Lecture Note
Basic Statistics: Part II - Probability Distributions
-
University:
University of Edinburgh -
Course:
MATH08051 | Statistics (Year 2) Academic year:
2017
-
Views:
272
Pages:
111
Author:
licheefdnm
Basic Statistics
Part II: Probability distribution
September 2017 Outline
1. Probability theory
I
I
Conditional probabilities and independence
Bayes’ theorem
2. Probability distributions
I
I
I
I
Discrete and continuous probability functions
Probability density function & cumulative distribution function
Binomial, Poisson and Normal distribution
E[X] and V[X]
3. Descriptive statistics
I
I
I
I
Sample statistics (mean, variance, percentiles)
Graphs (box plot, histogram)
Data transformations (log transformation, unit of measure)
Correlation vs. Causation
4. Statistical inference
I
I
I
I
I
Population vs. sample
Law of large numbers
Central limit theorem
Confidence intervals
Hypothesis testing and p-values
1 / 68 Random variables
Most of the outcomes or events we have considered so far have
been non-numerical, e.g. either head or tail. If the outcome of an
experiment is numerical, we call the variable that is determined by
the experiment a random variable.
Random variables may be either discrete (e.g. the number of days
the sun shines) or continuous (e.g. your salary after graduating
from the MSc). In contrast to a continuous random variable, we
can list the distinct potential outcomes of a discrete random
variable.
2 / 68 Random variables
Most of the outcomes or events we have considered so far have
been non-numerical, e.g. either head or tail. If the outcome of an
experiment is numerical, we call the variable that is determined by
the experiment a random variable.
Random variables may be either discrete (e.g. the number of days
the sun shines) or continuous (e.g. your salary after graduating
from the MSc). In contrast to a continuous random variable, we
can list the distinct potential outcomes of a discrete random
variable.
Notation
Random variables are usually denoted by capital letters, e.g. X .
The corresponding realisations are denote by small letters, e.g. x.
2 / 68 Should you make the bet?
Example III.1
I propose the following game. We toss a fair coin 10 times. If head
appears 4 times or less, I pay you £2. If head appears more than 4 times,
you pay me £1. Should you make the bet?
3 / 68 Should you make the bet?
Example III.1
I propose the following game. We toss a fair coin 10 times. If head
appears 4 times or less, I pay you £2. If head appears more than 4 times,
you pay me £1. Should you make the bet?
Let’s try to formalise the problem. Let the random variables
X1 , X2 , . . . , X10 be defined such that
1 if head appears on the ith toss
Xi =
for i = 1, . . . , 10.
0 if tail appears on the ith toss
3 / 68 Should you make the bet?
Furthermore, let the random variable Y denote the number of heads.
Clearly,
Y = X1 + X1 + . . . + X10 .
If the realisation of Y is greater than 4, I win.
4 / 68 Should you make the bet?
Furthermore, let the random variable Y denote the number of heads.
Clearly,
Y = X1 + X1 + . . . + X10 .
If the realisation of Y is greater than 4, I win.
Let P(Y = y) denote the probability that Y takes the value y.
Accordingly, P(Y ≤ 4) is the probability that we obtain 4 or less heads
and P(Y > 4) is the probability that we obtain more than 4 heads.
When would you make the bet?
4 / 68 Should you make the bet?
Furthermore, let the random variable Y denote the number of heads.
Clearly,
Y = X1 + X1 + . . . + X10 .
If the realisation of Y is greater than 4, I win.
Let P(Y = y) denote the probability that Y takes the value y.
Accordingly, P(Y ≤ 4) is the probability that we obtain 4 or less heads
and P(Y > 4) is the probability that we obtain more than 4 heads.
When would you make the bet?
Your expected value is
E[V ] = P(Y ≤ 4) · £2 + P(Y > 4) · (£−1)
where V is the money you get. If E[V ] > 0 (and you are risk neutral),
you’ll choose to play.
4 / 68 Should you make the bet?
Expected value
The expected value of a discrete random variable X is denoted by E[X ]
and given by
k
X
E[X ] = x1 P(X =x1 ) + x2 P(X =x2 ) + · · · + xk P(X =xk ) =
xi P(X =xi )
i=1
where k is the number of distinct outcomes.
5 / 68 Should you make the bet?
To solve the problem, we need to find P(Y ≤ 4) and P(Y > 4).
From the additive law (Rule 4), we know that
P(Y ≤4) = P(Y =0 ∪ Y =1 ∪ Y =2 ∪ Y =3 ∪ Y =4)
= P(Y =0) + P(Y =1) + P(Y =2) + P(Y =3) + P(Y =4)
P(Y >4) = P(Y =5) + P(Y =6) + P(Y =7) + P(Y =8) + P(Y =9) + P(Y =10)
Hence, we need to find P(Y = yi ) for i = 0, . . . , 10.
6 / 68 Discrete probability distribution
It is common to denote the probability distribution of a discrete random
variable Y by f (y).
Discrete probability distribution
The probability distribution or probability mass function of a discrete
random variable X associates with each of the distinct potential
outcomes xi (i = 1, . . . , k) a probability P(X = xi ). That is,
f (xi ) = P(X = xi ).
Pk
The sum of the probabilities add up to 1, i.e. i f (xi ) = 1.
7 / 68 Discrete probability distribution
Two examples:
8 / 68 Discrete probability distribution
Two examples:
Example III.2 (Discrete Uniform Distribution)
Let X be the result from rolling a fair dice. The probability distribution is
simply
1/6
for x = {1, 2, . . . , 6}
f (x) = P(X = x) =
.
0
otherwise
This probability distribution is an example for a discrete uniform distributions.
8 / 68 Discrete probability distribution
Two examples:
Example III.2 (Discrete Uniform Distribution)
Let X be the result from rolling a fair dice. The probability distribution is
simply
1/6
for x = {1, 2, . . . , 6}
f (x) = P(X = x) =
.
0
otherwise
This probability distribution is an example for a discrete uniform distributions.
Bernoulli distribution
It is said that a random variable X has a Bernoulli distribution with parameter
P(X = 1) = p (i.e. probability of success) if X can take only the values 1
(success) and 0 (failure). The probability distribution is given by
(
f (x) =
p
1−p
0
if x = 1
if x = 0
otherwise
8 / 68 Binomial coefficient & binomial distribution
Let’s start with f (0) = P(Y =0) which is the probability of obtaining no
heads. Using the multiplicative law,
10
P(Y =0) = P(X1 =0)P(X2 =0) . . . P(X10 =0) = (1/2)
= 0.00097656
9 / 68 Binomial coefficient & binomial distribution
Let’s start with f (0) = P(Y =0) which is the probability of obtaining no
heads. Using the multiplicative law,
10
P(Y =0) = P(X1 =0)P(X2 =0) . . . P(X10 =0) = (1/2)
= 0.00097656
Now, f (1) = P(Y =1). Since we are interested in the number of heads,
we have to take into account that there is more than one combination
that results in 1 head.
P(Y =1) = P(X1 =1)P(X2 =0) . . . P(X10 =0)
+ P(X1 =0)P(X2 =1) . . . P(X10 =0)
..
.
