Stat 330: Module 4 Homework Solution
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1. Classify each of the following stochastic processes as discrete-time or continuous-time, and discrete-space
or continuous-space.
Answer:
(a) continuous time and continuous state
(b) discrete time and continuous state
(c) continuous time and discrete state
(d) discrete time and discrete state
(e) discrete time and discrete state
2. A certain machine used in a manufacturing process can be in one of three states: Fully operational
(“full”), partially operational (“part”), or broken (“broken”). If the machine is fully operational today,
there is a .7 probability it will be fully operational again tomorrow, a .2 chance it will be partially
operational tomorrow, and otherwise tomorrow it will be broken. If the machine is partially operational
today, there is a .6 probability it will continue to be partially operational tomorrow and otherwise it
will be broken (because the machine is never repaired in its partially operational state). Finally, if the
machine is broken today, there is a .8 probability it will be repaired to fully operational status tomorrow;
otherwise, it remains broken. Let X = the state of the machine on day n.
Answer:
(a) {full, part, broken}
(b)
full part broken
full 0.7 0.2 0.1 !
P = part 0 0.6 0.4
broken 0.8 0 0.2
(c)
0.57 0.26 0.17
P 2 = P × P = 0.32 0.36 0.32
0.72 0.16 0.12
Thus P 2 [broken, broken] = 0.12.
3. Information bits (0s and 1s) in a binary communication system travel through a long series of relays. At
each relay, a “bit-switching” error might occur. Suppose that at each relay, there is a 4% chance of a 0
bit being switched to a 1 bit and a 5% chance of a 1 becoming a 0. Let X0 = a bit’s initial parity (0 or
1), and let Xn = the bit’s parity after traversing the nth relay.
Answer:
(a)
0 1
P = 0 0.96 0.04
1 0.05 0.95
(b) Let P0 = (0.8, 0.2), then
P0 × P = (0.778, 0.222)
and thus we see 77.8% 0s and 22.2% 1s exiting the first relay.
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Show all of your work, and upload this homework to Canvas.
1. Classify each of the following stochastic processes as discrete-time or continuous-time, and discrete-space
or continuous-space.
Answer:
(a) continuous time and continuous state
(b) discrete time and continuous state
(c) continuous time and discrete state
(d) discrete time and discrete state
(e) discrete time and discrete state
2. A certain machine used in a manufacturing process can be in one of three states: Fully operational
(“full”), partially operational (“part”), or broken (“broken”). If the machine is fully operational today,
there is a .7 probability it will be fully operational again tomorrow, a .2 chance it will be partially
operational tomorrow, and otherwise tomorrow it will be broken. If the machine is partially operational
today, there is a .6 probability it will continue to be partially operational tomorrow and otherwise it
will be broken (because the machine is never repaired in its partially operational state). Finally, if the
machine is broken today, there is a .8 probability it will be repaired to fully operational status tomorrow;
otherwise, it remains broken. Let X = the state of the machine on day n.
Answer:
(a) {full, part, broken}
(b)
full part broken
full 0.7 0.2 0.1 !
P = part 0 0.6 0.4
broken 0.8 0 0.2
(c)
0.57 0.26 0.17
P 2 = P × P = 0.32 0.36 0.32
0.72 0.16 0.12
Thus P 2 [broken, broken] = 0.12.
3. Information bits (0s and 1s) in a binary communication system travel through a long series of relays. At
each relay, a “bit-switching” error might occur. Suppose that at each relay, there is a 4% chance of a 0
bit being switched to a 1 bit and a 5% chance of a 1 becoming a 0. Let X0 = a bit’s initial parity (0 or
1), and let Xn = the bit’s parity after traversing the nth relay.
Answer:
(a)
0 1
P = 0 0.96 0.04
1 0.05 0.95
(b) Let P0 = (0.8, 0.2), then
P0 × P = (0.778, 0.222)
and thus we see 77.8% 0s and 22.2% 1s exiting the first relay.
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