Download Math/Stat 360-1: Probability and Statistics, Washington State

Math/Stat 360-1: Probability and Statistics,
Washington State University
Haijun Li
[email protected]
Department of Mathematics
Washington State University
Week 5
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 5
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Outline
1
Section 4-1: Probability Density Functions
2
Section 4-2: Cumulative Distribution Functions and Expected
Values
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
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Distribution of Continuous Random Variables
Random Variables (RV): Denoted by capital letters, X , Y ,
N, ...
Values of Random Variables: Denoted by small letters, x, y ,
n, ...
Distribution of a continuous RV X : Relative frequencies of
various values of X .
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Math/Stat 360-1: Probability and Statistics, Washington State University
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S&P 500 Monthly Returns (10/1965 - 7/2005)
S&P 500 Monthly Return = random variable
X.
http://www.optionetics.com/images/articles
The distribution = continuous version of the histogram of X .
9-6-06 2.gif (GIF Image, 572 × 419 pixels)
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Math/Stat 360-1: Probability and Statistics, Washington State University
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Probability Density Function
Probability Density Function (PDF)
f (x) describes the probability density of a continuous RV X at x.
1
f (x) ≥ 0
R∞
2
f (x)dx
= 1PROBABILITY DENSITY FUNCTIONS
RIBUTIONS
AND
−∞
Rb
3
P(a < X < b) = a f (x)dx = the area under f (x) between a
and b.
f (x)
P(a < X < b)
a
Haijun Li
b
Math/Stat 360-1: Probability and Statistics, Washington State University
x
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Probability Density Function
PDF describes relative frequency of a continuous random
variable X .
P(X = a) = 0.
P(a ≤ X ≤ b) = P(a < X ≤ b) = P(a ≤ X < b) = P(a <
X < b).
rb12_6.gif (GIF Image, 613 × 474 pixels)
Haijun Li
http://www.weibull.com/hotwire/issue12/rb12_6.gif
Math/Stat 360-1: Probability and Statistics, Washington State University
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Example
The PDF of the failure time X of an electronic component in a
copier (in hours) is
h
1
x i
f (x) =
exp −
, x ≥ 0,
3000
3000
and zero otherwise. Find the probability that
1
2
3
A component lasts
than 1000 hours before failure.
R ∞ more
x
x
1
P(X > 1000) = 1000 3000
e− 3000 dx = [−e− 3000 ]∞
1000 = 0.716.
A component fails between 1000 and 2000 hours.
R 2000 1 − x
x
P(1000 < X < 2000) = 1000 3000
e 3000 dx = [−e− 3000 ]2000
1000 =
0.203.
Determine the number of hours at which 10% of all
components have failed.
a
Find a such that P(X ≤ a) = 0.1. That is, 1 − e− 3000 = 0.1.
Solve this, we have a ≈ 316 hours.
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 5
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Cumulative Distribution Function =
Cumulative Frequency
Cumulative Distribution Function (CDF) of X
Z
x
F (x) = P(X ≤ x) =
f (u)du.
−∞
1
2
3
F (x) is non-decreasing.
0 ≤ F (x) ≤ 1.
F (x) → 1 as x → +∞.
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
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Figure : Comparison of PDF and CDF
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
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Rb
1. P(a ≤ X ≤ b) = F (b) −http://http-server.carleton.ca/~gkardos/88403/Reliability/RE
F (a) = a f (x)dx
2. f (x) = F 0(x)
2.gif (GIF Image, 600 × 427 pixels)
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 5
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Example
Consider a PDF
f (x) =
h
1
x i
exp −
, x ≥ 0,
3000
3000
and zero otherwise. Find the CDF.
Z x
u
x
1
F (x) =
e− 3000 du = 1 − e− 3000 , x ≥ 0.
0 3000
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
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Example
Consider the PDF of the uniform RV over [a, b].
f (x) =
1
, a ≤ x ≤ b,
b−a
and zero otherwise. Find the CDF.
Rx 1
F (x) = a b−a
du = x−a
, a ≤ x ≤ b.
b−a
F (x) = 0, x < a.
F (x) = 1, b < x.
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
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Mean and Variance
Mean of a continuous RV X
Z ∞
Z
E(X ) = µ =
xf (x)dx =
−∞
∞
x
−∞
Variance of a continuous RV X
Z ∞
Z
2
2
V (X ) = σ =
(x−µ) f (x)dx =
−∞
Standard deviation (SD) of X : σ =
R∞
σ 2 = −∞ x 2 f (x)dx − µ2 .
Haijun Li
f (x)dx
.
| {z }
prob. mass at x
∞
(x−µ)2
−∞
p
f (x)dx
| {z }
prob. mass at x
V (X ).
Math/Stat 360-1: Probability and Statistics, Washington State University
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Figure : σy = SD of Y .
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 5
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Example
Suppose that the CDF of the length (in millimeters) of computer
cables is

x ≤ 1200
 0
0.1x − 120 1200 < x ≤ 1210
F (x) =

1
x > 1210
1
Find P(X < 1208), and P(1195 < X < 1205).
P(X < 1208) = F (1208) = 0.8
P(1195 < X < 1205) = F (1205)−F (1195) = F (1205) = 0.5.
2
What is its PDF and mean?
f (x) = F 0 (x) = 0.1, 1200 < x ≤ 1210.
Z 1210
E(X ) =
0.1xdx = [0.05x 2 ]1210
1200 = 1205.
1200
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Math/Stat 360-1: Probability and Statistics, Washington State University
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