+ P(X1 =0)P(X2 =0) . . . P(X10 =1)
= 10 · (1/2)
10
= 0.00976563
9 / 68 Binomial coefficient & binomial distribution
Now, f (2) = P(Y =2). How many combinations are there that yield
2 heads out of 10 tosses? Given that the first toss produces a head, there
are 9 combinations that yield two heads in total. And so on...
10 / 68 Binomial coefficient & binomial distribution
combination
Now, f (2) = P(Y =2). How many combinations are there that yield
2 heads out of 10 tosses? Given that the first toss produces a head, there
are 9 combinations that yield two heads in total. And so on...
1
2
3
4
5
6
7
8
9
1
H
H
H
H
H
H
H
H
H
2
H
T
T
T
T
T
T
T
T
3
T
H
T
T
T
T
T
T
T
4
T
T
H
T
T
T
T
T
T
toss
5 6
T T
T T
T T
H T
T H
T T
T T
T T
T T
7
T
T
T
T
T
H
T
T
T
8
T
T
T
T
T
T
H
T
T
9
T
T
T
T
T
T
T
H
T
10
T
T
T
T
T
T
T
T
H
10 / 68 Binomial coefficient & binomial distribution
combination
Now, f (2) = P(Y =2). How many combinations are there that yield
2 heads out of 10 tosses? Given that the first toss produces a head, there
are 9 combinations that yield two heads in total. And so on...
1
2
3
4
5
6
7
8
9
1
H
T
T
T
T
T
T
T
T
2
H
H
H
H
H
H
H
H
H
3
T
H
T
T
T
T
T
T
T
4
T
T
H
T
T
T
T
T
T
toss
5 6
T T
T T
T T
H T
T H
T T
T T
T T
T T
7
T
T
T
T
T
H
T
T
T
8
T
T
T
T
T
T
H
T
T
9
T
T
T
T
T
T
T
H
T
10
T
T
T
T
T
T
T
T
H
10 / 68 Binomial coefficient & binomial distribution
combination
Now, f (2) = P(Y =2). How many combinations are there that yield
2 heads out of 10 tosses? Given that the first toss produces a head, there
are 9 combinations that yield two heads in total. And so on...
1
2
3
4
5
6
7
8
9
1
H
T
T
T
T
T
T
T
T
2
T
H
T
T
T
T
T
T
T
3
H
H
H
H
H
H
H
H
H
4
T
T
H
T
T
T
T
T
T
toss
5 6
T T
T T
T T
H T
T H
T T
T T
T T
T T
7
T
T
T
T
T
H
T
T
T
8
T
T
T
T
T
T
H
T
T
9
T
T
T
T
T
T
T
H
T
10
T
T
T
T
T
T
T
T
H
10 / 68 Binomial coefficient & binomial distribution
combination
Now, f (2) = P(Y =2). How many combinations are there that yield
2 heads out of 10 tosses? Given that the first toss produces a head, there
are 9 combinations that yield two heads in total. And so on...
1
2
3
4
5
6
7
8
9
1
H
T
T
T
T
T
T
T
T
2
T
H
T
T
T
T
T
T
T
3
T
T
H
T
T
T
T
T
T
4
H
H
H
H
H
H
H
H
H
toss
5 6
T T
T T
T T
H T
T H
T T
T T
T T
T T
7
T
T
T
T
T
H
T
T
T
8
T
T
T
T
T
T
H
T
T
9
T
T
T
T
T
T
T
H
T
10
T
T
T
T
T
T
T
T
H
10 / 68 Binomial coefficient & binomial distribution
combination
Now, f (2) = P(Y =2). How many combinations are there that yield
2 heads out of 10 tosses? Given that the first toss produces a head, there
are 9 combinations that yield two heads in total. And so on...
1
2
3
4
5
6
7
8
9
1
H
T
T
T
T
T
T
T
T
2
T
H
T
T
T
T
T
T
T
3
T
T
H
T
T
T
T
T
T
4
T
T
T
H
T
T
T
T
T
toss
5 6
H T
H T
H T
H T
H H
H T
H T
H T
H T
7
T
T
T
T
T
H
T
T
T
8
T
T
T
T
T
T
H
T
T
9
T
T
T
T
T
T
T
H
T
10
T
T
T
T
T
T
T
T
H
10 / 68 Binomial coefficient & binomial distribution
combination
Now, f (2) = P(Y =2). How many combinations are there that yield
2 heads out of 10 tosses? Given that the first toss produces a head, there
are 9 combinations that yield two heads in total. And so on...
1
2
3
4
5
6
7
8
9
1
H
T
T
T
T
T
T
T
T
2
T
H
T
T
T
T
T
T
T
3
T
T
H
T
T
T
T
T
T
4
T
T
T
H
T
T
T
T
T
toss
5 6
T H
T H
T H
T H
H H
T H
T H
T H
T H
7
T
T
T
T
T
H
T
T
T
8
T
T
T
T
T
T
H
T
T
9
T
T
T
T
T
T
T
H
T
10
T
T
T
T
T
T
T
T
H
10 / 68 Binomial coefficient & binomial distribution
combination
Now, f (2) = P(Y =2). How many combinations are there that yield
2 heads out of 10 tosses? Given that the first toss produces a head, there
are 9 combinations that yield two heads in total. And so on...
1
2
3
4
5
6
7
8
9
1
H
T
T
T
T
T
T
T
T
2
T
H
T
T
T
T
T
T
T
3
T
T
H
T
T
T
T
T
T
4
T
T
T
H
T
T
T
T
T
toss
5 6
T T
T T
T T
T T
H T
T H
T T
T T
T T
7
H
H
H
H
H
H
H
H
H
8
T
T
T
T
T
T
H
T
T
9
T
T
T
T
T
T
T
H
T
10
T
T
T
T
T
T
T
T
H
10 / 68 Binomial coefficient & binomial distribution
combination
Now, f (2) = P(Y =2). How many combinations are there that yield
2 heads out of 10 tosses? Given that the first toss produces a head, there
are 9 combinations that yield two heads in total. And so on...
1
2
3
4
5
6
7
8
9
1
H
T
T
T
T
T
T
T
T
2
T
H
T
T
T
T
T
T
T
3
T
T
H
T
T
T
T
T
T
4
T
T
T
H
T
T
T
T
T
toss
5 6
T T
T T
T T
T T
H T
T H
T T
T T
T T
7
T
T
T
T
T
T
H
T
T
8
H
H
H
H
H
H
H
H
H
9
T
T
T
T
T
T
T
H
T
10
T
T
T
T
T
T
T
T
H
10 / 68 Binomial coefficient & binomial distribution
combination
Now, f (2) = P(Y =2). How many combinations are there that yield
2 heads out of 10 tosses? Given that the first toss produces a head, there
are 9 combinations that yield two heads in total. And so on...
1
2
3
4
5
6
7
8
9
1
H
T
T
T
T
T
T
T
T
2
T
H
T
T
T
T
T
T
T
3
T
T
H
T
T
T
T
T
T
4
T
T
T
H
T
T
T
T
T
toss
5 6
T T
T T
T T
T T
H T
T H
T T
T T
T T
7
T
T
T
T
T
T
H
T
T
8
T
T
T
T
T
T
T
H
T
9
H
H
H
H
H
H
H
H
H
10
T
T
T
T
T
T
T
T
H
10 / 68 Binomial coefficient & binomial distribution
combination
Now, f (2) = P(Y =2). How many combinations are there that yield
2 heads out of 10 tosses? Given that the first toss produces a head, there
are 9 combinations that yield two heads in total. And so on...
