Discrete Random
Variables & Distributions
stat 430
Heike Hofmann
Outline
• (Discrete Random Variables)
• Expected Value,Variance
• Moment Generating Function
• Discrete Distributions: Binomial,
Geometric, Hypergeometric, Poisson
Ωk = {
00...00, 00...01, 0
Random Variables
...,
11...00, 11...01, 1
• Definition:
AAfunction
calleda arandom
random variab
function X : Ω �→ R is called
variable
� �
k
image of X:
i
im(X) = X(Omega) = set of all possible
� �
values X can take
n k
P (X = k) =
p (1
k
Examples:
•
#heads in 10 throws,
#of songs from 80s in 1h of LITE 104.1 (or KURE 88.5 FM),
(i)
0 ≤ P (A) ≤ 1
winnings in Darts
(ii)
P (∅) = 0
Probability
mass
function
�
2 �
V ar[X] =
(x
i−
(pmf)
VE[X])
ar[X] =· pX (x
(xi ) − E[X])
i
i
i
2
·
pX (x)
:=function
P (X = px)
is called the probability mass func
The
X (x) := P (X = x) is called the pro
unction
has mass
two main
properties:
of a pmf
p
probability
function
has two Properties
main properties:
Pro
X, if and only
if pX is a pmf, iff
Theorem:
ust(i)
be all
between
1 01between
≤
1 for
∈(x)
{x1≤
, x12
X (x)
values
be
0≤im(X)
and
1 0all≤xpX
0 ≤0must
pand
forpall
x in
X(x) ≤
�
�
ll (ii)
values
1 ofi pall
= {x1, x2, ... }
i ) = 1is 1for im(X)
X (x
theissum
values
i pX (xi ) = 1
•
•
•
E[h(X)] =
�
i
�
h(xE[h(X)]
) =: µh(xi ) · pX (xi
i ) · pX (xi=
i
Simple Dartboard
• red area is 1/9 of grey
area
• P(red) = 0.1
P(grey) = 0.9
payout: 25 cents for grey area
$1 for red area
How much should each game (of three darts) cost initially?
Expectation
y if
Expected Value
es must be between 0 and
� 1 0 ≤ pX (x) ≤ 1 for all x ∈ {
2
V ar[X]
(xi − E[X]) · pX (xi )
�=
m of all values is 1 i pXi(xi ) = 1
n pX (x) :=
(X = x)value
is called
the probability
TheP expected
of random
variable X ismass fu
� that we Properties
ss functionthe
haslong
twoterm
main
properties:
of a pm
average
will see,
E[h(X)] =
h(xi ) · pX (xi ) =: µ
when we repeat the same experiment over
if
i
and over:
must be between 0 and 1 0�
≤ pX (x) ≤ 1 for all x ∈ {x1 ,
xi · pX (xi ) =: µ
�E[X] =
of all values is 1 i pX (xi ) =i 1
for additional function h, we get:
�
=00...00,
h(x
(xi ) =: 00...11,
µ
ΩE[h(X)]
00...01,
i ) · pX 00...10,
k ={
•
•
..., i
Variance
Variance
• The variance is a measure of homogeneity:
V ar[X] =
�
i
(xi − E[X])2 · pX (xi )
pX (x) := P (X = x) is called the probability mass func
unction has two main properties: Properties of a pmf p
ust be between 0 and 1 0 ≤ p (x) ≤ 1 for all x ∈ {x , x
Rules for E[X] and Var[X]
• Let X and Y be two random variables, and
a,b two real values, then
• E[aX+bY] = a E[X] + b E[Y]
E[XY] = E[X]E[Y], if X,Y are independent
•
• Var[aX] = a Var[X]
Var[X] = E[X2] - (E[X])2
2
Var[X+Y] = Var[X] + Var[Y], if X,Y are ind.
Moment Generating
Function
→ Y = aX + b with a > 0, �
h(x, y)pX,Y (x, y)
→ Y = aX +E[h(X,
b with Ya )]
< :=
0,
Moment Generating
linear association between X and Y . ρ near ±1 indicate
Function
ar 0 indicates
lackYof
linear
Cov(X,
)=
E[(Xassociation.
− E[X])(Y − E[Y ])]
x,y
�
�
�
kth
Moment
of
k r.v. X:d E[Xk]
� (x, y)
M
(t)
E[h(X, YE[X
)] :=] = h(x,
y)p
X
X,Y
�
kt
d
x,y function MXt=0
Moment generating
(t):
� tX � � tx
MX (t) = E e
=
e i pX (xi )
Cov(X, Y ) = E[(X − E[X])(Y − E[Y ])]
i�
�
d
k
then E[X ] = k MX (t)��
d t4
t=0
� tX � � tx
i
M
(t)
=
E
e
=
e
pXfor
(xi mgf
)
hope:
X we find a “nice” expression
•
•
•
i
Special Distributions
Binomial Distribution
• X = #successes in n independent, identical
Bernoulli trials with P(success) = p
• sample space =
• P(X = k) =
E[X] =
Var[X] =
Geometric Distribution
• X = #attempts of independent, identical
Bernoulli trials with P(success) = p until
(including) first success
• sample space =
• P(X = k) =
E[X] =
Var[X] =
Hypergeometric
Distribution
• X = #attempts of independent, identical
Bernoulli trials with P(success) = p until
(including) rth success
• sample space =
• P(X = k) =
E[X] =
Var[X] =
Poisson Distribution
• X = #occurrences of event in 1 unit of
space/time,
with lambda = average #occurrences in
1unit space/time
• sample space =
• P(X = k) =
E[X] =
Var[X] =
Uniform Distribution
• all elements in the sample space have the
same probability, i.e. fair die:
• sample space = {1,2,3,4,5,6}
• Let X be the up-turned face of a die
E[X] = 3.5
Var[X] =