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2.
Discrete
Random
Variables
Part
II:
Expecta:on
ECE
302
Fall
2009
TR
3‐4:15pm
Purdue
University,
School
of
ECE
Prof.
Ilya
Pollak
Expected
value
of
X:
Defini:on
Expected
value
of
X:
Defini:on
Expected
value
of
X:
Defini:on
E[X] is also called the mean of X
€
Example:
mean
of
a
Bernoulli
random
variable
Example:
mean
of
a
discrete
uniform
random
variable
Example:
mean
of
a
discrete
uniform
random
variable
Example:
mean
of
a
discrete
uniform
random
variable
Example:
mean
of
a
discrete
uniform
random
variable
(Note: E[X] is not necessarily one of the values
that X can assume with a non-zero probability!)
Expected
value:
Center
of
gravity
of
PMF
•  Imagine that a box with weight pX(x) is
placed at each x.
c
x
Expected
value:
Center
of
gravity
of
PMF
•  Imagine that a box with weight pX(x) is
placed at each x.
•  Center of gravity c is the point at which
the sum of the torques is zero:
c
x
Expected
value:
Center
of
gravity
of
PMF
•  Imagine that a box with weight pX(x) is
placed at each x.
•  Center of gravity c is the point at which
the sum of the torques is zero:
c
x
Expected
value:
Center
of
gravity
of
PMF
•  Imagine that a box with weight pX(x) is
placed at each x.
•  Center of gravity c is the point at which
the sum of the torques is zero:
c
x
Expected
value:
Center
of
gravity
of
PMF
•  Imagine that a box with weight pX(x) is
placed at each x.
•  Center of gravity c is the point at which
the sum of the torques is zero:
c
x
Expected
value
and
empirical
mean
•  Many
independent
Bernoulli
trials
with
p=0.2
•  Bernoulli
random
variables
X1,
X2,
…
Expected
value
and
empirical
mean
•  Many
independent
Bernoulli
trials
with
p=0.2
•  Bernoulli
random
variables
X1,
X2,
…
•  E[Xn]=0.2
Expected
value
and
empirical
mean
• 
• 
• 
• 
Many
independent
Bernoulli
trials
with
p=0.2
Bernoulli
random
variables
X1,
X2,
…
E[Xn]=0.2
Empirical
average
from
many
experiments
close
to
the
expected
value
(law
of
large
numbers)
Expected
value
and
empirical
mean
• 
• 
• 
• 
Many
independent
Bernoulli
trials
with
p=0.2
Bernoulli
random
variables
X1,
X2,
…
E[Xn]=0.2
Empirical
average
from
many
independent
experiments
close
to
the
expected
value
(law
of
large
numbers)
What
E[X]
is
NOT
•  It’s
not
necessarily
the
most
likely
value
of
X
•  It’s
not
even
always
the
case
that
P(X=E[X])>0
•  It’s
not
guaranteed
to
equal
to
the
empirical
average
Two
ways
to
evaluate
E[g(X)]
Two
ways
to
evaluate
E[g(X)]
Two
ways
to
evaluate
E[g(X)]
Two
ways
to
evaluate
E[g(X)]
Two
ways
to
evaluate
E[g(X)]
Two
ways
to
evaluate
E[g(X)]
Two
ways
to
evaluate
E[g(X)]
Cau:on:
In
general,
E[g(X)]≠g(E[X])
•  Example:
average
speed
vs
average
:me.
•  Suppose
you
need
to
drive
60
miles.
A
very
bad
storm
will
hit
your
area
with
probability