1
2
3
4
5
6
7
8
9
1
H
T
T
T
T
T
T
T
T
2
T
H
T
T
T
T
T
T
T
3
T
T
H
T
T
T
T
T
T
4
T
T
T
H
T
T
T
T
T
toss
5 6
T T
T T
T T
T T
H T
T H
T T
T T
T T
7
T
T
T
T
T
T
H
T
T
8
T
T
T
T
T
T
T
H
T
9
T
T
T
T
T
T
T
T
H
10
H
H
H
H
H
H
H
H
H
10 / 68 Binomial coefficient & binomial distribution
combination
Now, f (2) = P(Y =2). How many combinations are there that yield
2 heads out of 10 tosses? Given that the first toss produces a head, there
are 9 combinations that yield two heads in total. And so on...
1
2
3
4
5
6
7
8
9
1
H
T
T
T
T
T
T
T
T
2
T
H
T
T
T
T
T
T
T
3
T
T
H
T
T
T
T
T
T
4
T
T
T
H
T
T
T
T
T
toss
5 6
T T
T T
T T
T T
H T
T H
T T
T T
T T
7
T
T
T
T
T
T
H
T
T
8
T
T
T
T
T
T
T
H
T
9
T
T
T
T
T
T
T
T
H
10
H
H
H
H
H
H
H
H
H
This gives us 10 · 9. This approach has the problem of double counting.
Each combination appears twice. So we have to divide by 2 and get
(10 · 9)/2 distinct combinations. Thus,
10
10 · 9 1
P(Y = 2) =
= 0.04394531.
2
2
10 / 68 Binomial coefficient & binomial distribution
For P(Y = 3), P(Y = 4), . . . this gets even more complicated.
11 / 68 Binomial coefficient & binomial distribution
For P(Y = 3), P(Y = 4), . . . this gets even more complicated.
Binomial coefficient
Suppose that there is a set of n distinct elements from which it is desired
to choose a subset of k elements (typically 1 ≤ k ≤ n). The binomial
coefficient gives the number of ways k elements can be selected from n
elements. The binomial coefficient is defined as
n
n!
n
Cn,k = Ck =
=
.
k
k!(n − k)!
where k! = k(k − 1)(k − 2) . . . 1 and 0! = 1.
11 / 68 Binomial coefficient & binomial distribution
For P(Y = 3), P(Y = 4), . . . this gets even more complicated.
Binomial coefficient
Suppose that there is a set of n distinct elements from which it is desired
to choose a subset of k elements (typically 1 ≤ k ≤ n). The binomial
coefficient gives the number of ways k elements can be selected from n
elements. The binomial coefficient is defined as
n
n!
n
Cn,k = Ck =
=
.
k
k!(n − k)!
where k! = k(k − 1)(k − 2) . . . 1 and 0! = 1.
Remark
Note that
n!
= n · (n − 1) · (n − 2) · . . . · (n − k + 1).
(n − k)!
For example,
7!
7!
7·6·5·4·3·2·1
=
=
= 7 · 6 · 5.
(7 − 3)!
4!
4·3·2·1
11 / 68 Binomial coefficient & binomial distribution
Let’s consider another example to get a better understanding of the
binomial coefficient.
Example III.3
Imagine a box with four distinct elements (n = 4) denoted as a, b, c, d.
We want to randomly pick two elements (k = 2). If the order of selecting
elements matters, there exist 4 · 3 different combinations. However, we
don’t want the order to matter, so we divide by 2, as there are two ways
of ordering two elements ({b, a} and {a, b}).
Therefore, there are
4!
4
4·3
=
=
=6
2
2!(4 − 2)!
2
different combinations.
{a, b} {a, c} {a, d} {b, c} {b, d} {c, d}.
12 / 68 Combination vs. permutation
Example III.3
Imagine a box with four distinct elements (n = 4) denoted as a, b, c, d.
We want to randomly pick two elements (k = 2). If the order of selecting
elements matters, there exist 4 · 3 different combinations. However, we
don’t want the order to matter, so we divide by 2, as there are two ways
of ordering two elements ({b, a} and {a, b}).
Therefore, there are
4!
4
4·3
=
=
=6
2
2!(4 − 2)!
2
different combinations.
{a, b} {a, c} {a, d} {b, c} {b, d} {c, d}.
Note the distinction between permutation (order matters) and
combination (order does not matter).
If order matters (e.g. we distinguish between {a, b} and {b, a}), the
solution to the above problem is simply 4 · 3 = 12.
13 / 68 Binomial coefficient & binomial distribution
Back to our problem: For P(Y = 3),
10
10
10!
1
1
=
2
3!(10 − 3)! 2
10
10 · 9 · 8 1
=
= 0.1171875.
3·2·1 2
f (3) = P(Y = 3) =
10
3
14 / 68 Binomial coefficient & binomial distribution
Back to our problem: For P(Y = 3),
10
10
10!
1
1
=
2
3!(10 − 3)! 2
10
10 · 9 · 8 1
=
= 0.1171875.
3·2·1 2
f (3) = P(Y = 3) =
10
3
The binomial coefficient allows us to find a general expression for f (y).
Binomial distribution
If the random variables X1 , . . . , Xn form n Bernoulli trials with parameter
p (i.e. probability of success), then Y = X1 + · · · + Xn follows a
binomial distribution. The binomial
distribution is given by
n y
f (y; n, p) =
p (1 − p)n−y
x
for y = 0, 1, . . . , n.
14 / 68 Binomial distribution
We now know the specific functional form of f (y) = P(Y = y). Hence,
we can obtain the probability that we draw 0, 1, 2, . . . , 10 heads.
f(y)
0.2
0.1
f (y)
0.00098
0.00977
0.04395
0.11719
0.20508
0.24609
0.20508
0.11719
0.04395
0.00977
0.00098
0.0
y
0
1
2
3
4
5
6
7
8
9
10
0.3
0.4
Binomial distribution (n=10,p=0.5)
0
1
2
3
4
5
6
7
8
9
10
y
15 / 68 Cumulative distribution function
However, we are interested in P(Y ≤ 4).
16 / 68 Cumulative distribution function
However, we are interested in P(Y ≤ 4).
Cumulative distribution function
The cumulative distribution function of a discrete random variable X is
denoted by F (x) and is defined as
F (x) = P(X ≤ x)
where −∞ ≤ x ≤ +∞. The cumulative distribution function F (x) gives
the probability that the outcome of X in a random trial will be less than
or equal to any specified value x.
16 / 68 Binomial distribution
y
0
1
2
3
4
5
6
7
8
9
10
f (y)
0.00098
0.00977
0.04395
0.11719
0.20508
0.24609
0.20508
0.11719
0.04395
0.00977
0.00098
F(y)
0.00098
0.01074
0.05469
0.17188
0.37695
0.62305
0.82813
0.94531
0.98926
0.99902
1.00000
For example,
F (2) = f (0) + f (1) + f (2) = 0.00098 + 0.00977 + 0.04395 = 0.05469.
17 / 68 Should you make the bet?
