MA280 Probability
1
Example 5c
Suppose there are days in a year, and that each person is
independently born on day r with probability pr , r = 1, . . . ,m,∑m
r=1 pr = 1. Let Ai,j be the event that persons i and j are born
on the same day. Show P(A1,3|A1,2) ≥ P(A1,3).
Solution.P(A1,3|A1,2) = P(A1,3∪A1,2)P(A1,2)
P(A1,2) = P(∪mr=1{1 & 2 born on day r}) =
∑m
r=1 p
2
r
P(A1,3) = P(∪mr=1{1 & 3 born on day r}) =
∑m
r=1 p
2
r
P(A1,3 ∩ A1,2) = P(∪mr=1{1,2 & 3 born on day r}) =
∑m
r=1 p
3
r
So P(A1,3|A1,2) =
∑m
r=1 p
3
r /
∑m
r=1 p
2
r
Let X be a random variable such that P(X = pr ) = pr .
E [X ] =
∑m
r=1 prP(X = pr ) =
∑m
r=1 p
2
r
E [X 2] =
∑m
r=1 p
2
r P(X = pr ) =
∑m
r=1 p
3
r
Because E [X 2] ≥ (E [X ])2 =⇒ ∑mr=1 p3r ≥ (∑mr=1 p2r )2.
Thus, P(A1,3|A1,2) =
∑m
r=1 p
3
r /
∑m
r=1 p
2
r ≥
(
∑m
r=1 p
2
r )
2/
∑m
r=1 p
2
r = P(A1,3).
2
Bernoulli and Binomial distributions - definition
Suppose
n independent trials are performed,
each succeeding with probability p
Let X count the number of successes within the n trials.
Then X has the Binomial distribution with parameters n
and p or, in short, X ∼ Binom(n,p).
Let X ∼ Binom(n,p). Then X = 0,1, . . . ,n, and
p(i) = P{X = i} =
(
n
i
)
pi(1− p)n−i , i = 0,1, . . . ,n
Note∑n
i=0 p(i) =
∑n
i=0
(
n
i
)
pi(1− p)n−i = [p + (1− p)]n = 1
Actvity Explain how to find p(i)?
3
Bernoulli and Binomial distributions - definition
The special case of n = 1 in Binomial distribution is called
the Bernoulli distribution with parameter p, with
p(0) = 1− p, p(1) = p
4
Example 6a
Five fair coins are flipped. Assume the outcomes are
independent. Find the prob. mass function of the no. of heads
obtained.
Solution.X : the number of heads (successes) that appear
X is a binomial random variable with parameters (n = 5,p = 12).
P(X = 0) =
(
5
0
)(
1
2
)0(1
2
)5
=
1
32
P(X = 1) =
(
5
1
)(
1
2
)1(1
2
)4
=
5
32
P(X = 2) =
(
5
2
)(
1
2
)2(1
2
)3
=
10
32
P(X = 3) =
(
5
3
)(
1
2
)3(1
2
)2
=
10
32
P(X = 4) =
(
5
4
)(
1
2
)4(1
2
)1
=
5
32
P(X = 5) =
(
5
5
)(
1
2
)5(1
2
)0
=
1
32
5
Expectation
Let X ∼ Binom(n,p). Then
E [X ] = np
Proof. We will use a cute trick: i = ddt t
i
∣∣
t=1.
E [X ] =
n∑
i=0
i
(
n
i
)
· pi(1− p)n−i
=
n∑
i=0
(
n
i
)
d
dt
t i
∣∣∣∣∣
t=1
· pi(1− p)n−i
=
d
dt
(
n∑
i=0
(
n
i
)
(tp)i(1− p)n−i
)∣∣∣∣∣
t=1
=
d
dt
(tp + 1− p)n
∣∣∣∣
t=1
= n(tp + 1− p)n−1 · p
∣∣∣
t=1
= np.
6
Variance
Let X ∼ Binom(n,p), that is, P(X = i) = (ni )pi(1− p)n−i .
Find Var(X ).
Solution.
Var(X ) = E [X 2]− (E [X ])2 = E [X (X − 1)] + E [X ]− (E [X ])2
Observe first i(i − 1) = d2dt2 t i
∣∣∣
t=1
E [X (X − 1)] =
n∑
i=0
(
n
i
)
i(i − 1) · pi(1− p)n−i
=
n∑
i=0
(
n
i
)
d2
dt2
t i
∣∣∣∣∣
t=1
· pi(1− p)n−i
=
d2
dt2
(
n∑
i=0
(
n
i
)
t i · pi(1− p)n−i
)∣∣∣∣∣
t=1
=
d2
dt2
(tp + 1− p)n
∣∣∣∣
t=1
= n(n − 1)(tp + 1− p)n−2 · p2∣∣t=1 = n(n − 1)p2.
