Download EC381/MN308 Probability and Some Statistics Lecture 6

EC381/MN308
Probability and Some Statistics
Lecture 6 - Outline
1. Derived Random Variables
Yannis Paschalidis
2. Conditional PMFs and Expectations
[email protected], http://ionia.bu.edu/
3. Random Vectors
Dept. of Manufacturing Engineering
Dept. of Electrical and Computer Engineering
Center for Information and Systems Engineering
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2.3 Derived Random Variables
Example (Example 2.29 on p. 73)
Given one random variable (RV) X, the function g mapping R Æ R
generates the derived RV
Y = g(x)
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• Prob. of number of pages in fax, X, given by
PX(x)
Y
For outcome s, X(s) is a real number,
Y(s) = g(X(s)) is the derived random variable
X
5 6 7 8
1 2
5 6 7 8
Y
• Price of pages:
A. Relation between the PMF of Y and the PMF of X
1 2 3 4
50
40
34
27
Y
19
10
X
A random variable can be: (1) the observation; (2) a function of the observation; (3) a function of another random variable
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3 4
X
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PY(y)
y
Figure 2.1 (p. 72)
Example 2.29 on p. 73
The derived random variable Y = g(X) for this example (Example 2.29).
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C. Special Case: Linear Transformation
B. Expectation of a Derived RV
Y = g(X)
Y = aX + b
E [Y] = ?
Expectation
Expectation is a linear operation!
Don’t need to compute PY(y)
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Variance
E [aX + b] = aE[X] + b
Var[aX + b] = a2 Var[X]
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Example (Quiz 2.7 on p. 76)
Special Example
Y = X – mX
Mean
– The number of memory chips M used by PC, depends on # of
application programs A user wants to run simultaneously
mX = E [X]
4 chips for 1 or 2 programs
6 chips for 3
8 chips for 4
E [Y ] = E [X ] – mx = mx– mx = 0
V ar[Y ] = E[Y 2] = E[(X − μX )2] = V ar[X]
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M = g(A)
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4 4
1 2 3 4
– Take A to be a RV with PMF
A
PA(a) = 0
0.1
1 (5 - a),
) a = 1,
1 2
2, 3
3, 4; and 0 otherwise
– Compute E [A]
– Write M = g(A) as a derived variable
– Compute E [M]. Is E [M] = g(E [A])?
True for any random variable.
0.4
PA(a)
0.3
0.2
0.1
1 2 3 4
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Solution
0.3
0.2
0.1
1 2 3 4
Problem: Generate a discrete RV Y with prescribed range:
a
SY = {0, 1, 2, .. k}, CDF FY(y), and PMF PY(y) = FY(y) – FY(y -1).
Example
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M = g(A)
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Generation of a Discrete Random Variable of Prescribed
PMF by Transformation of a Uniform Random Variable
0.4
PA(a)
a
FY(y)
6
4 4
1 2 3 4
A
nonlinear
y
PY(y)
-1
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0
1
2
3
4
5
6
y
12
2
Solution: Generate a random number X in [0,1] with uniform distribution
(MATLAB: rand(1,1)).
2.4 Conditional PMF
Find a transformation Y = g(X) that transforms X into Y = {0, 1, 2, .. k},
such that
The conditional PMF of X given an event B is
PX|B(x) = P [X = x | B]
Transformation: Lookup table for
determining the y corresponding to x
A = {X = x} is collection of outcomes
Since P [A|B] = P [A ∩ B] / P [B]
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FY(y)
5
x
y
4
3
2
Special case: B ⊂ SX
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Event B = outcomes of X with values in B
A value x ∈ SX is either in B or not x ∈ B Æ {X = x} ∩ B = {X = x}
0
-1
y
0
.1
.2
.3
.4
.5
.6
x
.7
.8
.9
1
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Example
PX|B(x) is a probability mass function with usual properties:
• Arrival time of a message is a RV X, nonnegative integer
B
valued, with uniform PMF
– PX(x) = 1/20, x = 1, …, 20
otherwise
– PX(x) = 0
1
20
5
X
• Consider the event B = {X > 6} (message does not
arrive by 6-th time slot)
– Compute PX|B(x|B)
Conditional Expectation
Conditional PMF PX|B(x) ≡ P [X = x|B]
E [X|B] = conditional expectation of RV X given that outcomes are in
event B.
Compute using the conditional PMF
E [X | X > 6] = (20+7)/2=13.5
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2.5 Multiple Discrete Random Variables
Conditional Expectations of Derived RVs
“Random Vectors”
Y = g(X)
Sample
space
sX
Example (cont.):
(x2,y2)
s2
sk
s1
(xk,yk)
(x1,y1)
k outcomes
Find variance
ariance of X given
gi en X > 6
Two Random
T
R d
V
Variables
i bl X
X, Y on
same probability space
Joint PMF
PX,Y (x, y)
– PX,Y (x, y) = P [X = x and Y = y]
Properties
∑x ∑y PX,Y (x, y) = 1
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PX (x) = ∑y PX,Y (x, y) Marginal PMF
or we can also use the formula for the variance of a uniform in {7, …, 20}
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PY (y) = ∑x PX,Y (x, y) Marginal PMF
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Conditional Probability
Independence
• Definition.
Two discrete random variables X, Y are independent if and
only if the pairs of events {X = x}, {Y = y} are independent
for all x ∈ SX, y ∈ SY
– PX |Y (x | y) = P [X = x | Y = y] = PX,Y (x, y) / PY (y)
– Interpretation: Observe outcomes associated with
event {Y = y} and compute conditional probability
mass function of X
P [X = x, Y = y] = P [X = x] P [Y = y], i.e., PX,Y(x, y) = PX(x) PY(y)
• Properties.
p
It is a PMF on X:
PX |Y (x | y) in [0,1]
∑x PX |Y (x | y) = 1
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Independent?
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