Example III.1 (continued)
I propose the following game. We toss a fair coin 10 times. If head
appears 4 times or less, I pay you £2. If head appears more than 4 times,
you pay me £1. Should you make the bet?
We can finally solve the problem. Your expected value is
E[V ] = P(Y ≤ 4) · £2 + P(Y > 4) · (£ − 1)
= F (4) · £2 + (1 − F (4)) · (£ − 1)
= 0.377 · £2 + 0.623 · (£ − 1) ≈ £0.131.
18 / 68 Should you make the bet?
Example III.1 (continued)
I propose the following game. We toss a fair coin 10 times. If head
appears 4 times or less, I pay you £2. If head appears more than 4 times,
you pay me £1. Should you make the bet?
We can finally solve the problem. Your expected value is
E[V ] = P(Y ≤ 4) · £2 + P(Y > 4) · (£ − 1)
= F (4) · £2 + (1 − F (4)) · (£ − 1)
= 0.377 · £2 + 0.623 · (£ − 1) ≈ £0.131.
You should make the bet (if you are risk neutral)!
What does E[V ] mean? If we repeat the game an infinite number of
times, your average payoff will be £0.131.
18 / 68 Binomial distribution (Simulation)
Suppose that we play the game m times. That is, we toss the coin 10 times,
write down the number of heads, and play again.
19 / 68 Binomial distribution (Simulation)
Suppose that we play the game m times. That is, we toss the coin 10 times,
write down the number of heads, and play again.
Let’s set m = 20. We get: 3 7 5 2 3 3 2 5 3 7 4 4 4 4 5 3 4 5 4 2.
19 / 68 Binomial distribution (Simulation)
Suppose that we play the game m times. That is, we toss the coin 10 times,
write down the number of heads, and play again.
Let’s set m = 20. We get: 3 7 5 2 3 3 2 5 3 7 4 4 4 4 5 3 4 5 4 2.
0
0
0.00
y
frequency
rel. frequency
1
0
0.00
2
3
0.15
3
5
0.25
4
6
0.30
5
4
0.20
6
0
0.00
Empirical binomial distribution (n=10,p=0.5)
20 repetitions
8
0
0.00
9
0
0.00
10
0
0.00
0.4
0.3
0.2
0.1
0.0
0.1
0.2
f(y)
0.3
0.4
Binomial distribution (n=10,p=0.5)
0.0
f(y)
7
2
0.10
0
1
2
3
4
5
0
1
6
7
8
9
10
0
1
2
3
4
5
6
7
8
9
10
y
19 / 68 Binomial distribution (Simulation)
Suppose that we play the game m times. That is, we toss the coin 10 times,
write down the number of heads, and play again.
Let’s set m = 50.
0
0
0.00
y
frequency
rel. frequency
1
0
0.00
2
3
0.06
3
7
0.14
4
11
0.22
5
8
0.16
6
7
0.14
Empirical binomial distribution (n=10,p=0.5)
50 repetitions
8
4
0.08
9
3
0.06
10
0
0.00
0.4
0.3
0.2
0.1
0.0
0.1
0.2
f(y)
0.3
0.4
Binomial distribution (n=10,p=0.5)
0.0
f(y)
7
7
0.14
0
1
2
3
4
5
y
6
7
8
9
10
0
1
2
3
4
5
6
7
8
9
10
y
19 / 68 Binomial distribution (Simulation)
Suppose that we play the game m times. That is, we toss the coin 10 times,
write down the number of heads, and play again.
Let’s set m = 100.
0
0
0.00
y
frequency
rel. frequency
1
2
0.02
2
3
0.03
3
17
0.17
4
26
0.26
5
22
0.22
6
14
0.14
Empirical binomial distribution (n=10,p=0.5)
100 repetitions
8
3
0.03
9
2
0.02
10
0
0.00
0.4
0.3
0.2
0.1
0.0
0.1
0.2
f(y)
0.3
0.4
Binomial distribution (n=10,p=0.5)
0.0
f(y)
7
11
0.11
0
1
2
3
4
5
y
6
7
8
9
10
0
1
2
3
4
5
6
7
8
9
10
y
19 / 68 Binomial distribution (Simulation)
Suppose that we play the game m times. That is, we toss the coin 10 times,
write down the number of heads, and play again.
Let’s set m = 10,000.
0
4
0.0004
y
frequency
rel. frequency
1
89
0.0089
2
429
0.0429
3
1171
0.1171
4
2045
0.2045
5
2470
0.2470
6
2075
0.2075
Empirical binomial distribution (n=10,p=0.5)
10000 repetitions
8
411
0.0411
9
103
0.0103
10
5
0.0005
0.4
0.3
0.2
0.1
0.0
0.1
0.2
f(y)
0.3
0.4
Binomial distribution (n=10,p=0.5)
0.0
f(y)
7
1198
0.1198
0
1
2
3
4
5
y
6
7
8
9
10
0
1
2
3
4
5
6
7
8
9
10
y
19 / 68 Binomial distribution
The binomial distribution has two parameters: n is the number of
(Bernoulli) trials and p is the probability of success in each trial. How
does the distribution look like for different values of n and p?
2
3
4
5
6
7
8
9
10
0.4
0.3
0.1
0.0
0
1
2
3
4
5
6
7
8
9
10
0
1
2
3
4
5
6
7
8
9
y
Binomial distribution (n=30,p=0.5)
Binomial distribution (n=30,p=0.7)
Binomial distribution (n=30,p=0.9)
10
15
y
20
25
30
0.2
f(y)
0.0
0.1
0.2
f(y)
0.0
0.1
0.2
0.1
5
10
0.3
y
0.3
y
0.0
0
0.2
f(y)
0.3
0.0
0.1
0.2
f(y)
0.3
0.2
f(y)
0.1
0.0
1
0.3
0
f(y)
Binomial distribution (n=10,p=0.9)
0.4
Binomial distribution (n=10,p=0.7)
0.4
Binomial distribution (n=10,p=0.5)
0
5
10
15
y
20
25
30
0
5
10
15
y
20
25
30
20 / 68 Binomial distribution
Expected value and variance
In the same way we summarise an observed dataset by the sample average
and the sample variance (or standard deviation), we can characterise a
probability distribution by its expected value and its variance. From the
figures we can see that expected value and variance change with n and p.
21 / 68 Binomial distribution
Expected value and variance
In the same way we summarise an observed dataset by the sample average
and the sample variance (or standard deviation), we can characterise a
probability distribution by its expected value and its variance. From the
figures we can see that expected value and variance change with n and p.
To find the expected value of Y , note that:
Linearity of expectation
If Y is the sum of random variables
1 , X2 , . . . , Xn , then:
" n X#
n
X
X
E[Y ] = E
Xi =
E[Xi ].
i=1
i=1
Furthermore, if c is a constant (i.e., non-random) and X a random
variable, then
E[X + c] = E[X ] + c
E[cX ] = cE[X ].
