.24$) and $1\to 2\to 2$ (prob $0.28$). So $\Pr(X_2=2)=0.52$. The one-step value $P_{12}=0.4$ is distractor (A); using only the single path $1\to 2\to 2 = 0.28$ or dropping a term lands near (B).
- Chapman-KolmogorovA weather chain on states Sunny ($S$) and Rainy ($R$) has $P=\begin{pmatrix} 0.75 & 0.25 \\ 0.40 & 0.60 \end{pmatrix}$ (row/column order $S,R$). Today ($n=0$) is Sunny. Calculate $\Pr(X_3=R)$. (A) $0.250$ (B) $0.310$ (C) $0.350$ (D) $0.368$ (E) $0.632$**Answer: (D).** Start from the row vector $\alpha=(1,0)$ and propagate forward, $\alpha P^{n}$: $\alpha P=(0.75,\,0.25)$. $\alpha P^2=(0.75,0.25)P=\big(0.75(0.75)+0.25(0.40),\ 0.75(0.25)+0.25(0.60)\big)=(0.6625,\,0.3375)$. $\alpha P^3=(0.6625,0.3375)P=\big(0.6625(0.75)+0.3375(0.40),\ 0.6625(0.25)+0.3375(0.60)\big)=(0.631875,\,0.368125)$. So $\Pr(X_3=R)\approx 0.368$. The one-step value $0.25$ is distractor (A); $0.632$ is $\Pr(X_3=S)$, the complement.
- Chapman-KolmogorovA chain on $\{1,2,3\}$ has $P=\begin{pmatrix} 0.5 & 0.3 & 0.2 \\ 0.2 & 0.5 & 0.3 \\ 0.3 & 0.3 & 0.4 \end{pmatrix}$. Given $X_0=1$, calculate $\Pr(X_2=3)$. (A) $0.20$ (B) $0.27$ (C) $0.30$ (D) $0.31$ (E) $0.36$**Answer: (B).** $\Pr(X_2=3\mid X_0=1)=(P^2)_{13}=\sum_k P_{1k}P_{k3}$. $=P_{11}P_{13}+P_{12}P_{23}+P_{13}P_{33}$ $=0.5(0.2)+0.3(0.3)+0.2(0.4)$ $=0.10+0.09+0.08=0.27$. So $\Pr(X_2=3)=0.27$. The one-step entry $P_{13}=0.20$ is distractor (A); $0.31=(P^2)_{23}$, the wrong starting row.
- Chapman-KolmogorovA chain on $\{1,2\}$ has $P=\begin{pmatrix} 0.9 & 0.1 \\ 0.5 & 0.5 \end{pmatrix}$ and initial distribution $\alpha=(0.3,\,0.7)$. Calculate $\Pr(X_2=1)$. (A) $0.620$ (B) $0.700$ (C) $0.748$ (D) $0.800$ (E) $0.820$**Answer: (C).** The distribution after $n$ steps is the row vector $\alpha P^{n}$. $\alpha P=(0.3,0.7)P=\big(0.3(0.9)+0.7(0.5),\ 0.3(0.1)+0.7(0.5)\big)=(0.62,\,0.38)$. $\alpha P^2=(0.62,0.38)P=\big(0.62(0.9)+0.38(0.5),\ 0.62(0.1)+0.38(0.5)\big)=(0.748,\,0.252)$. So $\Pr(X_2=1)=0.748$. Stopping after one step gives the distractor $0.62$ (A); $\Pr(X_2=2)=0.252$.
