Transcript ppt

ECEN4503 Random Signals
Lecture #12
10 February 2014
Dr. George Scheets
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
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Problems 4.20 - 4.23 (1st Edition)
Problems 4.38 – 4.41 (2nd Edition)
Quiz < Expected Values
ECEN4503 Random Signals
Lecture #13
12 February 2014
Dr. George Scheets






Read 5.1 - 5.2
Problems: 2011 Exam #1
Quiz #4, 14 February ( < Expected Values)
Exam #1, 28 February
Quiz 2 Results
Hi = 10.0, Low = 3.3, Average = 8.10
Standard Deviation = 1.80
Quiz 3 Results
Hi = 10.0, Low = 2.4, Average = 8.56
Standard Deviation = 2.13
Expected Values
+∞
 E[
*]=
∫
(*) fX(x) dx
-∞
 E[
* ] = ∑(*i)P(*i)
all possible
outcomes
(discrete)
Lottery Assistance
Histogram
Histogram
Histogram #3
As the number
of die rolls
increases, this
histogram will
tend to look
flatter if viewed
on full vertical
scale.
Histogram #3, Version 2
If viewed
on partial
vertical scale,
it will still
show a lot of
random
variability.
Default Values for this Class
Function Bounds
-∞ to +∞
 Function Value
Zero outside specified bounds
 Ex) PDF fx(x) = 1/2
Authorities will assume you think exists
everywhere.
 Ex) fx(x) = 1/2; 0 < x < 2
Authorities will assume you think
function = zero outside specified bounds.
(i.e. no need to add fx(x) = 0; elsewhere)

Mean vs. Median
4 household incomes
 Case 1
$55K, $60K, $70K, $80K
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 Mean
= $66,250
 Median = $65,000
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Case 2) Bill Gates Moves In
$55K, $60K, $70K, $10B
 Mean
= $2.5 Billion
 Median = $65,000
PDF's
Case 1
fX(x)
1/4
55
60
70
80
$K
fX(x)
Case 2
1/4
55
60
70
1 inch
$K
15.78 miles
10B
Mean
(or Median)
is only Part of the Story
Variance = σ2 = E[X2] - E[X]2
 Case 1
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 E[X2]
= (3.025 + 3.6 + 4.9 + 6.4)109/4
= 4.481 109
 E[X]2 = 4.389 109
 σ2 = 91.94 106
 σ = $9,588
Mean
(or Median)
is only Part of the Story
Variance = σ2 = E[X2] - E[X]2
 Case 2
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 E[X2]
= (3.025 + 3.6 + 4.9 + 109)109/4
= 2.500 1019
 E[X]2 = 6.250 1018
 σ2 = 1.875 1019
 σ = $4.330 B