21 / 68 Binomial distribution
Expected value and variance
Recall that Y = X1 + X2 + X3 + · · · + Xn . Therefore,
E[Y ] = E[X1 ] + E[X2 ] + E[X3 ] + · · · + E[Xn ]
22 / 68 Binomial distribution
Expected value and variance
Recall that Y = X1 + X2 + X3 + · · · + Xn . Therefore,
E[Y ] = E[X1 ] + E[X2 ] + E[X3 ] + · · · + E[Xn ]
Recall that Xi follows a Bernoulli distribution. The expected value of a
Bernoulli variable is
E[Xi ] = p · 1 + (1 − p) · 0 = p.
22 / 68 Binomial distribution
Expected value and variance
Recall that Y = X1 + X2 + X3 + · · · + Xn . Therefore,
E[Y ] = E[X1 ] + E[X2 ] + E[X3 ] + · · · + E[Xn ]
Recall that Xi follows a Bernoulli distribution. The expected value of a
Bernoulli variable is
E[Xi ] = p · 1 + (1 − p) · 0 = p.
Therefore,
E[Y ] = E[X1 ] + E[X2 ] + E[X3 ] + . . . + E[Xn ] = np.
22 / 68 Binomial distribution
Expected value and variance
Variance
The variance of a discrete random variable X is denoted by V[X ] and
given by
k
X
(xi − E[X ])2 P(X = xi ).
V[X ] =
i
23 / 68 Binomial distribution
Expected value and variance
Variance
The variance of a discrete random variable X is denoted by V[X ] and
given by
k
X
(xi − E[X ])2 P(X = xi ).
V[X ] =
i
Variance of the sum of uncorrelated random variables
If Y is the sum of independent (!) random variables X1 , X2 , . . . , Xn ,
then:
" n
#
n
X
X
V[X ] = V
Xi =
V[Xi ]
i=1
i=1
23 / 68 Binomial distribution
Expected value and variance
Variance
The variance of a discrete random variable X is denoted by V[X ] and
given by
k
X
(xi − E[X ])2 P(X = xi ).
V[X ] =
i
Variance of the sum of uncorrelated random variables
If Y is the sum of independent (!) random variables X1 , X2 , . . . , Xn ,
then:
" n
#
n
X
X
V[X ] = V
Xi =
V[Xi ]
i=1
i=1
The variance of Xi is given by
V[X ] = (1 − E[X ])2 p + (0 − E[X ])2 (1 − p) = p(1 − p)
Therefore
V[Y ] = np(1 − p)
23 / 68 Poisson distribution
Example III.4
Let X be the number of cars passing by in an hour. On average λ cars
pass by. What is the probability that 3 cars pass by?
24 / 68 Poisson distribution
Example III.4
Let X be the number of cars passing by in an hour. On average λ cars
pass by. What is the probability that 3 cars pass by?
To simplify the problem, we divide each hour into 60 minutes. Since
E[X ] = λ, the probability that one car passes by in any particular minute
is λ/60.
24 / 68 Poisson distribution
Example III.4
Let X be the number of cars passing by in an hour. On average λ cars
pass by. What is the probability that 3 cars pass by?
To simplify the problem, we divide each hour into 60 minutes. Since
E[X ] = λ, the probability that one car passes by in any particular minute
is λ/60.
Using this simplification, we can work with the binomial distribution.
f (3) = P(X = 3) ≈
60−3
3
60
λ
λ
1−
60
60
3
24 / 68 Poisson distribution
Example III.4
Let X be the number of cars passing by in an hour. On average λ cars
pass by. What is the probability that 3 cars pass by?
To simplify the problem, we divide each hour into 60 minutes. Since
E[X ] = λ, the probability that one car passes by in any particular minute
is λ/60.
Using this simplification, we can work with the binomial distribution.
f (3) = P(X = 3) ≈
60−3
3
60
λ
λ
1−
60
60
3
However, this approach does not take into account that more than one
car may pass by in a minute.
24 / 68 Poisson distribution
We can address this problem by dividing each hour into 3,600 seconds,
3,600,000 milliseconds, and so on. More general, with n being the
number of units we divide the hour into:
x
n−x
n
λ
λ
f (x) = P(X = x) ≈
1−
x
n
n
25 / 68 Poisson distribution
We can address this problem by dividing each hour into 3,600 seconds,
3,600,000 milliseconds, and so on. More general, with n being the
number of units we divide the hour into:
x
n−x
n
λ
λ
f (x) = P(X = x) ≈
1−
x
n
n
Let n → ∞, do some maths and you’ll arrive at:
Poisson distribution
If X follows a poisson process, then
e −λ λx
x!
n
for x = 0, 1, 2, 3, . . . , ∞. Note that e = limn→∞ 1 + n1 = 2.718 . . . .
f (x) = P(X = x) =
25 / 68 Poisson distribution
Poisson distribution
lambda=3
f(x)
0.0
0.00
0.05
0.10
0.2
0.1
f(x)
0.15
0.3
0.20
Poisson distribution
lambda=1
0
5
10
x
15
20
0
5
10
15
20
x
The poisson distribution is asymmetric (or right-skewed) for
E[X ] = λ = 1 and λ = 3
26 / 68 Poisson distribution
Poisson distribution
lambda=1000
f(x)
0.006
0.004
0.06
0.00
0.000
0.02
0.002
0.04
f(x)
0.08
0.008
0.10
0.010
0.12
0.012
Poisson distribution
lambda=10
0
5
10
x
15
20
850
900
950
1000
1050
1100
1150
x
The higher λ, the more symmetric is the poisson distribution. Also, the
distribution looks very similar to the normal distribution!
27 / 68 Continuous distributions
Probability density function (PDF)
If the random variable X is continuous, we use f (x) to denote the
probability density function (PDF) of X . The PDF satisfies two
requirements:
Z
+∞
(1) f (x) ≥ 0
and
(2)
f (x)dx = 1
−∞
Remark (!)
If the random variable X is continuous, then the probability that X takes
a particular value x is zero. That is,
f (x) 6= P(X = x) = 0.
28 / 68 Continuous distributions
Uniform distribution
Let a and b be two real numbers (a < b) and consider an experiment
where a number X is randomly selected from the interval [a, b]. If the
probability that X belongs to any subinterval of [a, b] is proportional to
the length of the subinterval, we say that X is uniformly distributed. The
PDF of X is given by
1
for x ∈ [a, b]
b−a
f (x) =
0
otherwise
We write that X ∼ u(a, b).
29 / 68 Continuous distributions
Example III.5
Anna and Achim arrange to meet ‘between 1pm and 2pm’ at Old College.
Their arrival time is uniformly distributed and they arrive independently of each
other. What is the probability that no one will have to wait more than 15
minutes?
Let X denote Anna’s arrival time and Y denote Achim’s arrival time.
I
Express the event ‘no one will wait for more than 15 minutes’ in
terms of X and Y .
I
What is the joint distribution of X and Y ?
Solution here.
30 / 68 Continuous distributions
The normal distribution is by far the single most important probability
distribution. Many natural phenomena are (approximately) normally
distributed. Another reason for its importance comes from the central
limit theorem (to be discussed in the next lecture).
Normal distribution
If the random variable X follows normal distribution with mean µ
(−∞ < µ < ∞) and variance σ 2 (σ > 0), its PDF is given by
f (x) = √
(x−µ)2
1
−
e 2σ2
σ 2 2π
for −∞ < x < ∞. We write that X ∼ N (µ, σ 2 ).
