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Unformatted text preview: ln 1
f0 (u)du = e−1
2
2 √ pmiss =
−∞ 2 1
f1 (u)du = e−1
2 ∞
ln √ e−u du =
2
√ ln −∞ 2 1
√ ≈ 0.1301
2e 2 1
eu du = √ ≈ 0.2601.
e2 118 CHAPTER 3. CONTINUOUSTYPE RANDOM VARIABLES The average error probability for the MAP rule using the given prior distribution is
√
pe = (2/3)pf alse alarm + (1/3)pmiss = 3e2 2 = 0.1734. Chapter 4 Jointly Distributed Random Variables
This chapter focuses on dealing with multiple random variables that may not be independent.
Earlier in these notes, we’ve sometimes worked with multiple random variables. For example,
a Bernoulli process, as discussed in Section 2.6, is composed of independent Bernoulli random
variables, and other random variables are also associated with the processes, such as the number
of trials required for each one, and the cumulative number of ones up to somc time n. Similarly,
Poisson processes, discussed in Section 2.7, involve multiple random variables deﬁned on the same
space. But typically we considered independent random variables. In many applications, there are
multiple statistically dependent random variables...
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This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Institute of Technology.
 Spring '08
 Zahrn
 The Land

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