LECTURE 5 Random variables every possible outcome Lecture outline

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LECTURE 5
Random variables
• Readings: Sections 2.1-2.3, start 2.4
• An assignment of a value (number) to
every possible outcome
Lecture outline
• Mathematically: A function
from the sample space Ω to the real
numbers
• Random variables
• Probability mass function (PMF)
– discrete or continuous values
• Expectation
• Can have several random variables
defined on the same sample space
• Variance
• Notation:
– random variable X
– numerical value x
How to compute a PMF pX (x)
– collect all possible outcomes for which
X is equal to x
– add their probabilities
– repeat for all x
Probability mass function (PMF)
• (“probability law”,
“probability distribution” of X)
• Notation:
• Example: Two independent rools of a
fair tetrahedral die
pX (x) = P(X = x)
= P({ω ∈ Ω s.t. X(ω) = x})
• pX (x) ≥ 0
F : outcome of first throw
S: outcome of second throw
X = min(F, S)
!
x pX (x) = 1
• Example: X=number of coin tosses
until first head
4
– assume independent tosses,
P(H) = p > 0
3
S = Second roll
2
pX (k) = P(X = k)
= P(T T · · · T H)
= (1 − p)k−1p,
1
k = 1, 2, . . .
1
2
3
F = First roll
– geometric PMF
pX (2) =
1
4
Binomial PMF
Expectation
• Definition:
• X: number of heads in n independent
coin tosses
E[X] =
$
x
• P(H) = p
• Interpretations:
– Center of gravity of PMF
– Average in large number of repetitions
of the experiment
(to be substantiated later in this course)
• Let n = 4
pX (2) = P(HHT T ) + P(HT HT ) + P(HT T H)
+P(T HHT ) + P(T HT H) + P(T T HH)
= 6p2(1 − p)2
=
"4#
2
• Example: Uniform on 0, 1, . . . , n
p2(1 − p)2
In general:
"n#
pX (k) =
pk (1−p)n−k ,
k
pX(x )
1/(n+1)
...
k = 0, 1, . . . , n
0
E[X] = 0×
1
– Easy: E[Y ] =
y
$
x
Recall:
E[g(X)] =
$
x
ypY (y)
g(x)pX (x)
• Second moment: E[X 2] =
g(x)pX (x)
• Variance
%
! 2
x x pX (x)
var(X) = E (X − E[X])2
• Caution: In general, E[g(X)] %= g(E[X])
=
$
x
Prop erties:
x
n
Variance
• Let X be a r.v. and let Y = g(X)
– Hard: E[Y ] =
n- 1
1
1
1
+1×
+· · ·+n×
=
n+1
n+1
n+1
Properties of expectations
$
xpX (x)
&
(x − E[X ])2pX (x)
= E[X 2] − (E[X])2
If α, β are constants, then:
• E[α] =
Properties:
• E[αX] =
• var(X) ≥ 0
• var(αX + β) = α2var(X)
• E[αX + β] =
2
MIT OpenCourseWare
http://ocw.mit.edu
6.041SC Probabilistic Systems Analysis and Applied Probability
Fall 2013
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