1/2.
Cau:on:
In
general,
E[g(X)]≠g(E[X])
•  Example:
average
speed
vs
average
:me.
•  Suppose
you
need
to
drive
60
miles.
A
very
bad
storm
will
hit
your
area
with
probability
1/2.
–  If
the
storm
hits,
you
will
drive
30
miles/hour
during
the
en:re
trip;
–  If
the
storm
does
not
hit,
you
will
do
60
miles/hour
during
the
en:re
trip.
Cau:on:
In
general,
E[g(X)]≠g(E[X])
•  Example:
average
speed
vs
average
:me.
•  Suppose
you
need
to
drive
60
miles.
A
very
bad
storm
will
hit
your
area
with
probability
1/2.
–  If
the
storm
hits,
you
will
drive
30
miles/hour
during
the
en:re
trip;
–  If
the
storm
does
not
hit,
you
will
do
60
miles/hour
during
the
en:re
trip.
•  What’s
the
expected
value
of
your
speed?
Cau:on:
In
general,
E[g(X)]≠g(E[X])
•  Example:
average
speed
vs
average
:me.
•  Suppose
you
need
to
drive
60
miles.
A
very
bad
storm
will
hit
your
area
with
probability
1/2.
–  If
the
storm
hits,
you
will
drive
30
miles/hour
during
the
en:re
trip;
–  If
the
storm
does
not
hit,
you
will
do
60
miles/hour
during
the
en:re
trip.
•  What’s
the
expected
value
of
your
speed?
–  (1/2)
∙
30
+
(1/2)
∙
60
=
45
mph
Cau:on:
In
general,
E[g(X)]≠g(E[X])
•  Example:
average
speed
vs
average
:me.
•  Suppose
you
need
to
drive
60
miles.
A
very
bad
storm
will
hit
your
area
with
probability
1/2.
–  If
the
storm
hits,
you
will
drive
30
miles/hour
during
the
en:re
trip;
–  If
the
storm
does
not
hit,
you
will
do
60
miles/hour
during
the
en:re
trip.
•  What’s
the
expected
value
of
your
speed?
–  (1/2)
∙
30
+
(1/2)
∙
60
=
45
mph
•  Is
the
expected
value
of
your
travel
:me
60/45
=
1
hr
20
min?
Cau:on:
In
general,
E[g(X)]≠g(E[X])
•  Example:
average
speed
vs
average
:me.
•  Suppose
you
need
to
drive
60
miles.
A
very
bad
storm
will
hit
your
area
with
probability
1/2.
–  If
the
storm
hits,
you
will
drive
30
miles/hour
during
the
en:re
trip;
–  If
the
storm
does
not
hit,
you
will
do
60
miles/hour
during
the
en:re
trip.
•  What’s
the
expected
value
of
your
speed?
–  (1/2)
∙
30
+
(1/2)
∙
60
=
45
mph
•  Is
the
expected
value
of
your
travel
:me
60/45
=
1
hr
20
min?
–  NO:
T
=
60/V,
therefore
E[T]
=
(1/2)
∙
60/30
+
(1/2)
∙
60/60
=
1
hr
30
min
Cau:on:
In
general,
E[g(X)]≠g(E[X])
•  Example:
average
speed
vs
average
:me.
•  Suppose
you
need
to
drive
60
miles.
A
very
bad
storm
will
hit
your
area
with
probability
1/2.
–  If
the
storm
hits,
you
will
drive
30
miles/hour
during
the
en:re
trip;
–  If
the
storm
does
not
hit,
you
will
do
60
miles/hour
during
the
en:re
trip.
•  What’s
the
expected
value
of
your
speed?
–  (1/2)
∙
30
+
(1/2)
∙
60
=
45
mph
•  Is
the
expected
value
of
your
travel
:me
60/45
=
1
hr
20
min?
–  NO:
T
=
60/V,
therefore
E[T]
=
(1/2)
∙
60/30
+
(1/2)
∙
60/60
=
1
hr
30
min
–  Because
you
have
equal
chances
to
spend
1
hour
or
2
hours
driving.
Cau:on:
In
general,
