Download Distribution P ertinen t Info/Inputs Ω, the sample space (i.e. the set of

1
Pertinent
Info/Inputs
# of trials N
and probability
of success p
N trials, n
different types
of success,
p1 =
probability of
type 1 success,
p2 = ... , and
pn =
probability of
type n-success.
Also, number
of success for
each type, i.e.
ki = # of i
successes with
the
Pn condition
i=1 ki = N .
Probability of
termination p
Distribution
Binomial
(Discrete)
Multinomial
(Discrete)
Geometric
(Discrete)
We use the
Russian
Roullette
example. An
outcome would
be death on
the third shot.
Another
outcome is
living forever.
There are
infinitely many
outcomes.
Different
outcomes after
N trials, e.g.
success of type
1 for all trials.
The number of
outcomes is
kN
Different
outcomes for
N trials, e.g.
all successes
for all N trials
is one outcome
in our sample
space. The
number of
elements in our
sample space is
2N
Ω, the sample
space (i.e. the
set of
outcomes)
X(outcome ) =
length of the
game
X(outcome) =
(# success of
type 1, success
of type 2, . . .,
success of
types n). Note
our random
variable
actually spits
out a n-tuple.
X(outcome)
=# of
successes
X, the random
variable
(1) Flip a coin
N times
(p = 1/2 )
(2)Take a Test
with N
Questions
(probability of
answering right
p) (3) Meet N
people that
have a disease
(probability of
contraction p)
(1) have N
clients, n
workouts and
the probability
that any client
wants to do
workout i is pi
(2) Have urn of
balls colored in
n different
ways and pick
N of them.
Also, have the
probability is
pi for each
color i ranging
from 1 through
n.
(1) Russian
Roullette with
six shooter and
spin after each
shot (p = 1/6)
(2) Quiz show
and you stop
playing when
you get one
wrong
(probability of
getting
question
WRONG is p)
(3) Roll a die
until you get a
six (p = 1/6)
Situations
modeled by
Random
Variable
N
k
«
pk (1 −
P (X = k) =
(p − 1)k−1 p
n
p1 1 . . . pk
n
k
P (X =
(k
=
„ 1 , . . . , kn )) «
N
k1 . . . kn
p)N −k
„
PDF fX (x)
x ∈ R or PMF
P (X = k)
k∈N
P (X ≥ k) =
(1 − p)k−1 (for
Russian.
Roulette,
means game
lasts until
k-shot) while
F (X) =
P (X ≤ k) =
1 − (1 − p)k
(means
someone gets
shot before or
on kth turn)
Not applicable
(1 − p)k−i
Continuous
CDF F (x) =
P (X ≤ x)
x ∈ R/
Discrete CDF
P (X ≤ k)
k∈N „
«
Pk
N
pi
i=0
k
1
p
We often only
want to know
the average #
of success of
type i, in this
case we have
N pi
Np
E(X)
1−p
p2
Variance of #
of success of
type i is
N pi (1 − pi ).
Np(1-p)
V(X)
In theory, note
that the
geometric
random
variable takes
on value
infinity if the
game lasts
forever.
However, this
has probability
0 since
P (X = ∞) =
limn→∞ pn =
0 as 0 < p < 1.
A variant of
the geometric
distribution is
stopping a
game after
some # of
sucesses.
Finding the
PMF is the
content of
12.4.76.
If N is very
large, then
P (X = k) ≈
P (Xλ = k),
where Xλ is
Poisson
random
variable with
parameter λ
and λ is
obtained by
E(X) = λ =
N p.
Might be best
to not
memorize
binomial and
multinomial,
but to be able
to think of
them combinatorially. See
workout
example below.
Comments
2
Parameter λ.
