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Magazine Chapter 1: Q17E (page 1)
Return to Example 3.10.6. Assume that the state at time\({\bf{n - 1}}\)is\(\left\{ {{\bf{Aa,aa}}} \right\}\).
a. Suppose that we learn that\({{\bf{X}}_{{\bf{n + 1}}}}\,\,{\bf{is}}\,\,\left\{ {{\bf{AA,aa}}} \right\}\). Find the conditional distribution of\({{\bf{X}}_{\bf{n}}}\). (That is, find all the probabilities for the possible states at time n given that the state at the time\({\bf{n + 1}}\,\,{\bf{is}}\,\,\left\{ {{\bf{AA,aa}}} \right\}\).)
b. Suppose that we learn that\({{\bf{X}}_{{\bf{n + 1}}}}\,\,{\bf{is}}\,\,\left\{ {{\bf{AA,aa}}} \right\}\). Find the conditional distribution of\({{\bf{X}}_{\bf{n}}}\).
Short Answer
a.\(\Pr \left( {{X_n} = {x_n}\left| {{X_{n - 1}} = \left\{ {Aa,aa} \right\},{X_{n + 1}} = \left\{ {AA,aa} \right\}} \right.} \right) = \frac{{\Pr \left( {{X_n} = {x_n},{X_{n - 1}} = \left\{ {Aa,aa} \right\}\left| {{X_{n - 1}} = \left\{ {Aa,aa} \right\}} \right.} \right)}}{{\Pr \left( {{X_{n + 1}} = \left\{ {AA,aa} \right\}\left| {{X_{n - 1}} = \left\{ {Aa,aa} \right\}} \right.} \right)}}\)
b. \(\Pr \left( {{X_n} = {x_n}\left| {{X_{n - 1}} = \left\{ {Aa,aa} \right\},{X_{n + 1}} = \left\{ {Aa,aa} \right\}} \right.} \right) = \frac{{\Pr \left( {{X_n} = {x_n},{X_{n - 1}} = \left\{ {Aa,aa} \right\}\left| {{X_{n - 1}} = \left\{ {Aa,aa} \right\}} \right.} \right)}}{{\Pr \left( {{X_{n + 1}} = \left\{ {aa,aa} \right\}\left| {{X_{n - 1}} = \left\{ {Aa,aa} \right\}} \right.} \right)}}\)
Step by step solution
Given information
Referring to example 3.10.6, assume that the state at the time \({\rm{n - 1}}\,\,{\rm{is}}\,\,\left\{ {{\rm{Aa,aa}}} \right\}\).
(a) Finding the conditional distribution
The conditional distribution of\({{\rm{X}}_{\rm{n}}}\,\,{\rm{given}}\,\,{{\rm{X}}_{{\rm{n - 1}}}}{\rm{ = }}\left\{ {{\rm{Aa,aa}}} \right\}\,\,{\rm{and}}\,\,{{\rm{X}}_{{\rm{n + 1}}}}{\rm{ = }}\left\{ {{\rm{AA,aa}}} \right\}\)
For each possible state\({x_n}\)
Then,
\(\Pr \left( {{X_n} = {x_n}\left| {{X_{n - 1}} = \left\{ {Aa,aa} \right\},{X_{n + 1}} = \left\{ {AA,aa} \right\}} \right.} \right) = \frac{{\Pr \left( {{X_n} = {x_n},{X_{n - 1}} = \left\{ {Aa,aa} \right\}\left| {{X_{n - 1}} = \left\{ {Aa,aa} \right\}} \right.} \right)}}{{\Pr \left( {{X_{n + 1}} = \left\{ {AA,aa} \right\}\left| {{X_{n - 1}} = \left\{ {Aa,aa} \right\}} \right.} \right)}}\)
Then the dominator is 0.0313 from the 2-step transition matrix.
The numerator is the product of two terms from the 1-step transition probability matrix: one from\(\left\{ {{\rm{Aa,aa}}} \right\}\,\,{\rm{to}}\,\,{{\rm{x}}_{\rm{n}}}\)and the other form\({{\rm{x}}_{\rm{n}}}\,\,{\rm{to}}\,\,\left\{ {{\rm{AA,aa}}} \right\}\).
So,
\(\begin{array}{l}\left\{ {AA,AA} \right\} = 0\\\left\{ {AA,Aa} \right\} = 0\\\left\{ {AA,aa} \right\} = 0\\\left\{ {Aa,Aa} \right\} = 0.25 \times 0.125\\\left\{ {Aa,aa} \right\} = 0\\\left\{ {aa,aa} \right\} = 0\end{array}\)
\(\Pr \left( {{X_{n + 1}} = \left\{ {AA,aa} \right\}\left| {{X_{n - 1}} = \left\{ {Aa,aa} \right\}} \right.,{X_{n - + }} = \left\{ {Aa,aa} \right\}} \right) = 1\)
And all other states have a probability of 0.
(b) Finding the conditional distribution
\(\Pr \left( {{X_n} = {x_n}\left| {{X_{n - 1}} = \left\{ {Aa,aa} \right\},{X_{n + 1}} = \left\{ {Aa,aa} \right\}} \right.} \right) = \frac{{\Pr \left( {{X_n} = {x_n},{X_{n - 1}} = \left\{ {Aa,aa} \right\}\left| {{X_{n - 1}} = \left\{ {Aa,aa} \right\}} \right.} \right)}}{{\Pr \left( {{X_{n + 1}} = \left\{ {aa,aa} \right\}\left| {{X_{n - 1}} = \left\{ {Aa,aa} \right\}} \right.} \right)}}\)
The denominator is 0.3906.
The numerator products and their ratio to the denominator are,
This time,
\(\Pr \left( {{X_n} = {x_n}\left| {{X_{n - 1}} = \left\{ {Aa,aa} \right\},{X_{n + 1}} = \left\{ {Aa,Aa} \right\}} \right.} \right) = \left\{ {\begin{array}{*{20}{c}}{0.04}&{if}&{{x_n} = \left\{ {Aa,Aa} \right\}}\\{0.32}&{if}&{{x_n} = \left\{ {Aa,aa} \right\}}\\{0.64}&{if}&{{x_n} = \left\{ {aa,aa} \right\}}\end{array}} \right.\)
And all others are 0.
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