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The solutions to problem 1 through problem 14 of the ece 863 - analysis of stochastic systems homework set #1 for the fall 2001 semester. The problems cover various concepts related to stochastic systems, including sample spaces, probabilities, and independence.
Typology: Exercises
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1
Problem 1
2.4 (a)
The sample space S = set of ordered pairs which can be expressed as: (i, 1) (i, 2) …… (i, i), for i = 1, 2, 3, 4, 5, 6
Therefore: S 1, 1 2,1 2, 3,1 3,2 3,
2.4 (b)
2.4 (c)
2.4 (d)
2
Problem 2
2.5 (a)
Each testing of a pen has two possible outcomes: “pen is good” (g) or “pen is bad” (b). The experiment consists of testing the pens until a good pen is found. Therefore, each outcome of the experiment consists of a string of “b”s ended by a “g.” Assuming that each tested pen is not put back in the
2.5 (b)
In this case, we simply record the number of pens that have been tested.
2.5 (c)
In this case,
2.5 (d)
Problem 3
2.
A « C = ∆ lifetime is greater than 5 and not greater than 3
4
Problem 6
2.
P A B C P A B C P A B P C P A B C P A P B P A B P C P A C B C P A P B P C P A B P A C P B C P A C B C P A P B P C P A B P A C P B C P A B C
Problem 7
2.
j 6 2
j 6
6 5
=
Similarly P B 1 2
10
c
10
10
5
(You can also solve this using P A » B = P A + P B - P A « B)
Problem 8
2.48 (a)
5
Using the definition P A /B
when A « B = ∆ fi P A « B = 0 fi P A /B = 0
when A B A B A P A /B P A P B
à fi « = fi =
when B Ã A fi A « B = B fi P A /B = 1
2.48 (b)
If P A /B > P A fi P A « B >P A P B
fi P B / A P A >P A P B
fi P B / A >P B
Therefore, P A /B > P A (which is also equivalent to P B / A > P B as we just showed) implies that A & B tend to occur jointly.
Problem 9
2.
P A B C P A / B C P B C P A / B C P B / C P C
Problem 10
2.
Let A = total number of dots is even B = both tosses are even
7
Similarly, let A 2 = The event that the professor will arrive between 8:50 and 9:00.
fi P B / A = = =
2 2
Problem 13
2.
Let’s use the following diagram (i.e. instead of the one in the book)
Input Output 1 - e 0 0 0 p 0 e (^0)
e (^1) 1 1 p 1 1 – e (^1)
2.57 (a)
Find the probability that the output is O.
Let: O 0 represents the event that the output is 0. O 1 represents the event that the output is 1. I 0 represents the event that the output is 0. I 1 represents the event that the output is 1.
Since I 0 and I 1 represent a Partition (i.e. I 0 « I 1 = ∆ & I 0 » I 1 = S)
fi = + = - +
1 e P e p
0 0 0 0 0 1 1
8
2.57 (b)
Find the probability that the input is 0 given that the output is 1.
e p e p 1 e p
0 1
1 0 0 1 0 0 1 1 1 0 0 0 0 1 1
2.57 (c)
Find the probability that the input is 1 given that the output is 1.
1 e p e p 1 e p
1 1
1 1 1 1 0 0 1 1 1 1 1 0 0 1 1
2.57 (d)
Under what conditions it is more likely that the input is 1 given that the output is 1.
P I / O P I / O 1 e p e p 1 e p e 1 p
e
e 1 p p
1 1 0 1 1 1 0 0 1 1 0 1
1
0 1 1
fi - >
fi - >
e 1 e
1 p (^1 0) p
1 1
Problem 14
2.
10
1 - P A c^ P B c^ P Cc
I think you got the idea (d) and (e) can be solved the same way.
Problem 17
2.
In terms of relative frequencies, we expect
Problem 18
p , p , p 0 1 2 & p 3 , respectively. The receiver detects the transmitted signal and makes a decision that one of the four symbols has been sent. The
transmitted symbols consists of two bits: s 0 = 00; s 1 = 01; s 2 = 10; s 3 = 11. If the probability of a bit-error is e, what is the probability that the transmitter has sent s 0 given that the receiver detects r 3?
P s / r P s^ r P r
P r / s P s (^0 3) P r / s P s P r / s P s P r / s P s P r / s P s 0 3 3
3 0 0 3 0 0 3 1 1 3 2 2 3 3 3
e p e p e 1 e p 1 e e p 1 e p
2 0
11
Probability of a symbol error
= « + « + «
P s r P s r P s r P s r P s r P s r P s r P s r P s r P s r P s r P s r
p 1 e e p 1 e e p e p 1 e e p 1 e e p e p 1 e e p 1 e e p e p 1 e e p 1 e e p e 1 e e
0 1 0 2 0 3 1 0 1 2 0 3 2 0 2 1 2 3 3 0 3 1 3 2
0 0 0
2
1 1 1
2
2 2 2 2 3 3 3
2
2e 2e e 2e e
2 2 2 2 2
ECE.863.Homework.Set.1.Solutions.Fall2000.slh