Sometimes you
will be given
the average
lifespan
L = E(X) and
so λ = 1/L
Must be given
σ (standard
deviation), µ
(mean)
λ(t) the hazard
rate function
Normal
Distribution
(continuous)
Hazard-Rate
(continuous)
interval [a, b]
Uniform
(Continuous)
Exponential
(continuous)
λ denotes the
expected value
of success per
some fixed
interval unit of
time. We are
also given an
interval of
time, say t
Poisson
(Discrete)
We use the
example of
distributions of
height. The
sample space is
the world and
each person’s
heights. The µ
and σ are
mean and SD
of the heights.
Same as
exponential
We use the
lightbulb
example. The
sample space
would be the
various times
at which the
lightbulb could
go out once it
is plugged in.
Sample space
would be all
the numbers
between [a, b]
Sample space
would be the
number of
“hits” in some
unit time.
Note we can
have an
arbitrarily high
number of hits,
so our sample
space is infinite
Same as
exponential
X(outcome) =
the number of
people of a
particular
height
X( outcome)=
the time when
the lightbulb
goes out.
X(outcome)=
the number
picked
X(outcome) =
# of hits
(1) Aging (2)
More exotic
lightbulbs
(1) Test Scores
(2) Heights
(3)Wealth
(1) Pick a
number on
[a, b] at
random (2)
You arrive at a
bus stop
between 10 and
11am such that
your arrival is
uniformly
distributed.
(1) Radioactive
particle decay
(2) Lightbulb/
battery
durations
(1) Patients
coming into an
ER (λ =
patients per
hour) (2) Calls
into a police
station (λ =
calls per day).
(3) When
lifetime of bulb
exponentially
distributed
with parameter
λ, and
interested # of
times we have
to replace in
some time.
Same λ’s for
exp. and
poisson here!
e−λt .
S(x) = P (X >
x). Can show
f (x) = −S 0 (x)
when x ≥ 0
and 0
otherwise (on
homework
using FTC)
e
2πσ
−(x−µ)2 /(2σ)2
f (x) = √ 1
f (x) = λe−λx
when x ≥ 0,
and f (x) = 0
when x < 0
1
f (x) = b−a
when x in [a, b]
otherwise
f (x) = 0.
(λt)k
k!
e−λ or if
we are given
some unit of
time t, then we
can write:
P (N (t) = k) =
λk
k!
P (X = k) =
F (x) = 1 − S(t)
e
√
2πσ
−(x−µ)2 /(2σ)2
−∞
F (x) =
1 − e−λx (for
the lightbulb
example this
means the
probability the
lightbulb will
last at most x
units of time)
and also
S(x) =
1 − F (x) (e.g.
the probability
the lightbulb
will last more
than x units of
time).
F
R x(x) =
f (x) =
R−∞
x
1
x
F (x) = b−a
when x is in
[a, b].
F (x) = 0,
when x ≤ a
and F (x) = 1
when x ≥ b.
P (X ≤ k) =
Pk
λi −λ
i=0 i! e
and similarly,
for
P (N (t) ≤ k).
+ b)
0 S(t)dt (on
homework
using
integration by
parts)
R∞
µ
1
λ
1 (a
2
λ
− a)2
Not relevant
σ
1
λ2
1 (b
12
λ
When λ(t) = λ
this is just the
exponential
distribution.
68% occur
within one SD,
95% within
two SD’s and
99% within 3
SD’s. You use
the table after
making the
substitution
u = (x − µ)/σ
The
Memoryless
Property:
P (X >
t + h|X > t) =
P (X > h). Use
the lightbulb
example to see
what this
means
intuitively.
Also, see
Poisson
Distribution
Comments.
You computed
expectations of
max and min’s
of independent
uniform
distributions
on your
homework
Notice that
P (N (t) = 0) =
e−λt , which is
the survival
function of an
exponetial
random
variable, thus
enabling us to
relate the two
distributions
(see the (3) in
situations
modeled by
exponential).
For examples
(3), if Tk
denotes the
waiting time
for
exponentially
distributed
random
variables, then
P (N (t) < k) =
P (Tk > t).