随机过程英语讲义-3

Total Probability Theorem:
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Thomas Bayes
1 No earlier portrait or claimed portrait survives. Born Died Nationality
Conditioning, Marginals and Total Probability Combining conditioning, marginals, and total probability, we get a variety of useful identities:
Conditional probability density function
Solution:
P[chip defective] = P[def . | A]P[ A] + P[def . | B]P[ B] + P[def . | C ]P[C ] = (10−3 ) p A + 5(10−3 ) pB + 10(10−3 ) pC
Continue…
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10 −3 p A P[ def . | A]P[ A] = P[ A | chip defective] = P[ def .] (10 −3 ) p A + 5(10 −3 ) pB + 10(10 −3 ) pC pA = p A + 5 pB + 10 pC
σ = Var ( X ) = E[( X − μ X ) ] = ∫ ( x − μ X ) 2 f X ( x)dx
2 x 2 −∞
∞
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1.5 Moment Generating , Characteristic Function and Laplace Transform
吧The moment generating function of X is defined by
ψ (t )=E[etX ]
= ∫ etx dF ( x).
psi
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When a moment generating function exists , it uniquely determines the distribution . This is quite important because it enables us to characterize the probability distribution of a random variable by its generating function . Example：: Normal Distribution Function with mean Probability Density Function and variance
P[ H ] = P[ H | coin 1]P[coin 1] + P[ H | coin 2]P[coin 2] 1 1 1 = p1 ⋅ + p2 ⋅ = ( p1 + p2 ) 2 2 2 Now using Bayes' Rule,
1 P[ H | coin 2]P[coin 2] p2 2 p2 P[coin 2 | H ] = = 1 = P[ H ] ( p1 + p2 ) 2 ( p1 + p 2 )
Signature
c. 1701 London, England 7 April 1761) (aged 59) Tunbridge Wells, Kent, England English
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Bayes’ Rule Suppose we are given Prior probabilities: Transition probabilities:
Moment Generating Function
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Example：
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吧 Characteristic functions
phi
吧
1
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吧
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Example：
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Laplace Transform 吧
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1.6 Conditional Expectation
Conditional Probability mass function
How can we compute
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Problem 1: A nonsymmetrical binary communication channel is shown below. Assume inputs are equiprobable. Find the probability that the output is 0. Find the probability that input was 0 given that the output is 1. Find the probability that input is 1 given that the output is a 1.
input 0
1-ε1
0
output
ε1 ε2
1-ε2
1
1
Solution Continues…
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Solution : Let X denote the input and Y the output. a) P[Y = 0] = P[Y = 0 | X = 0]P[ X = 0] + P[Y = 0 | X = 1]P[ X = 1]
similarly 10 pC P[C | chip defective] = p A + 5 pB + 10 pC
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Problem 3: One of the two coins is selected at random and tossed. The first coin comes up heads with probability p1 and second coin with probability p2. What is the probability that coin 2 was used given that heads occurred? Solution :
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Problem 2: A computer manufacturer uses chips from three sources. Chips from source A, B and C are defectives with the probability .001, .005, .01, respectively. If randomly selected chip found to be defective , find the probability that the manufacturer was A; that the manufacture was C.
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1.4 Moments and Central Moments
Definition : The kth moment is The first moment = Average = Expectation:
μ X = E[ X ] = ∫ xf X ( x)dx
−∞
∞
Definition : The kth central moment is The second moment = variance:
Visualizing Conditioning View joint PMF along slice Y = y and renormalize.
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•Probability of A given B occurred •B is now our new universe
Total Probability Formula Definition : A partition of such that is a collection of disjoint events
1 fX
( x) = ∫
∞
−∞
f X Y ( x y ) fY ( y )dy
∞ −∞
f X ,Y ( x, y ) = ∫
f X ,Y Z ( x, y z ) f Z ( z )dz
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Prior distribution Transition PDFs
fY X ( y x )
Question: given an observation {Y = y} what information do we have about X?
1 1 = (1 − ε1 ) + ε 2 2 2
b)
1 P[Y = 1 | X = 0]P[ X = 0] 2 ε1 = 1 P[ X = 0 | Y = 1] = 1 − 2 (1 − ε1 ) − 1 ε 2 P[Y = 1] 2
ε1 = (1 − ε 2 ) + ε1
P[ X = 1 | Y = 1] = 1 − P[ X = 0 | Y = 1] = 1− ε 2 (1 − ε 2 ) + ε1