II. Ìb (‚M) ¸‰æb

(1)

œ0μ3 (Review of Probability)

Min Wang

Industrial Engineering & Management Chaoyang University of Technology

Min Wang

œ0μ3 (Review of Probability)

œ0ƒb¸œ0}º Ìb¸‰æb :¯œ0ƒb

I. œ0ƒb¸œ0}º

(Probability Function & Probability Distribution)

œ0ƒb (probability function)

Óœ‰b X íœ0ƒb (probability function) ·Hv‰bª?ß Þ X 5Míœ0

J X Ñ×àÓœ‰b(discrete random variable), †wœ0 ƒb˚Ñœ0”¾ƒb(probability mass function, pmf);

J X Ñ©/Óœ‰b(continuous random variable), †w œ0ƒb˚Ñœ0òƒb(probability density function, pdf)

Min Wang

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œ0”¾ƒb (probability mass function, pmf), p.70

For a discrete random variable X with possible values x₁, x2, . . . , xn, a probability mass function is a function such that

(1) f (xi) ≥ 0 (2) ⁿ

i=1f (xi) = 1

(3) f (xi) = P(X = xi)

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(2)

Example 3-4, p.69

This is a chance that a bit transmitted trough a digital transmission channel is received in error. Let X equal the number of bits in error in the next four bits transmitted. The possible values for X are {0, 1, 2, 3, 4, }. Based on a model for the errors that is presented in the following section, probabilities for these values will be determined. Suppose that the probabilities are

P(X =0)=0.6561 P(X =1)=0.2916 P(X =2)=0.0486 P(X =3)=0.0036 P(X =4)=0.0001

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Example 3-4 (Cont’)

0 1 2 3 4 x

0.2916 0.0036

0.0001 0.0486

0.6561 f (x)

Figure 3-1 Probability distribution for bits in error.

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Example (Exercise 3-14), p.71

The sample space of a random experiment is

{a, b, c, d, e, f },and each outcome is equally likely.

A random variable is deﬁned as follows:

outcome a b c d e f

x 0 0 1.5 1.5 2 3 Determine the probability mass function of X .

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Solution

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(3)

Úœ0}ºƒb (cumulative distribution function), p.72

The cumulative distribution function of a discrete random variable X , denoted as F (x), is

F (x) = P(X ≤ x) =

xi≤x

f (xi)

For a discrete random variable X , F (x) satisﬁes the following properties.

(1) F (x) = P(X ≤ x) =

xi≤x f (xi) (2) 0 ≤ F(x) ≤ 1

(3) If x ≤ y, then F (x) ≤ F (y)

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Example

Continued from Exercise 3-14. Determine the cumulative distribution function of X .

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Solution

-Çéý¤ CDF, F(x).

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Example 3-7, p.72

Determine the probability mass function of X from the following cumulative distribution function:

F (x) =

⎧⎪

⎪⎨

⎪⎪

⎩

0 x < −2

0.2 −2 ≤ x < 0 0.7 0 ≤ x < 2

1 2 ≤ x

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(4)

Solution

-Çéý¤ CDF, F(x).

0 0.2

2 –2

0.7 1.0

x F(x)

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œ0òƒb (probability density function), p.111

For a continuous random variable X , a probability density function is a function such that

(1) f (x) ≥ 0 (2) _∞

−∞f (x)dx = 1 (3) P(a ≤ X ≤ b) = _b

a f (x)dx = area under f (x) from a to b for any a and b

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©/Óœ‰bíœ0 (Probability of a continuous random variable)

If X is a continuous random variable, for any x1 and x2,

P(x1 ≤ X ≤ x2) = P(x1 < X ≤ x2)

= P(x1 ≤ X < x2)

= P(x1 < X < x2)

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Example 4-1, p. 112

Let the continuous random variable X denote the current measured in a thin copper wire in

milliamperes. Assume that the range of X is

[0, 20mA], and assume that the probability density function of X is f (x) = 0.05 for 0 ≤ x ≤ 20. What is the probability that a current measurement is less than 10 milliamperes?

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(5)

Example 4-1(Cont’), p. 112

The probability density function is shown in Fig. 4-4. It is assumed that f (x) = 0 wherever it is not speciﬁcally deﬁned.

The probability requested is indicated by the shaded area in Fig. 4-4.

P(X < 10) = 10

0 f (x)dx = 10

0

0.05dx = 0.5

0 10 20 x

0.05 f (x)

Figure 4-4 Probability density function for Example 4-1.

Min Wang

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Úœ0}ºƒb (cumulative distribution function), p.114

The cumulative distribution function of a continuous random variable X is

F (x) = P(X ≤ x) = _x

−∞f (u) du for −∞ < x < ∞.

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II. Ìb (‚M) ¸‰æb

(Mean (Expected Value) and Variance)

Min Wang

Ìb (‚M) ¸‰æb–×àÓœ‰b, p.74

The mean or expected value of the discrete random variable X , denoted as μ or E (X ), is

μ = E(X) =

x

xf (x)

The variance of X , denoted as σ² or V (X ), is

σ² = V (X ) = E (X − μ)²=

x

(x − μ)²f (x) =

x

x²f (x) − μ²

√ ¹⁷

(6)

Ìb (‚M) ¸‰æb–×àÓœ‰b, p.77

If X is a discrete random variable with probability mass function f (x),

E [h(X )] =

x

h(x)f (x)

Example: I X Ñø×àÓœ‰b, /I h(X) = X², † E [h(X )] = E [X²] =

x

x²f (x)

¤¹Fø X² 5‚M FJ σ² = E [X²]− μ²

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Example (Exercise 3-39, p.77)

If the range of X is the set {0, 1, 2, 3, 4} and P(X = x) = 0.2 determine the mean and variance of the random variable.

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Solution (Cont’)

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Suppose X is a continuous random variable with probability density function f (x).

The mean or expected value of X , denoted as μ or E (X ), is μ = E(X) =

_∞

−∞xf (x)dx The variance of X , denoted as V (X ) or σ², is

σ² = V (X ) = _∞

−∞(x − μ)²f (x)dx = _∞

−∞x²f (x)dx − μ² The standard deviation of X is σ =√

σ².

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(7)

If X is a continuous random variable with probability mass function f (x),

E [h(X )] = _∞

−∞h(x)f (x)

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xample (Exercise 4-23, p. 118)

Suppose f (x) = 0.25 for 0 < x < 4. Determine the mean and variance of X .

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