E[g(X)]≠g(E[X])
•  Example:
average
speed
vs
average
:me.
•  Suppose
you
need
to
drive
60
miles.
A
very
bad
storm
will
hit
your
area
with
probability
1/2.
–  If
the
storm
hits,
you
will
drive
30
miles/hour
during
the
en:re
trip;
–  If
the
storm
does
not
hit,
you
will
do
60
miles/hour
during
the
en:re
trip.
•  What’s
the
expected
value
of
your
speed?
–  (1/2)
∙
30
+
(1/2)
∙
60
=
45
mph
•  Is
the
expected
value
of
your
travel
:me
60/45
=
1
hr
20
min?
–  NO:
T
=
60/V,
therefore
E[T]
=
(1/2)
∙
60/30
+
(1/2)
∙
60/60
=
1
hr
30
min
–  Because
you
have
equal
chances
to
spend
1
hour
or
2
hours
driving.
•  Interpreta:on:
in
N
independent
repe::ons
of
the
trip,
–  You
would
expect
to
spend
a
total
of
1.5N
hours
driving
Cau:on:
In
general,
E[g(X)]≠g(E[X])
•  Example:
average
speed
vs
average
:me.
•  Suppose
you
need
to
drive
60
miles.
A
very
bad
storm
will
hit
your
area
with
probability
1/2.
–  If
the
storm
hits,
you
will
drive
30
miles/hour
during
the
en:re
trip;
–  If
the
storm
does
not
hit,
you
will
do
60
miles/hour
during
the
en:re
trip.
•  What’s
the
expected
value
of
your
speed?
–  (1/2)
∙
30
+
(1/2)
∙
60
=
45
mph
•  Is
the
expected
value
of
your
travel
:me
60/45
=
1
hr
20
min?
–  NO:
T
=
60/V,
therefore
E[T]
=
(1/2)
∙
60/30
+
(1/2)
∙
60/60
=
1
hr
30
min
–  Because
you
have
equal
chances
to
spend
1
hour
or
2
hours
driving.
•  Interpreta:on:
in
N
independent
repe::ons
of
the
trip
–  You
would
expect
to
spend
a
total
of
1.5N
hours
driving
–  Your
average
speed
per
trip
(not
per
unit
:me
traveled!)
would
be
about
45mph
What
is
the
meaning
of
E[b]
if
b
is
a
non‐random
number?
What
is
the
meaning
of
E[b]
if
b
is
a
non‐random
number?
Linearity
of
expecta:on
Variance
and
Standard
Devia:on
of
X:
Defini:ons
[
var(X ) = E ( X − E[X ])
2
]
Variance
and
Standard
Devia:on
of
X:
Defini:ons
[
var(X ) = E ( X − E[X ])
2
]
Variance
and
Standard
Devia:on
of
X:
Defini:ons
[
var(X ) = E ( X − E[X ])
2
]
Calcula:ng
var(X)
[
2
]
2
var(X ) = E ( X − E[X ]) = ∑ (x − E[X ]) pX (x)
x
Calcula:ng
var(X)
var(X ) = E [( X − E[X ]) ] = ∑ (x − E[X ]) p (x)
2
2
X
x
= ∑ ( x 2 − 2xE[X ] + (E[X ]) 2 ) pX (x)
x
Calcula:ng
var(X)
var(X ) = E [( X − E[X ]) ] = ∑ (x − E[X ]) p (x)
2
2
X
x
= ∑ ( x 2 − 2xE[X ] + (E[X ]) 2 ) pX (x)
x
= ∑ x 2 pX (x) − 2∑ xE[X ]pX (x) + ∑ (E[X ]) 2 pX (x)
x
€
x
x
Calcula:ng
var(X)
var(X ) = E [( X − E[X ]) ] = ∑ (x − E[X ]) p (x)
2
2
X
x
= ∑ ( x 2 − 2xE[X ] + (E[X ]) 2 ) pX (x)
x
= ∑ x 2 pX (x) − 2∑ xE[X ]pX (x) + ∑ (E[X ]) 2 pX (x)
x
x
x
= ∑ x 2 pX (x) − 2E[X]∑ xpX (x) + (E[X ]) 2 ∑ pX (x)
x
€
€
x
x
Calcula:ng
var(X)
var(X ) = E [( X − E[X ]) ] = ∑ (x − E[X ]) p (x)
2
2
X
x
= ∑ ( x 2 − 2xE[X ] + (E[X ]) 2 ) pX (x)
x
= ∑ x 2 pX (x) − 2∑ xE[X ]pX (x) + ∑ (E[X ]) 2 pX (x)
x
x
x
= ∑ x 2 pX (x) − 2E[X]∑ xpX (x) + (E[X ]) 2 ∑ pX (x)
x 
x 
x






