Download Presentation Thursday January 31

Show that the function defined by Benford’s Law is in fact a
probability distribution function on the set
D = {1, 2, 3, 4, 5, 6, 7, 8, 9}.
k +1⎞
⎛
i.e., Show that the function P(k) = log10
on D satisfies
⎝ k ⎠
9
(1). 0 ≤ P(k) ≤ 1
(2).
∑ P(k) = 1
k =1
⎛ d + 1⎞
0 ≤ p(d) = P(X = d) = log ⎜
≤ log ( 2 ) < 1
⎟
⎝ d ⎠
Show that the function defined by Benford’s Law is in fact a
probability distribution function on the set
D = {1, 2, 3, 4, 5, 6, 7, 8, 9}.
k +1⎞
⎛
i.e., Show that the function P(k) = log10
on D satisfies
⎝ k ⎠
9
(2).
∑ P(k) = 1
k =1
⎛ d + 1⎞
∑ p(d) = ∑ log ⎜⎝ d ⎟⎠ =
d=1
d=1
9
9
⎛ 3⎞
⎛ 4⎞
⎛ 10 ⎞
log ( 2 ) + log ⎜ ⎟ + log ⎜ ⎟ ++ log ⎜ ⎟
⎝ 2⎠
⎝ 3⎠
⎝ 9⎠
⎛ d + 1⎞
∑ p(d) = ∑ log ⎜⎝ d ⎟⎠ =
d=1
d=1
9
9
⎛ 3⎞
⎛ 4⎞
⎛ 10 ⎞
log ( 2 ) + log ⎜ ⎟ + log ⎜ ⎟ ++ log ⎜ ⎟
⎝ 2⎠
⎝ 3⎠
⎝ 9⎠
⎛ 2 3 4 5 6 7 8 9 10 ⎞
log ⎜ × × × × × × × × ⎟
⎝1 2 3 4 5 6 7 8 9 ⎠
= log (10 ) = 1
Actually(Benford’s(Law(can(be(extended(to(the(two(
most(signi:icant,(or(three(most(signi:icant,(etc.(digits(
in(a(randomly(chosen(river(length((or(similar(such(
data)(in(the(following(way.(
Let(D(denote(the(set(of(all(2Adigit(numbers.(Then(
⎛ d + 1⎞
p(d) = P(X = d) = log ⎜
⎝ d ⎟⎠
10 ≤ d ≤ 99
is(in(fact(a(probability(distribution(function(on(D.
By the same trick as before,
⎛ d + 1⎞
∑ log ⎜⎝ d ⎟⎠ = log (100 ) − log (10 ) = 2 − 1 = 1
d=10
99
Actually(Benford’s(Law(can(be(extended(to(the(two(
most(signi:icant,(or(three(most(signi:icant,(etc.(digits(
in(a(randomly(chosen(river(length((or(similar(such(
data)(in(the(following(way.(
Let(D(denote(the(set(of(all(2Adigit(numbers.(Then(
⎛ d + 1⎞
p(d) = P(X = d) = log ⎜
⎝ d ⎟⎠
10 ≤ d ≤ 99
is(in(fact(a(probability(distribution(function(on(D.
Given(that(a(randomly(selected(river(length(begins(
with(a(1(what(is(the(probability(that(the(second(digit(is(
a(3?
Let(D(denote(the(set(of(all(2Adigit(numbers.(Then(
⎛ d + 1⎞
p(d) = P(X = d) = log ⎜
⎝ d ⎟⎠
10 ≤ d ≤ 99
Let(F1(denote(the(event(that(the(length(begins(with(a(1(
and(S3(the(event(that(the(second(digit(is(a(3.
(
Let(L13(denote(the(event(that(the(length(begins(with(the(
digits(13.
(
)
14
log
P ( F1 ∩ S3) P ( L13)
13
P ( S3 | F1) =
=
=
= 0.10692
P ( F1)
P(F1)
log(2)
What(is(the(probability(that(the(second(digit(in(a(
randomly(chosen(river(length(is(a(0?
