07. midterm practice

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Probability

07. Midterm Practice

ConditionIndependence

郭俊利 2009/04/13

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ProbabilityConcept

Homework

p(n) > 1 Discrete

f (x) > 1 Continuous

F(x) > 1

2.1 ~ 4.1

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ProbabilityBasic Probability

Set P(A∪B∪C) =

P(A) + P(AC∩B) + P(AC∩BC∩C) Condition

P(A∩B∩C) = P(A) P(B|A) P(C|A∩B) Independence

P(A∩B) = P(A) P(B)

Problem 1.31 – error bit

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ProbabilityExpectation

E[a] = a E[aX + b] = aE[X] + b E[g(X)] = Σx g(x) px(x) E[X] = Σi E[Xi] = np (p is uniform!)

E[X] = Σi P(Ai) E[X | Ai] E[X] = E[E[X|Y]]

var(aX + b) = a2 var(X)

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ProbabilityGraph

Mean Center (not find f(x))

Variance E[X2] – E[X]2 (need f(x) or formula)

f(x)

x

1/3

2/3

1 2

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Probability

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ProbabilityOther E.V.

Independence

E[XY] = E[X] E[Y] var(X + Y) = var(X) + var(Y)

Sum

E[T] = E[X] E[N] var(T) = var(X) E[N] + E2[X] var(N)

T = X1 + … + XN

f(x, y) = f(x) f(y)

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ProbabilityConditional Sum of Independence 1st

Let X1, X2 and X3 be independent and identical binomial random variables, that is, P(Xi = k) = Cn

k pk (1 − p)n−k, 0 ≤ k ≤ n. Compute the P(Z = X1 + X2 + X3) Compute E[Z], var(Z)

E[] = np ; var() = np(1-p)

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Probability

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ProbabilityJoint

fX,Y(x,y) = fY(y) fX|Y(x|y)

X, Y are independent, X and Y are in [0, 2].fXY(x, y) = xy / 4, find E[f(x, y)]

Double integration or…∵ f(x, y) = f(x) f(y)

∴ f(x) = x / 2 f(y) = y / 2 or

　 f(x) = x f(y) = y / 4

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Probability

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ProbabilityImportant Random Variable

Bernoulli pX(k) = p, 1-p

Binomial pX(k) = Cn

k pk (1 – p)n – k

Geometric pX(k) = (1 – p)k-1 p

E[X] = p var(X) = p(1-p)

E[X] = np var(X) = np(1-p)

E[X] = 1/p var(X) = (1-p)/p2

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ProbabilityGeometric random number

Xiao-Quan plays a game rock-paper-scissors with another. He plays until he loses. Find the expectation of the number of rounds. If Xiao-Quan has won 3 times and drawn 2 times,

how many rounds will he expect to play?

Problem 2.22, 2.23 Non-memoryless

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ProbabilityConditional Sum of Independence 2nd

Suppose that X and Y are independent and identical geometrical random variables with parameter p, that is,P(X = k) = P(Y = k) = qk−1p, k ≥ 1.

Compute P(X = i | X + Y = n), i = 1, 2, ..., n − 1. Compute E(X | X + Y = n), var(X | X + Y = n).

P(X = i | X + Y = n) = P(X = i ∩ X + Y = n)

P(X + Y = n)

=P(X = i ∩ Y = n – i)

P(X + Y = n)

=P(X = i) P(Y = n – i)

P(X + Y = n)= 1 / n-1

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ProbabilityExponential random number

f(x) = λe–λx

P(x a) =∫≧ a∞ λe–λx dx

= –e–λx | a∞ = e–λa

F(x) = 1 – e–λx (Geometric F(n) = 1 – (1–p)n)

E[X] = 1 / λ var(X) = 1 / λ2 (E[X2] = 2 / λ2)

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ProbabilityExponential Examples

The spent time of work is modeled as an exponential random variable. The average time that Xiao-Ming completes the task is 10 hours. What is the probability that Xiao-Ming has done this task early (in advance)?

f(x) = ½ λe–λx, x ≥ 0 ½ λe+λx, x < 0

F(x) = ?

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Probability

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ProbabilityCondition

P(A|B) = P(A∩B) / P(B) pX|A(x) = P({X = x}∩A) / P(A) fX|A(x) = f(x) / P(A)

Roll a fair die shown k points, what is the probability p(k) given some roll is even number?

Xiao-Wang arrivals is a uniform random variable from 7:10 to 7:30. The bus comes at 7:15 and 7:30. What is the waiting time f(x) of Xiao-Wang?

Show exponential random number is memoryless.

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Probability

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ProbabilityNormal random number

Standard normal distribution N(–a) = P(Y ≦ –a) = P(Y ≧ a) = 1 – P(Y ≦ a)

N(–a) = 1 – N(a) CDF

P(X ≦ a) = P(Y ≦ ) = N( )a – μσ

a – μσ

The grades of the exam is suitable for a normal random variable. The average of grades = 60 and the standard deviation = 20. What is the probability that Xiao-Kuo’s grade will be higher over 70?

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ProbabilityF() and f()

Derived

Linear Y = aX + b

General y = g(x), x = h(y) y = g(h(y))

fY(y) = fX( h(y) ) |h’(y)|

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ProbabilityDerived Distribution

Find the PDF of Z = g(X, Y) = Y/X

FZ(z) = 0 ≤ z ≤ 1 FZ(z) = z > 1

fZ(z) =

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ProbabilityLinear Mapping

X is an exponential random variable,Y = –λX + 2, Find the PDF of Y

f(x) = λe–λx

f(y) = e–λ(y-2 / –λ)

= e y-2

λ

|-λ|

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ProbabilityTravel Problem

Xiao-Hua is driving from Boston to New York 180 miles. His speed is uniformly distributed between 30 and 60 mph. What is the distribution of the duration of the trip?

fV(v) = 1 / 30 30 ≤ v ≤ 60

T(v) = 180 / v

fT(t) = fV(v) T(t)’ = 6 / t2

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Probability

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ProbabilityConvolution

W = X + Y PW(w) = Σ PX(x) PY(w – x)

W = |X| + 2Y

p(x) = 1/3, if x = –1, 0, 1

p(y) = 1/2, y = 0 1/3, y = 1 1/6, y = 2

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ProbabilityMaximum

W = max {X, Y} (X, Y = [0, 1] uniformly)

W = max {X, Y} (X, Y = 0.1, 0.2, …, 1.0 uniformly)

P(X≦w) = P(Y≦w) = w

FW(w) = P(X≦w) P(Y≦w) = w2

fW(w) = 2w

pW(w) = FW(w) – FW(w – 0.1)

= w2 – (w–0.1)2