€
€
E[X 2 ]
E[X ]
1
Calcula:ng
var(X)
var(X ) = E [( X − E[X ]) ] = ∑ (x − E[X ]) p (x)
2
2
X
x
= ∑ ( x 2 − 2xE[X ] + (E[X ]) 2 ) pX (x)
x
= ∑ x 2 pX (x) − 2∑ xE[X ]pX (x) + ∑ (E[X ]) 2 pX (x)
x
€
€
x
x
= ∑ x 2 pX (x)−2E[X]∑ xpX (x) + (E[X ]) 2 ∑ pX (x)
x 
x 
x






E[ X ]
E[ X 2 ]

1
−( E[ X ]) 2
Calcula:ng
var(X)
var(X ) = E [( X − E[X ]) ] = ∑ (x − E[X ]) p (x)
2
2
X
x
= ∑ ( x 2 − 2xE[X ] + (E[X ]) 2 ) pX (x)
x
= ∑ x 2 pX (x) − 2∑ xE[X ]pX (x) + ∑ (E[X ]) 2 pX (x)
x
€
x
= ∑ x 2 pX (x)−2E[X]∑ xpX (x) + (E[X ]) 2 ∑ pX (x)
x 
x 
x






E[ X ]
E[ X 2 ]

1
−( E[ X ]) 2
= E[X 2 ] − (E[X]) 2
€
€
x
Calcula:ng
var(X)
var(X ) = E [( X − E[X ]) ] = ∑ (x − E[X ]) p (x)
2
2
X
x
= ∑ ( x 2 − 2xE[X ] + (E[X ]) 2 ) pX (x)
x
= ∑ x 2 pX (x) − 2∑ xE[X ]pX (x) + ∑ (E[X ]) 2 pX (x)
x
€
x
x
= ∑ x 2 pX (x)−2E[X]∑ xpX (x) + (E[X ]) 2 ∑ pX (x)
x 
x 
x






E[ X ]
E[ X 2 ]