- Chapman-KolmogorovA chain on $\{1,2\}$ has $P=\begin{pmatrix} 0.6 & 0.4 \\ 0.3 & 0.7 \end{pmatrix}$. Given $X_0=1$, calculate the three-step probability $\Pr(X_3=1)$. (A) $0.396$ (B) $0.417$ (C) $0.444$ (D) $0.480$ (E) $0.556$**Answer: (C).** We need $(P^3)_{11}$. First $P^2=\begin{pmatrix} 0.48 & 0.52 \\ 0.39 & 0.61 \end{pmatrix}$ (e.g. $(P^2)_{11}=0.6(0.6)+0.4(0.3)=0.48$). Then $(P^3)_{11}=(P^2)_{11}P_{11}+(P^2)_{12}P_{21}=0.48(0.6)+0.52(0.3)=0.288+0.156=0.444$. So $\Pr(X_3=1)=0.444$. The two-step value $(P^2)_{11}=0.48$ is distractor (D); $(P^3)_{21}=0.417$ uses the wrong starting row (B).
- Chapman-KolmogorovA no-claims-discount (NCD) chain on levels $\{0\%,20\%,40\%\}$ (states $1,2,3$) has $P=\begin{pmatrix} 0.2 & 0.8 & 0 \\ 0.1 & 0 & 0.9 \\ 0 & 0.1 & 0.9 \end{pmatrix}$. A new policyholder starts at $0\%$ ($X_0=1$). Calculate the probability the policyholder is at the $40\%$ level after $2$ years. (A) $0.00$ (B) $0.12$ (C) $0.16$ (D) $0.72$ (E) $0.90$**Answer: (D).** Start from $\alpha=(1,0,0)$. After one year, $\alpha P=(0.2,\,0.8,\,0)$ — just row 1 of $P$. After two years, $(0.2,0.8,0)P$, taking the third (state $3$) component: $\Pr(X_2=3)=0.2(0)+0.8(0.9)+0(0.9)=0.72$. So the policyholder is at the $40\%$ level with probability $0.72$. The one-year value of $0$ for state $3$ is distractor (A); $0.12$ and $0.16$ are the year-$2$ probabilities of the $0\%$ and $20\%$ levels respectively.
- Transition matricesIn the weather chain $P=\begin{pmatrix} 0.75 & 0.25 \\ 0.40 & 0.60 \end{pmatrix}$ (states $S,R$), today is Sunny ($X_0=S$). Calculate the probability of the exact sequence Sunny, Rainy, Rainy over days $0,1,2$. (A) $0.100$ (B) $0.150$ (C) $0.250$ (D) $0.338$ (E) $0.375$**Answer: (B).** For a *fixed* path the joint probability is the product of one-step transitions along it (no summing). With $X_0=S$ given: $\Pr(S\to R\to R)=P_{SR}\cdot P_{RR}=0.25\times 0.60=0.15$. So the answer is $0.15$. The endpoint probability $\Pr(X_2=R\mid X_0=S)=0.3375$ (distractor D) sums over *all* intermediate paths, not just $S\to R\to R$; $0.25$ is the single step $P_{SR}$.
- Transition matricesA row of a purported transition matrix on $\{1,2,3\}$ is $(0.3,\,x,\,0.45)$, and a candidate claims the matrix entry $P_{23}=-0.1$. Which statement is correct about a valid stochastic matrix? (A) $x=0.35$, and $P_{23}=-0.1$ is allowed since columns need not sum to $1$ (B) $x=0.25$, and $P_{23}=-0.1$ is invalid because every entry must be $\ge 0$ (C) $x=0.55$, and $P_{23}=-0.1$ is invalid because rows must sum to $0$ (D) $x=0.25$, and $P_{23}=-0.1$ is allowed since only rows must sum to $1$ (E) $x=0.75$, and the matrix can never be stochastic**Answer: (B).** A stochastic matrix has two requirements: every entry $P_{ij}\ge 0$, and **each row sums to $1$** (from any state the chain must go somewhere). Columns need not sum to anything in particular. Row sum: $0.3+x+0.45=1\Rightarrow x=0.25$. An entry $P_{23}=-0.1<0$ is a probability and cannot be negative, so it is invalid. Choice (D) gets $x$ right but wrongly permits a negative entry; (C) wrongly states rows sum to $0$.