⎛ 10d + 1 ⎞
∑ p (10d ) = ∑ log ⎜⎝ 10d ⎟⎠ = 0.1197
d =1
d =1
9
9
Suppose(that(a(bag(contains(20(red(marbles(and(10(
blue(marbles.(The(marbles(are(chosen(one(at(a(time(
without(replacement?
What(is(the(probability(that(the(6th(blue(marble(is(
the(12th(marble(drawn?(
So(that(means(that(there(are(5(blue(marbles(among(
the(:irst(11(drawn(and(then(the(12th(marble(is(blue.
Hint:(Let(A(and(B(be(the(events:
A(A(there(are(5(blue(marbles(among(the(:irst(11(drawn.
B(A(the(12th(marble(is(blue.
Then(we(want(the(value(of(((((((((((((((((((.
P(A ∩ B)
Suppose(that(a(bag(contains(20(red(marbles(and(10(
blue(marbles.(The(marbles(are(chosen(one(at(a(time(
without(replacement?
Hint:(Let(A(and(B(be(the(events:
A(A(there(are(5(blue(marbles(among(the(:irst(11(drawn.
B(A(the(12th(marble(is(blue.
Then(we(want(the(value(of(((((((((((((((((((.
P(A ∩ B)
⎡ ⎛ 20 ⎞ ⎛ 10 ⎞ ⎤
⎢⎜ ⎟ ⎜ ⎟ ⎥
5 ⎢⎝ 6 ⎠ ⎝ 5 ⎠ ⎥
P(A ∩ B) = P(B | A)P(A) = ⋅
19 ⎢ ⎛ 30 ⎞ ⎥
⎢ ⎜ ⎟ ⎥
⎣ ⎝ 11 ⎠ ⎦
The(probability(of(winning(the(pick(three(
lottery(in(SC(on(any(given(day(is(.001
What(is(the(probability(of(winning(at(
least(once(in(1000(games?
1 ⎞
⎛
1 − ⎜1 −
⎝ 1000 ⎟⎠
1000
1
≈1−
e
which(is(approximately(0.6321
The(probability(of(winning(the(pick(three(
lottery(in(SC(on(any(given(day(is(.001
What(is(the(probability(of(winning(at(
least(once(in(100(games?
1 ⎞
⎛
1− ⎜ 1−
⎝ 1000 ⎟⎠
100
⎡⎛
1 ⎞
= 1− ⎢⎜ 1−
⎟⎠
⎝
1000
⎣
which(is(approximately(0.095
1000
1
10
⎤
⎛ 1⎞
⎥ ≈ 1− ⎜⎝ ⎟⎠
e
⎦
1
10
= 1−
1
e
1
10
Suppose that we have a box that contains
96 blue marbles and 4 red marbles.
We perform 200 independent trials of picking
a marble from the box.
What is the probability we never pick a red
marble?
The probability we do not pick a red marble
on any given trial is
96
4
= 1−
100
100
We perform 200 independent trials of picking
a marble from the box.
What is the probability we never pick a red
marble?
The probability we do not pick a red marble
on any given trial is
96
4
= 1−
100
100
The probability we never pick a red marble is then
4 ⎞
⎛
⎜⎝ 1−
⎟⎠
100
200
⎡⎛
4 ⎞
= ⎢⎜ 1−
⎟⎠
⎝
100
⎣
100
2
⎤ ⎛ 1 ⎞2 1
⎥ ≈ ⎜⎝ 4 ⎟⎠ = 8
e
e
⎦
We perform 200 independent trials of picking
a marble from the box.
What is the probability we never pick a red
marble?