1
−( E[ X ]) 2
= E[X 2 ] − (E[X]) 2
€
€
[
Sometimes this is easier to compute than E ( X − E[X ])
2
]
Example
2.4:
Variance
of
a
Bernoulli
random
variable
Example
2.4:
Variance
of
a
Bernoulli
random
variable
Example
2.4:
Variance
of
a
Bernoulli
random
variable
Example
2.4:
Variance
of
a
Bernoulli
random
variable
€
Another
Bernoulli
random
variable
Toss a coin with P(H) = p, and let
 a if H
Y =
 b if T
 p if k = a
Then pY (k) = 
1− p if k = b
€
Another
Bernoulli
random
variable
Toss a coin with P(H) = p, and let
 a if H
Y =
 b if T
 p if k = a
Then pY (k) = 
1− p if k = b
How to compute the expectation and
variance of this random variable?
€
€
Another
Bernoulli
random
variable
Toss a coin with P(H) = p, and let
 a if H
Y =
 b if T
 p if k = a
Then pY (k) = 
1− p if k = b
Note :
Y − b 1 if H
=
=X
a − b 0 if T
How to compute the expectation and
variance of this random variable?
€
€
Another
Bernoulli
random
variable
Toss a coin with P(H) = p, and let
 a if H
Y =
 b if T
 p if k = a
Then pY (k) = 
1− p if k = b
Note :
Y − b 1 if H
=
=X
a − b 0 if T
€
How to compute the expectation and
variance of this random variable?
Therefore,
Y = X(a − b) + b
€
€
Another
Bernoulli
random
variable
Toss a coin with P(H) = p, and let
 a if H
Y =
 b if T
 p if k = a
Then pY (k) = 
1− p if k = b
Note :
Y − b 1 if H
=
=X
a − b 0 if T
€
€
How to compute the expectation and
variance of this random variable?
Therefore,
Y = X(a − b) + b
E[Y] = E[X](a − b) + b = p(a − b) + b
€
€
€
Another
Bernoulli
random
variable
Toss a coin with P(H) = p, and let
 a if H
Y =
 b if T
 p if k = a
Then pY (k) = 
1− p if k = b
Note :
Y − b 1 if H
=
=X
a − b 0 if T
How to compute the expectation and
variance of this random variable?
Therefore,
Y = X(a − b) + b
E[Y] = E[X](a − b) + b = p(a − b) + b
E[Y 2 ] = E[X 2 (a − b) 2 + 2X(a − b)b + b 2 ]
€
€
€
€
€
Another
Bernoulli
random
variable
Toss a coin with P(H) = p, and let
 a if H
Y =
 b if T
 p if k = a
Then pY (k) = 
1− p if k = b
Note :
Y − b 1 if H
=
=X
a − b 0 if T
How to compute the expectation and
variance of this random variable?
Therefore,
Y = X(a − b) + b
E[Y] = E[X](a − b) + b = p(a − b) + b
E[Y 2 ] = E[X 2 (a − b) 2 + 2X(a − b)b + b 2 ]
€
= E[X 2 ](a − b) 2 + 2E[X](a − b)b + b 2
€
€
€
€
Another
Bernoulli
random
variable
Toss a coin with P(H) = p, and let
 a if H
Y =
 b if T
 p if k = a
Then pY (k) = 
1− p if k = b
Note :
Y − b 1 if H
=
=X
a − b 0 if T
How to compute the expectation and
variance of this random variable?
Therefore,
Y = X(a − b) + b
E[Y] = E[X](a − b) + b = p(a − b) + b
E[Y 2 ] = E[X 2 (a − b) 2 + 2X(a − b)b + b 2 ]
€
= E[X 2 ](a − b) 2 + 2E[X](a − b)b + b 2
= p(a − b)€2 + 2 p(a − b)b + b 2
€
€
€
Another
Bernoulli
random
variable
Toss a coin with P(H) = p, and let
 a if H
Y =
 b if T
 p if k = a
Then pY (k) = 
1− p if k = b
Note :
Y − b 1 if H
=
=X
a − b 0 if T
How to compute the expectation and
variance of this random variable?
Therefore,
Y = X(a − b) + b
E[Y] = E[X](a − b) + b = p(a − b) + b
E[Y 2 ] = p(a − b) 2 + 2 p(a − b)b + b 2
2 €
var(Y ) = E[Y ] − (E[Y ]) 2 = ( p − p 2 )(a − b)
€
Standard
devia:on
as
measurement
error
•  Run
a
series
of
independent,
iden:cal
experiments,
e.g.,
Bernoulli
trials.
•  Empirically
es:mate
the
probability
of
an
event,
say,
the
success
in
a
Bernoulli
trial,
as
the
number
of
successes
divided
by
the
number
of
experiments.
•  Standard
devia:on
answers
the
following
ques:on:
In
many
experiments,
by
how
much
would
we
expect
our
es:mate
to
deviate
from
the
actual
probability
of
success?
•  For
example,
if
p=0.2,
then
E[X]
=
0.2
and
var(X)=0.16.
Root
mean‐square
devia:on
of
the
es:mate
of
p
from
0.2,
as
a
func:on
of
the
number
of
trials
Standard
devia:on
as
risk
•  Suppose
you
have
two