Var(X ) = n(n − 1)p2 + np − (np)2 = np(1− p). 7
Bernoulli distribution
For X ∼ Binom(n,p). The special case of n = 1 is called
the Bernoulli distribution with parameter p, with
p(0) = 1− p, p(1) = p
Actvity Find E [X ] and Var(X ).
8
Poisson Random Variable-Definition
The Poisson distribution is of central importance in
Probability.
For a constant λ > 0. The random variable X is Poisson
distributed with parameter λ, in short X ∼ Poi(λ), if
it is non-negative integer valued
its mass function is
p(i) = P{X = i} = e−λ · λ
i
i!
, i = 0,1,2, . . .
Actvity Verify that
∑∞
i=0 p(i) = 1.
9
Poisson - Expectation
For X ∼ Poi(λ), E [X ] = Var(X ) = λ.
Proof.
E [X ] =
∞∑
i=0
ip(i) =
∞∑
i=1
i · e−λλ
i
i!
= λ
∞∑
i=1
e−λ
λi−1
(i − 1)! = λ
∞∑
j=0
e−λ
λj
j!
= λ.
(Note the Taylor expansion for eλ)
E [X (X − 1)] =
∞∑
i=0
i(i − 1)p(i) =
∞∑
i=2
i(i − 1) · e−λλ
i
i!
= λ2
∞∑
i=2
e−λ
λi−2
(i − 2)! = λ
2
∞∑
j=0
e−λ
λj
j!
= λ2.
Var(X ) = E [X 2]− (E [X ])2
= E [X (X − 1)] + E [X ]− (E [X ])2
= λ2 + λ− λ2 = λ. Actvity Find E [X 3].
10
Poisson approximation of Binomial
Poisson approximation of Binomial:
Fix λ > 0, and suppose that Yn ∼ Binom(n,p) with p = p(n)
in such a way that n · p → λ. Then the distribution of Yn
converges to Poisson (λ) :
∀i ≥ 0 P {Yn = i} −→
n→∞ e
−λλ
i
i!
.
That is, take Y ∼ Binom(n,p) with large n, small p, such
that np ≃ λ. Then Y is approximately Poisson (λ)
distributed.
11
Geometric random variable - definition
Suppose that independent trials, each succeeding with
probability p, are repeated until the first success.
The total number X of trials made has theGeometric (p)
distribution
(in short, X ∼ Geom(p) ).
Suppose X ∼ Geom(p) ).
X can take on positive integers, with probabilities
p(i) = (1− p)i−1 · p, i = 1,2, . . .
Actvity Explain how to find p(i).
The above is a mass function by noting
∞∑
i=1
p(i) =
∞∑
i=1
(1− p)i−1 · p = p
1− (1− p) = 1
12
Some properties of Geometric random variables
Suppose X is a Geometric (p) random variable
P{X ≥ k} = (1− p)k−1 for k ≥ 1(The prob that we have at
least k − 1 failures).
Actvity Show P{X ≥ k} = (1− p)k−1.
The Geometric random variable is (discrete) memoryless:
for every k ≥ 1,n ≥ 0
P{X ≥ n + k | X > n} = P{X ≥ k}
Proof.
13
Geometric random variable - Expectation
For a Geometric( p) random variable X ,
E [X ] =
1
p
Proof.
E [X ] =
∞∑
i=1
i · (1− p)i−1p
=
∞∑
i=0
d
dt
t i
∣∣∣∣∣
t=1
· (1− p)i−1p = d
dt
( ∞∑
i=0
t i · (1− p)i−1p
)∣∣∣∣∣
t=1
=
p
1− p ·
d
dt
1
1− (1− p)t
∣∣∣∣
t=1
=
p
1− p ·
1− p
(1− (1− p))2 =
1
p
.
14
Geometric random variable - Variance
For a Geometric( p) random variable X , Var(X ) = 1−pp2
Proof.
E [X (X − 1)] =
∞∑
i=1
i(i − 1) · (1− p)i−1p
=
∞∑
i=0
d2
dt2
t i
∣∣∣∣∣
t=1
· (1− p)i−1p = d
dt
( ∞∑
i=0
t i · (1− p)i−1p
)∣∣∣∣∣
t=1
=
p
1− p ·
d2
dt2
1
1− (1− p)t
∣∣∣∣
t=1
=
p
1− p ·
2(1− p)2
(1− (1− p))3 =
2(1− p)
p2
.
Var(X ) = E [X 2]− (E [X ])2
= E [X (X − 1)] + E [X ]− (E [X ])2
=
2(1− p)
p2
+
1
p
−
(
1
p
)2
=
1− p
p2
.
15