Question 4: The weather in Columbus is either good, indifferent, or bad on any given day. If the weather is good today, there is a 60% chance the weather will be good tomorrow, a 30% chance the weather will be indifferent, and a 10% chance the weather will be bad. If the weather is indifferent today, it will be good tomorrow with probability .40 and indifferent with probability .30. Finally, if the weather is bad today, it will be good tomorrow with probability .40 and indifferent with probability .50.

What is the stochastic matrix for this situation?

Suppose there is a 50% chance of good weather today and a 50% chance of indifferent weather. What are the chances of bad weather tomorrow?

Suppose the predicted weather for Monday is 40% indifferent weather and 60% bad weather. What are the chances for good weather on Wednesday?

a)

Let \(G\) represent the good weather, \(I\) represent the indifferent weather, and \(B\) represent the bad weather.

The following table represents the changes in the weather.

Therefore, the stochastic matrix is \(P = \left( {\begin{array}{*{20}{c}}{.6}&{.4}&{.4}\\{.3}&{.3}&{.5}\\{.1}&{.3}&{.1}\end{array}} \right)\).

b)

Theorem 18states that if \(P\) is a \(n \times n\) regular stochastic matrix,then it has a unique steady-state vector \({\mathop{\rm q}\nolimits} \).Further, if \({{\mathop{\rm x}\nolimits} _0}\) is any initial state and \({{\mathop{\rm x}\nolimits} _{k + 1}} = P{{\mathop{\rm x}\nolimits} _k}\) for \(k = 0,1,2,....\) then the Markov chain \(\left\{ {{{\mathop{\rm x}\nolimits} _k}} \right\}\) converges to \({\mathop{\rm q}\nolimits} \)as \(k \to \infty \).

The initial state vector is \({{\mathop{\rm x}\nolimits} _0} = \left( {\begin{array}{*{20}{c}}{.5}\\{.5}\\0\end{array}} \right)\) since, today, there is 50% chance of good weather and 50% chance of indifferent weather.

Compute \({{\mathop{\rm x}\nolimits} _1}\) as shown below:

\(\begin{array}{c}{{\mathop{\rm x}\nolimits} _1} = P{{\mathop{\rm x}\nolimits} _0}\\ = \left( {\begin{array}{*{20}{c}}{.6}&{.4}&{.4}\\{.3}&{.3}&{.5}\\{.1}&{.3}&{.1}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{.5}\\{.5}\\0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{0.3 + 0.2 + 0}\\{0.15 + 0.15 + 0}\\{0.05 + 0.15 + 0}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{0.5}\\{0.3}\\{0.2}\end{array}} \right)\end{array}\)

Hence, the chance of bad weather for tomorrow is 20%.

c)

Since the predicted weather for Monday is 40% indifferent and 60% bad, the initial state vector is \({{\mathop{\rm x}\nolimits} _0} = \left( {\begin{array}{*{20}{c}}0\\{.4}\\{.6}\end{array}} \right)\).

Compute \({{\mathop{\rm x}\nolimits} _2}\) as shown below:

\(\begin{array}{c}{{\mathop{\rm x}\nolimits} _1} = P{{\mathop{\rm x}\nolimits} _0}\\ = \left( {\begin{array}{*{20}{c}}{.6}&{.4}&{.4}\\{.3}&{.3}&{.5}\\{.1}&{.3}&{.1}\end{array}} \right)\left( {\begin{array}{*{20}{c}}0\\{.4}\\{.6}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{0 + 0.16 + 0.24}\\{0 + 0.12 + 0.3}\\{0 + 0.12 + 0.06}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{0.4}\\{0.42}\\{0.18}\end{array}} \right)\end{array}\)

\(\begin{array}{c}{{\mathop{\rm x}\nolimits} _2} = P{{\mathop{\rm x}\nolimits} _1}\\ = \left( {\begin{array}{*{20}{c}}{.6}&{.4}&{.4}\\{.3}&{.3}&{.5}\\{.1}&{.3}&{.1}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{0.4}\\{0.42}\\{0.18}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{0.24 + 0.168 + 0.072}\\{0.12 + 0.126 + 0.09}\\{0.04 + 0.126 + 0.018}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{0.48}\\{0.336}\\{0.184}\end{array}} \right)\end{array}\)

Hence, the chance of good weather on Wednesday is 48%.