investment
opportuni:es:
–  Opportunity
1:
invest
$1000,
have
equal
chances
of
a
total
wipeout
or
of
$2000
profit
–  Opportunity
2:
invest
$1000,
earn
$500
profit
guaranteed.
Standard
devia:on
as
risk
•  Suppose
you
have
two
investment
opportuni:es:
–  Opportunity
1:
invest
$1000,
have
equal
chances
of
a
total
wipeout
or
of
$2000
profit
–  Opportunity
2:
invest
$1000,
earn
$500
profit
guaranteed.
•  Expected
profit
for
1
is
0.5(‐$1000)
+
0.5($2000)
=
$500.
Standard
devia:on
as
risk
•  Suppose
you
have
two
investment
opportuni:es:
–  Opportunity
1:
invest
$1000,
have
equal
chances
of
a
total
wipeout
or
of
$2000
profit
–  Opportunity
2:
invest
$1000,
earn
$500
profit
guaranteed.
•  Expected
profit
for
1
is
0.5(‐$1000)
+
0.5($2000)
=
$500.
•  Expected
profit
for
2
is
$500.
Standard
devia:on
as
risk
•  Suppose
you
have
two
investment
opportuni:es:
–  Opportunity
1:
invest
$1000,
have
equal
chances
of
a
total
wipeout
or
of
$2000
profit
–  Opportunity
2:
invest
$1000,
earn
$500
profit
guaranteed.
•  Expected
profit
for
1
is
0.5(‐$1000)
+
0.5($2000)
=
$500.
•  Expected
profit
for
2
is
$500.
•  But
clearly
the
two
opportuni:es
are
very
different:
you
risk
a
lot
under
the
first
one,
whereas
the
second
one
is
riskless.
Standard
devia:on
as
risk
•  Suppose
you
have
two
investment
opportuni:es:
–  Opportunity
1:
invest
$1000,
have
equal
chances
of
a
total
wipeout
or
of
$2000
profit
–  Opportunity
2:
invest
$1000,
earn
$500
profit
guaranteed.
•  Expected
profit
for
1
is
0.5(‐$1000)
+
0.5($2000)
=
$500.
•  Expected
profit
for
2
is
$500.
•  But
clearly
the
two
opportuni:es
are
very
different:
you
risk
a
lot
under
the
first
one,
whereas
the
second
one
is
riskless.
•  Standard
devia:on
characterizes
the
risk:
–  Opportunity
1:
st.dev.
=
(0.5(500+1000)2
+
0.5(500‐2000)2)1/2
=
1500
Standard
devia:on
as
risk
•  Suppose
you
have
two
investment
opportuni:es:
–  Opportunity
1:
invest
$1000,
have
equal
chances
of
a
total
wipeout
or
of
$2000
profit
–  Opportunity
2:
invest
$1000,
earn
$500
profit
guaranteed.
•  Expected
profit
for
1
is
0.5(‐$1000)
+
0.5($2000)
=
$500.
•  Expected
profit
for
2
is
$500.
•  But
clearly
the
two
opportuni:es
are
very
different:
you
risk
a
lot
under
the
first
one,
whereas
the
second
one
is
riskless.
•  Standard
devia:on
characterizes
the
risk:
–  Opportunity
1:
st.dev.
=
(0.5(500+1000)2
+
0.5(500‐2000)2)1/2
=
1500
–  Opportunity
2:
st.dev.
=
(1(500‐500)2)1/2
=
0
Standard
devia:on
as
risk
•  Suppose
you
have
two
investment
opportuni:es:
–  Opportunity
1:
invest
$1000,
have
equal
chances
of
a
total
wipeout
or
of
$2000
profit
–  Opportunity
2:
invest
$1000,
earn
$500
profit
guaranteed.
•  Expected
profit
for
1
is
0.5(‐$1000)
+
0.5($2000)
=
$500.
•  Expected
profit
for
2
is
$500.
•  But
clearly
the
two
opportuni:es
are
very
different:
you
risk
a
lot
under
the
first
one,
whereas
the
second
one
is
riskless.
•  Standard
devia:on
characterizes
the
risk:
–  Opportunity
1:
st.dev.
=
(0.5(500+1000)2
+
0.5(500‐2000)2)1/2
=
1500
–  Opportunity
2:
st.dev.
=
(1(500‐500)2)1/2
=
0
•  Standard
devia:on
characterizes
the
spread
of
possible
profits
around
the
expected
profit.
Problem
2.20:
Expecta:on
and
variance
of
a
geometric
random
variable
As
an
ad
campaign,
a
chocolate
factory
places
golden
:ckets
in
some
of
its
candy
bars,
with
the
promise
that
a
golden
:cket
is
worth
a
trip
through
the
chocolate
factory,
and
all
the
chocolate
you
can
eat
for
life.
If
the
probability
of
finding
a
gold
:cket
is
p,
find
the
mean
and
the
variance
of
the
number
of
candy
bars
you
need
to
eat
to
find
a
:cket.
Mean
of
a
geometric
random
variable
•  Let
C
=
#
candy
bars
un:l
1st
success
•  Model
C
as
geometric
with
parameter
p:
(1− p) k−1 p, k = 1,2,…
pC (k) = 
0, otherwise