The probability we do not pick a red marble
on any given trial is
n
⎛ a⎞
a
lim ⎜ 1+ ⎟ = e
n→∞ ⎝
n⎠
a=–4
The probability we never pick a red marble is then
4 ⎞
⎛
⎜⎝ 1−
⎟⎠
100
200
⎡⎛
4 ⎞
= ⎢⎜ 1−
⎟⎠
⎝
100
⎣
100
2
⎤ ⎛ 1 ⎞2 1
⎥ ≈ ⎜⎝ 4 ⎟⎠ = 8
e
e
⎦
The(Gamecocks(and(Clemson(will(play(each(other(in(
football(once(a(year(over(the(next(10Ayear(period.(Given(
that(the(Gamecocks(are(60%(likely(to(win(each(game,(
What(is(the(probability(that(the(gamecocks(win(
exactly(7(games?
How many ways are there for the Gamecocks to win 7 games?
⎛ 10 ⎞
⎜⎝ 7 ⎟⎠
What is the probability for any specific way that the
Gamecocks can win 7 games,
0.6 7 × 0.4 3
The(Gamecocks(and(Clemson(will(play(each(other(in(
football(once(a(year(over(the(next(10Ayear(period.(Given(
that(the(Gamecocks(are(60%(likely(to(win(each(game,(
What(is(the(probability(that(the(gamecocks(win(
exactly(7(games?
⎛ 10 ⎞
7
3
0.6
×
0.4
⎜⎝ 7 ⎟⎠
Problems such as this occur frequently.
Suppose that we have a sequence of n independent Bernoulli
trials each with probability of success p and of course
probability q = 1 – p probability of failure.
Let X denote the number of successes in the n trials.
X is said to be a binomial random variable.
We may also express this by saying that X is b(n,p).
For example, we may roll a die 12 times and let X denote the
number of times a 3 appears.
For a given 0 ≤ k ≤ n, what is P(X = k)?
⎛ n ⎞ k n−k
P(X = k) = ⎜ ⎟ p q
⎝ k⎠
1
n = 12, p =
6
The mean or Expected Value of a random variable X is
E(X) = ∑ kP ( X = k )
k∈S
We roll a die and let X = number that appears.
6
6
6
1
1
21
E(X) = ∑ kP ( X = k ) = ∑ k = ∑ k =
= 3.5
6 k=1
6
k=1
k=1 6
The pdf of this X is the constant function:
1
p(k) = P ( X = k ) = , k ∈{1, 2, 3, 4, 5, 6} .
6
The mean or Expected Value of a random variable X is
E(X) = ∑ kP ( X = k )
k∈S
We roll a die and let X = number that appears. E(X) = 3.5
6
6
6
1
1
21
E(X) = ∑ kP ( X = k ) = ∑ k = ∑ k =
= 3.5
6 k=1
6
k=1
k=1 6
A random variable is discrete if it can take on only a finite or
countably infinite number of values.
The Probability Distribution Function or pdf of a discrete
random variable is p(k) = P(X = k)
We draw six cards from a deck of 52, X = number of hearts.
What is E(X)?
⎛ 13⎞ ⎛ 39 ⎞
6
⎜⎝ k ⎟⎠ ⎜⎝ 6 − k ⎟⎠
6
µ = E(X) = ∑ kP ( X = k ) = ∑ k
= 1.5
⎛ 52 ⎞
k=0
k=0
⎜⎝ k ⎟⎠
⎛ 13⎞ ⎛ 39 ⎞
⎜⎝ k ⎟⎠ ⎜⎝ 6 − k ⎟⎠
for 0 ≤ k ≤ 13
The pdf of X is p(k) =
⎛ 52 ⎞
⎜⎝ k ⎟⎠
If X is a Bernoulli Random Variable with
X = 1 for a success with probability p
X = 0 for a failure with probability q = 1– p.
E(X) = p
If X is a binomial b(n,p) random variable, what is E(X)?
It’s what you’d expect – and we’ll verify that soon.
E(X) = np
⎛ n ⎞ k n−k
(x + y) = ∑ ⎜ ⎟ x y
k=0 ⎝ k ⎠
n
Binomial Theorem
n
⎛ n ⎞ k n−k
1 = ( p + q) = ∑ ⎜ ⎟ p q
k=0 ⎝ k ⎠
n
n