31 / 68 Continuous distributions
1
0.5
u(1, 3)
u(−3, 0)
N (0, 1)
0.3
0.6
f (x)
f (x)
0.4
0.8
0.2
0.4
0.1
0.2
0
−4
−2
0
x
2
4
0
−4
−2
0
2
4
x
32 / 68 E[X ] for continuous distributions
Expected value
The expected value of a continuous
random variable is
Z +∞
E[X ] =
xf (x)dx.
−∞
I
E[X ] is the balance point of the probability mass. The probability mass to
the left of E[X ] is in balance to the probability mass on the right of E[X ].
33 / 68 E[X ] for continuous distributions
a
I
I
Thus, the expected value of the normal distribution is simply at its
highest point (due to symmetry) and the expected value of a
uniform distribution is half way between a and b.
b
34 / 68 E[X ] for the uniform distribution
Let’s do this formally for the uniform distribution.
Z +∞
E[X ] =
xf (x)dx
−∞
Z +∞
=
x
−∞
1
dx
b−a
b
1
xdx
b−a a
1 2 b
1
x
b−a 2
a
1 1 2
b − a2
b−a2
1 (b2 − a 2 )(b + a)
2 (b − a)(b + a)
1
(b + a)
2
(1)
(2)
Z
=
=
=
=
=
(3)
(4)
(5)
(6)
(7)
35 / 68 E[X ] for the uniform distribution
(1) By the definition of the expected value.
(2) By the definition of the uniform distribution.
(3) If k is a constant, then
Z
Z
kf (x)dx = k
f (x)dx.
(4) Since
Z
x n dx =
1
x n+1 + c
n+1
(n 6= −1).
(4-5) Since
Z b
b
= F (b) − F (a)
f (x)dx = F (x)
a
where
d
dx F (x)
a
= f (x).
36 / 68 V[X ] for continuous distributions
Variance
The variance of a continuous
random variable X is
Z
+∞
V[X ] =
−∞
(x − E[X ])2 f (x)dx.
(8)
0.5
N (0, 1)
N (0, 2)
N (0, 3)
0.4
f (x)
0.3
0.2
0.1
0
−4
−2
0
2
4
x
37 / 68 V[X ] for uniform distribution
Instead of using the definition above, we can make use of the
following:
38 / 68 V[X ] for uniform distribution
Instead of using the definition above, we can make use of the
following:
Variance
h
V[X ] = E (X − E[X ])2
h
i
= E X 2 − 2X E[X ] + (E[X ])2
i
= E[X 2 ] − 2(E[X ])2 + (E[X ])2
= E[X 2 ] − (E[X ])2
where the third line uses the fact that E[A + B] = E[A] + E[B]
and E[E[A]] = E[A].
We know E[X ]. Hence, we only need to find E[X 2 ].
38 / 68 V[X ] for uniform distribution
We treat X 2 as a new random variable which follows the same PDF as X .
Z +∞
E[X 2 ] =
x 2 f (x)dx
−∞
1
=
b−a
Z
b
x 2 dx
a
b
1
1 3
x
b−a 3
a
1 1 3
=
b − a3
b−a3
1 (b3 − a 3 )(a 2 + ab + b2 )
=
3 (b − a)(a 2 + ab + b2 )
1
= (a 2 + ab + b2 )
3
=
39 / 68 V[X ] for uniform distribution
We treat X 2 as a new random variable which follows the same PDF as X .
Z +∞
E[X 2 ] =
x 2 f (x)dx
−∞
1
=
b−a
Z
b
x 2 dx
a
b
1
1 3
x
b−a 3
a
1 1 3
=
b − a3
b−a3
1 (b3 − a 3 )(a 2 + ab + b2 )
=
3 (b − a)(a 2 + ab + b2 )
1
= (a 2 + ab + b2 )
3
=
2
1 2
1
1
2
V [X ] = E[X ]−(E[X ]) = (a +ab +b )−
(b + a) =
(b −a)2
3
2
12
2
2
39 / 68 PDF and probability
Consider the standard normal distribution depicted in the figure. As you
can see, f (0) ≈ 0.4. To be precise, f (0) = 0.3989423 . . . . It is important
to understand that this does not mean that P(X = 0) = 0.3989423 . . . !
If a random variable is continuous, there are infinite distinct values that
the random variable can take. Thus, the probability that the random
variable takes a specific value is zero.
0.5
N (0, 1)
0.4
f (x)
0.3
0.2
0.1
0
−4
−2
0
2
4
x
40 / 68 PDF and probability
However, we can say that the probability that X is below 0 is equal to
the shaded gray area. Since we know that the area under f (x) is 1, we
know (due to symmetry) that P(X ≤ 0) = 0.5.
0.5
N (0, 1)
0.4
f (x)
0.3
0.2
0.1
0
−4
−2
0
2
4
x
41 / 68 PDF and probability
But what is the probability that, say, P(X ≤ −1)?
0.5
N (0, 1)
0.4
f (x)
0.3
0.2
0.1
0
−4
−2
−1
0
2
4
x
42 / 68 CDF and probability
Cumulative density function
The cumulative density function (CDF) of a continuous random variable
X is denoted by F (x) and given by
Z x
F (x) = P(X ≤ x) =
f (u)du for − ∞ ≤ x ≤ +∞.
−∞
The CDF gives the probability that the outcome of X in a random
experiment is less than or equal to x.
43 / 68 CDF and probability
0.5
N (0, 1)
0.4
f (x)
0.3
0.2
0.1
We can read from the CDF that
0
−4
F (−1) = P(X ≤ −1) ≈ 0.159 . . .
−2
−1
0
2
4
0
2
4
1.2
and
N (0, 1)
1
F (0) = P(X ≤ 0) = 0.5.
F (x)
0.8
0.6
0 .5
0.4
0.2
0 .159 ..
0
−4
−2
−1
x
44 / 68 CDF and probability
0.5
N (0, 1)
0.4
f (x)
0.3
0.2
What is the probability that X
lies between −1 and 0? That is,
what is
0.1
0
−4
P(−1 ≤ X ≤ 0)?
−2
−1
0
2
4
0
2
4
1.2
N (0, 1)
It is simply
1
0.8
F (x)
F (0)−F (−1) = 0.5−0.159 ≈ 0.341.
0.6
0 .5
0.4
0.2
0 .159 ..
0
−4
−2
−1
x
45 / 68 CDF and probability
0.5
N (0, 1)
0.4
f (x)
0.3
0.2
What is the probability that X
is below +1? Due to symmetry
0.1
0
−4
F (1) = 1 − F (−1).
−2
−1
0
1
2
4
0
1
2
4
1.2
N (0, 1)
Thus,
1
1 − F (−1) = 1 − 0.159 ≈ 0.841.
0 .841 ..
F (x)
0.8
0.6
0 .5
0.4
0.2
0 .159 ..
0
−4
−2
−1
x
46 / 68 Inverse functions and CDF
We will often use the inverse function of the CDF.
Inverse function
In general, if g(x) is an invertible function, then the inverse function is
given by
g −1 (g(x)) = x.
Intuition: A function works like a machine. It takes x as an input and
returns the output g(x) = a. An inverse function works the over way
around: g −1 (a) = x.
Suppose we are interested in the following question: What is the value of
x such that P(X ≤ x) = 0.95.