€
Mean
of
a
geometric
random
variable
•  Let
C
=
#
candy
bars
un:l
1st
success
•  Model
C
as
geometric
with
parameter
p:
(1− p) k−1 p, k = 1,2,…
pC (k) = 
0, otherwise

∞
E[C] = ∑ k(1− p) k−1 p
k=1
€
€
Mean
of
a
geometric
random
variable
•  Let
C
=
#
candy
bars
un:l
1st
success
•  Model
C
as
geometric
with
parameter
p:
(1− p) k−1 p, k = 1,2,…
pC (k) = 
0, otherwise

∞
E[C] = ∑ k(1− p) k−1 p
k=1
d
k−1
€
Useful fact : k(1− p) = − {(1− p) k }
dp
€
Mean
of
a
geometric
random
variable
•  Let
C
=
#
candy
bars
un:l
1st
success
•  Model
C
as
geometric
with
parameter
p:
(1− p) k−1 p, k = 1,2,…
pC (k) = 
0, otherwise

∞
E[C] = ∑ k(1− p) k−1 p
k=1
d
k−1
€
Useful fact : k(1− p) = − {(1− p) k }
dp
€
Mean
of
a
geometric
random
variable
•  Let
C
=
#
candy
bars
un:l
1st
success
•  Model
C
as
geometric
with
parameter
p:
(1− p) k−1 p, k = 1,2,…
pC (k) = 
0, otherwise

∞
E[C] = ∑ k(1− p) k−1 p
k=1
d
k−1
€
Useful fact : k(1− p) = − {(1− p) k }
dp
∞
d
€
Therefore, E[C] = − p∑ {(1− p) k }
dp
k =1
Mean
of
a
geometric
random
variable
•  Let
C
=
#
candy
bars
un:l
1st
success
•  Model
C
as
geometric
with
parameter
p:
(1− p) k−1 p, k = 1,2,…
pC (k) = 
0, otherwise

∞
E[C] = ∑ k(1− p) k−1 p
k=1
d
k−1
€
Useful fact : k(1− p) = − {(1− p) k }
dp
∞
∞


d
d
k
k
€
Therefore,
E[C] = − p∑ {(1− p) } = − p ∑ (1− p) 
dp  k =1

k =1 dp
Mean
of
a
geometric
random
variable
•  Let
C
=
#
candy
bars
un:l
1st
success
•  Model
C
as
geometric
with
parameter
p:
(1− p) k−1 p, k = 1,2,…
pC (k) = 
0, otherwise

∞
E[C] = ∑ k(1− p) k−1 p
k=1
d
k−1
€
Useful fact : k(1− p) = − {(1− p) k }
dp
∞
∞


d
d
k
k
€
Therefore,
E[C] = − p∑ {(1− p) } = − p ∑ (1− p) 
dp  k =1

k =1 dp
d 1− p 
= −p 

dp  p 
Mean
of
a
geometric
random
variable
•  Let
C
=
#
candy
bars
un:l
1st
success
•  Model
C
as
geometric
with
parameter
p:
(1− p) k−1 p, k = 1,2,…
pC (k) = 
0, otherwise

∞
E[C] = ∑ k(1− p) k−1 p
k=1
d
k−1
€
Useful fact : k(1− p) = − {(1− p) k }
dp
∞
∞


d
d
k
k
€
Therefore,
E[C] = − p∑ {(1− p) } = − p ∑ (1− p) 
dp  k =1

k =1 dp
 1 1
d 1− p 
= −p 
 = − p− 2  =
dp  p 
 p  p
How
many
candy
bars
un:l
first
success?
•  If
there
are
five
golden
:ckets
per
1,000,000
bars,
then
p=1/200,000.
How
many
candy
bars
un:l
first
success?
•  If
there
are
five
golden
:ckets
per
1,000,000
bars,
then
p=1/200,000.
•  [Note:
we
assume
an
infinite
number
of
bars
so
that
C
is
truly
geometric.]
How
many
candy
bars
un:l
first
success?
•  If
there
are
five
golden
:ckets
per
1,000,000
bars,
then
p=1/200,000.
•  [Note:
we
assume
an
infinite
number
of
bars
so
that
C
is
truly
geometric.]
•  Then
E[C]
=
200,000.
How
many
candy
bars
un:l
first
success?
•  If
there
are
five
golden
:ckets
per
1,000,000
bars,
then
p=1/200,000.
•  [Note:
we
assume
an
infinite
number
of
bars
so
that
C
is
truly
geometric.]
•  Then
E[C]
=
200,000.
•  On
average,
have
to
buy
200,000
chocolate
bars
un:l
first
success.
€
2
E[C ]
∞
E [C ] = ∑ k 2 (1− p) k−1 p
2
k=1
€
€
2
E[C ]
∞
E [C ] = ∑ k 2 (1− p) k−1 p
2
k=1
2
Another useful fact : k (1− p)
k−1
d2
d
k +1
= 2 {(1− p) } + {(1− p) k }
dp
dp
€
€
2
E[C ]
∞
E [C ] = ∑ k 2 (1− p) k−1 p
2
k=1
2
Another useful fact : k (1− p)
k−1
d2
d
k +1
= 2 {(1− p) } + {(1− p) k }
dp
dp
[Because the right - hand side is (k + 1)k(1− p) k−1 − k(1− p) k−1 .]
€
€
€
2
E[C ]
∞
E [C ] = ∑ k 2 (1− p) k−1 p
2
k=1
2
Another useful fact : k (1− p)
k−1
d2
d
k +1
= 2 {(1− p) } + {(1− p) k }
dp
dp
[Because the right - hand side is (k + 1)k(1− p) k−1 − k(1− p) k−1 .]
∞
2  ∞