47 / 68 Inverse functions and CDF
0.5
N (0, 1)
0.4
Suppose we are interested in the
following question: What is the
value of x such that P(X ≤
x) = 0.95. Using the inverse
CDF:
0.2
0.1
0
−4
(0.95) ≈ 1.64 . . .
This implies that, due to symmetry, 90% of the probability
mass is approximately in the
±1.64 interval.
−2
−1
0
1
1 .64 2
4
0
1
1 .64 2
4
1.2
N (0, 1)
1
0 .95
0.8
F (x)
F
−1
f (x)
0.3
0.6
0.4
0.2
0
−4
−2
−1
x
48 / 68 Standard normal distribution
Standard normal distribution
If X ∼ N (µ, σ 2 ), then
X −µ
Z=
∼ N (0, 1).
σ
We say that Z follows a standard normal distribution. The PDF and CDF
of the standard normal distribution are often denoted by φ(z) and Φ(z).
0.5
Z ∼ N(0, 1)
X ∼ N(10, 4)
0.4
f (x)
0.3
0.2
0.1
0
0
5
10
15
49 / 68 Standard normal distribution
Example III.6
√
Suppose X ∼ N (10, 4) and Z = (X − 10)/ 4 ∼ N (0, 1). What is P(X ≤ 8)?
X −µ
8−µ
≤
σ
σ
=P
8 − 10
X − 10
√
≤ √
4
4
= P (Z ≤ −1)
0.5
Z ∼ N(0, 1)
X ∼ N(10, 4)
0.4
0.3
f (x)
P(X ≤ 8) = P
0.2
0.1
0
−1 0
5
8
10
15
50 / 68 Multivariate distributions (discrete)
Joint probability function
The joint probability function of two discrete random variables X and Y is
given by
f (x, y) = P(X = x and Y = y)
and
XX
i
f (xi , yj ) = 1.
j
Table of Probabilities
X \Y
1
2
3
4
1
0.1
0.3
0
0
2
0
0
0.2
0
3
0.1
0.1
0
0
4
0
0.2
0
0
51 / 68 Multivariate distributions (discrete)
Joint probability function
The joint probability function of two discrete random variables X and Y is
given by
f (x, y) = P(X = x and Y = y)
XX
and
i
f (xi , yj ) = 1.
j
f (x, y)
0.4
0.3
Table of Probabilities
X \Y
1
2
3
4
1
0.1
0.3
0
0
2
0
0
0.2
0
3
0.1
0.1
0
0
0.2
4
0
0.2
0
0
0.1
1
2
3
4
1
2
3
4
y
x
51 / 68 Multivariate distributions (continuous)
Joint probability density function
The joint probability density function (or joint PDF) of two continuous random
variables X and Y is given by
Z
f (x, y)
+∞
Z
f (x, y)dxdy = 1.
−∞
f (x, y)
+∞
and
−∞
0.1
0
−1
4
−0.5
3.5
0
3
0.5
2.5
1
x
2
1.5
1.5
2
y
1
2.5
0.5
30
52 / 68 Multivariate distributions (continuous)
Joint probability density function
The joint probability density function of two continuous random variables X
and Y is given by
Z
f (x, y)
+∞
Z
+∞
and
f (x, y)dxdy = 1.
−∞
−∞
Example III.7
If X and Y have joint PDF f (x, y) then the probability that X lies between 0
and 2 and that, at the same time, Y lies between 0 and 1 is
Z
1
Z
P(0 ≤ X ≤ 2 and 0 ≤ Y ≤ 1) =
2
f (x, y)dxdy
0
0
53 / 68 Marginal distributions
Marginal probability function (discrete)
If X and Y are two discrete random variables for which the joint probability
function is f (x, y), then the marginal probability function for X is
fX (x) = P(X = x) =
X
y
P(X = x and Y = y) =
X
f (x, y)
y
The marginal probability gives the probability of observing a specific value of X
(say X = x). To calculate the probability of observing x, we need to add the
probabilites of all events that correspond to X = x: That is,
P(X = x) = f (x, y1 ) + f (x, y2 ) + . . . + f (x, yn ).
54 / 68 Marginal distributions
Example III.8 (Discrete Marginal Distribution)
Table of Probabilities
X\Y
1
2
3
4
1
0.1
0.3
0
0
2
0
0
0.2
0
3
0.1
0.1
0
0
4
0
0.2
0
0
What is the marginal probability function for
X?
55 / 68 Marginal distributions
Example III.8 (Discrete Marginal Distribution)
Table of Probabilities
X\Y
1
2
3
4
1
0.1
0.3
0
0
2
0
0
0.2
0
3
0.1
0.1
0
0
What is the marginal probability function for
X?
4
0
0.2
0
0
fX (1) = 0.2, fX (2) = 0.6, fX (3) = 0.2,
fX (4) = 0.
Marginal probability density function (continuous)
If X and Y are two continuous variables for which the joint probability density
function is f (x, y), then the marginal probability density function for X is
Z
+∞
fX (x) =
f (x, y)dy
−∞
55 / 68 Joint distributions and independence
Recall from the last lecture that, if X and Y are two independent events,
then P(X and Y ) = P(X )P(Y ). We can generalize this statement:
56 / 68 Joint distributions and independence
Recall from the last lecture that, if X and Y are two independent events,
then P(X and Y ) = P(X )P(Y ). We can generalize this statement:
Independence
Two independent continuous (or discrete) random variables are
independent if and only if
f (x, y) = fX (x)fY (y) ⇐⇒ F (x, y) = FX (x)FY (y)
where fX (x) and fY (y) are marginal PDF’s. FX (x) and FY (y) denote
marginal CDF’s.
56 / 68 Conditional distributions
Recall from the last lecture that the conditional probability is defined as
)
P(X |Y ) = P(X,Y
P(Y ) . Furthermore, recall that if X and Y are two
independent events, then P(X |Y ) = P(X ).
Conditional probability density function
Suppose that X and Y are two continuous (or discrete) random variables
for which the joint PDF is f (x, y) and the marginal PDF’s are fX (x) and
fY (y). Suppose also that the value y has already been observed. The
conditional probability density function of X given that Y = y is given by
f (x, y)
fX (x|y) =
.
fY (y)
Note that if X and Y are independent, we get the relation
fX (x|y) = fX (x).
57 / 68 f (x, y)
Conditional distributions
0.1
0
−1
4
−0.5
3.5
0
3
0.5
2.5
1
x
2
1.5
1.5
2
y
1
2.5
0.5
30
The black, thick line shows f (x, 1) which is proportional to the
conditional distribution of X given Y = 1. More specific:
fX (x|Y = 1) =
f (x, 1)
.
fY (1)
58 / 68 (Conditional) Expectations and Covariance
Example III.9
Let X and Z be two independently distributed standard normal random
variables and let Y = X 2 + Z . Hence, V(X ) = V(Z ) = 1 and
E(X ) = E(Z ) = 0.
a) Derive E[Y |X ].
b) Derive E[Y ]. [Hint: Use V[X ] = E[X 2 ] − (E[X ])2 .]
c) Derive E[XY ]. [Hint: Since X is standard normal, E[X 3 ] = 0.]
d) Find Cov(X , Y ) = E[(X − E(X ))(Y − E(Y ))].