d
d
2
k +1
k
E [C ] = p 2 ∑ (1− p)  + p ∑ (1− p) 
dp  k =1
dp  k =1


€
€
€
2
E[C ]
∞
E [C ] = ∑ k 2 (1− p) k−1 p
2
k=1
2
Another useful fact : k (1− p)
k−1
d2
d
k +1
= 2 {(1− p) } + {(1− p) k }
dp
dp
[Because the right - hand side is (k + 1)k(1− p) k−1 − k(1− p) k−1 .]
∞
2  ∞



d
d
2
k +1
k
E [C ] = p 2 ∑ (1− p)  + p ∑ (1− p) 
dp  k =1
dp

 k =1

(1− p )2 / p
−1/ p
2
E[C ]
∞
E [C ] = ∑ k 2 (1− p) k−1 p
2
k=1
2
Another useful fact : k (1− p)
€
k−1
d2
d
k +1
= 2 {(1− p) } + {(1− p) k }
dp
dp
[Because the right - hand side is (k + 1)k(1− p) k−1 − k(1− p) k−1 .]
∞
2  ∞



d
d
2
k +1
k
E [C ] = p 2 ∑ (1− p)  + p ∑ (1− p) 
dp  k =1
dp

 k =1

€
(1− p )2 / p
d2  1
= p 2  −2+
dp  p
€
€
 1
p −
 p
−1/ p
2
E[C ]
∞
E [C ] = ∑ k 2 (1− p) k−1 p
2
k=1
2
Another useful fact : k (1− p)
€
k−1
d2
d
k +1
= 2 {(1− p) } + {(1− p) k }
dp
dp
[Because the right - hand side is (k + 1)k(1− p) k−1 − k(1− p) k−1 .]
∞
2  ∞



d
d
2
k +1
k
E [C ] = p 2 ∑ (1− p)  + p ∑ (1− p) 
dp  k =1
dp

 k =1

€
(1− p )2 / p
d2  1
= p 2  −2+
dp  p
€
€
−1/ p
 1
 1
d  1
p − = p − 2 + 1 −
dp  p
 p
 p
2
E[C ]
∞
E [C ] = ∑ k 2 (1− p) k−1 p
2
k=1
2
Another useful fact : k (1− p)
€
k−1
d2
d
k +1
= 2 {(1− p) } + {(1− p) k }
dp
dp
[Because the right - hand side is (k + 1)k(1− p) k−1 − k(1− p) k−1 .]
∞
2  ∞



d
d
2
k +1
k
E [C ] = p 2 ∑ (1− p)  + p ∑ (1− p) 
dp  k =1
dp

 k =1

€
(1− p )2 / p
d2  1
= p 2  −2+
dp  p
€
€
−1/ p
 1
 1 2p 1 2 1
d  1
p − = p − 2 + 1 − = 3 − = 2 −
dp  p
p p
p
 p
 p p
Variance
of
a
geometric
random
variable
2


2 1
1
1 1
2
2
var(C) = E [C ] − ( E[C]) = 2 − −   = 2 −
p
p  p
p
p
€
Variance
of
a
geometric
random
variable
2


2 1
1
1 1
2
2
var(C) = E [C ] − ( E[C]) = 2 − −   = 2 −
p
p  p
p
p
€
€
If p = 1/200,000, the standard deviation is
200,000 2 − 200,000 ≈ 200,000
Thus, your actual number candy bars until first success
may be quite far from the mean!