What is interesting about the results?
Solution here.
59 / 68 Independence and Covariance
Covariance is defined as
Cov(X , Y ) = E[(X − E(X ))(Y − E(Y ))] = E[XY ] − E[X ]E[Y ].
If X and Y are independent, E[XY ] = E[X ]E[Y ]. Therefore, if X and
Y are independent, Cov(X , Y ) = 0.
However, Cov(X , Y ) = 0 (and Corr(X , Y ) = 0) does not imply
independence as demonstrated in the previous example.
60 / 68 Summary
I
Random variables are either discrete or continuous. Random
variables are usually denoted by capital letters, e.g. X , and
realisations by small letters, e.g. x.
I
For a continuous random variable, P(X = x) = 0 and
f (x) 6= P(X = x) where f (x) denotes the probability density
function.
I
Many probability distributions are closely related. For example, we
can derive the Poisson distribution from the Binomial distribution
and the Poisson distribution behaves similar to the Normal
distribution as λ → ∞.
I
Independence implies Cov(X , Y ) = 0 (and Corr(X , Y ) = 0), but
not the other way around. Cov(X , Y ) and Corr(X , Y ) measure the
strength of the linear relation between two variables.
61 / 68 Appendix: Solution to Example III.5
Since X and Y are independent, the joint density distribution is given by
f (x, y) = f (y)f (x). Both X and Y are uniformly distributed over the [0, 60]
interval. Hence,
1
1
1
f (x, y) = f (x)f (y) =
=
.
60 − 0 60 − 0
3600
Note that the absolute waiting time is given by |Y − X |. We are interested in
the event that |Y − X | ≤ 15. The problem is easily solved graphically:
Y
Since the probability is constant over the
entire square, the probability of an event
is proportional to its area. P(|Y − X | ≤
15) is equal to the shaded region in the
diagram divided by the total area. Thus,
60
45
30
P(|Y −X | > 15) =
15
2025
452
=
= 0.5625
3600
3600
and
X
0
15
30
45
P(|Y −X | ≤ 15) = 1−0.5625 = 0.4375
60
Back
62 / 68 Appendix: Alternative Solution to Example III.5
Recall X denotes Anna’s arrival time and Y denotes Achim’s arrival time.
Both X and Y are uniformly distributed over the [0, 60] interval. We
distinguish between four cases.
1. Anna arrives first (X < Y ) and Anna arrives between 0 and 45
(0 ≤ X ≤ 45). Note that P(0 ≤ X ≤ 45) = 3/4.
2. Anna arrives first (X < Y ) and Anna arrives between 45 and 60
(45 ≤ X ≤ 60). Note that P(45 ≤ X ≤ 60) = 1/4.
3. Achim arrives first (Y < X ) and Achim arrives between 0 and 45
(0 ≤ Y ≤ 45). Note that P(0 ≤ Y ≤ 45) = 3/4.
4. Achim arrives first (Y < X ) and Achim arrives between 45 and 60
(45 ≤ Y ≤ 60). Note that P(45 ≤ Y ≤ 60) = 1/4.
Due to symmetry, it is sufficient to consider the first two cases.
Case 1: Anna arrives at time X and Achim arrives later. The probability that
Achim arrives between time X and time X + 15 is simply 15/60 = 1/4. Since
X and Y are independent, it does not matter when Anna will arrive. The
probability that Achim arrives between 0 and 15 minutes after her is always the
same.
Back
63 / 68 Appendix: Alternative Solution to Example III.5
Case 2: Anna arrives between 45 and 60. Her conditional expected value is
E[X |45 ≤ X ≤ 60] = 12 (45 + 60) = 52.5. That is, since Achim arrives after
Anna by assumption, there is on average a 7.5 minute window left for Achim.
Thus, the probability is simply 7.5/60 = 1/8.
Hence,
P(|Y − X | ≤ 15) = P(Y −X <15)P(0 ≤ X ≤ 45) + P(Y −X <15)P(45 ≤ X ≤ 60)
+ P(X −Y <15)P(0 ≤ Y ≤ 45) + P(X −Y <15)P(45 ≤ Y ≤ 60)
3
1
1
1
3
1
1
1
= × + × + × + × = 0.4375
4
4
8
4
4
4
8
4
which is in line with the graphical solution.
Back
64 / 68 Appendix: Solution to Example III.9
a) Derive E[Y |X].
E(Y |X) = E(X 2 + Z |X)
= E(X 2 |X) + E(Z |X)
2
(1)
(2)
= X + E(Z |X)
(3)
= X 2 + E(Z )
(4)
= X2
(5)
(1) By definition of Y . (2) Since in general E[A + B] = E[A] + E[B]. (3) Since we
condition on X, we can take X out of the expectation operator. (4) Since X and Z
are independent E[Z |X] = E[Z ]. (5) Since E[Z ] = 0.
b) Derive E[Y ].
E[X 2 ] = V[X] + (E[X])2
= 1 + (0)2 = 1
(1)
(2)
(1) Using the hint. (2) Since V[X] = 1 and E[X] = 0. Furthermore,
E[Y ] = E[X 2 + Z ] = E[X 2 ] + E[Z ] = 1 + 0 = 1.
Back
65 / 68 Appendix: Solution to Example III.9
c) Derive E[XY ].
E(XY ) = E[X(X 2 + Z )]
(1)
= E[X 3 + XZ ]
= E[X 3 ] + E[XZ ]
(2)
= 0 + E[XZ ]
(3)
= 0 + E[X]E[Z ] = 0
(4)
(1) By definition of Y . (2) Since in general E[A + B] = E[A] + E[B]. (3) Using the
hint. (4) In general, if A and B are independent, E[AB] = E[A]E[B].
d) Show that Cov(X, Y ) = 0.
Cov(X, Y ) = E[(X − E(X))(Y − E(Y ))]
(1)
= E[(X − 0)(Y − 1)] = E[XY − X]
= E[XY ] − E[X] = 0
(1) By definition of covariance.
Back
66 / 68 Appendix: Derivation of Poisson Distribution
P(X = x) = lim
n→∞
= lim
n→∞
= lim
n→∞
x
n
x
λ
n
λx
n!
(n − x)!x! n x
1−
λ
n
1−
n−x
λ
n
n−x
n!
λx
λ
lim
lim 1 −
x
(n − x)!n n→∞ x! n→∞
n
(1)
n
lim
1−
n→∞
λ
n
−x
(2)
(1) By definition. (2) Since lim f (x)g(x) = lim f (x) lim g(x).
Taking the limits in turn:
lim
n→∞
n(n − 1) . . . (n − x + 1)
n!
= lim
n→∞
(n − x)!n x
nx
x
n + ...
= lim
=1
n→∞
nx
where we expand n(n − 1) . . . (n − x + 1).
67 / 68 Appendix: Derivation of Poisson Distribution
The second term does not depend on n. Thus,
lim
n→∞
λx
λx
=
.
x!
x!
By definition of e:
lim
1−
n→∞
λ
n
n
= e −λ
As n → ∞, λ/n → 0. Thus,
lim
n→∞
1−
λ
n
−x
=1
Thus, we get the result:
f (x) = P(X = x) =
e −λ λx
x!
Watch: Khan Academy 1, 2
Back
68 / 68
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