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ECMT5007
The Economics of Financial Markets:
Descriptive Statistics and Probability
Grace Ye
The University of Sydney
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Overview
Descriptive Statistics
and Probability
Introduction to econometrics
Measures of central tendency
Measures of variation
Shapes of distributions
Probability
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Let 1, 2, 3 and 4 be independent, identically distributed random
variables from a population with a mean and variance 2. Let ത =
1
4
(1 + 2 + 3 + 4) note the average of these four random variables:
• What is the expected value and variance of ത in terms of and
variance 2?
Example – Mean and Variance
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• Since we have:
1 = 2 = ⋯ = =
1 = 2 = ⋯ = =
2
ത can be calculated as:
ത =
1 + 2 + 3 + 4
4
=
1 + 2 + 3 + 4
4
=
4 ∗
4
=
• Some useful properties of the expectation operator:
= ∗ ,ℎ
± = ± ,ℎ
Example – Mean and Variance
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• Since we have 1, 2 = 0
• ത can be calculated as:
ത =
1 + 2 + 3 + 4
4
=
1 + 2 + 3 + 4
16
=
4 ∗ 2
16
=
2
4
• Some useful properties of the variance operator:
= 2 ∗ ,ℎ
± = 2 + 2 ± 2 , ,
ℎ
Example – Mean and Variance
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Now consider a different estimator of
=
1
8
1 +
1
8
2 +
1
4
3 +
1
2
4
This is an example of a weighted average of the . Show that is also a
weighted average of . Find the variance of W.
Example – Mean and Variance
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• can be calculated as:
=
1
8
+
2
8
+
3
4
+
4
2
=
1
8
+
2
8
+
3
4
+
4
2
=

8
+

8
+

4
+

2
=
Example – Mean and Variance
• The variance of W is then:
=
1
8
+
2
8
+
3
4
+
4
2
=
1
64
+
2
64
+
3
16
+
4
4
=
2
64
+
2
64
+
2
16
+
2
4
=
11
32
2
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Based on the above, which estimator of do you prefer, ത or ?
• Both estimators are unbiased, i.e. on average my guess is correct:
ത = =
• So, in this respect, both of them are good. We need another criterion to judge it.
We then look at the variance of these estimators:
ത =
2
4
=
8
32
2
=
11
32
2
• Notice that the variance (measure of precision of the guess) is smaller in the case
of ഥY. So, we prefer the estimator with the smallest variance (higher precision), i.e.
we prefer to use ഥY as opposed to W.
Example – Mean and Variance
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Example 1 - Probability
If = 0.65, = 0.76, and ∪ = 0.80, then ∩ is?
(0.61)
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Example 2 - Probability
A table of joint probabilities is shown below:
• Calculate 2 1 .
• Are the events 2 and 1 independent? Explain.
1 2 3
1 0.15 0.25 0.20
2 0.10 0.15 0.15
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Example 2 - Probability
A table of joint probabilities is shown below:
• Calculate 2 1 .
• Are the events 2 and 1 independent? Explain.
• 2 1 =
(2∩1)
(1)
=
0.1
0.25
= 0.4
• 2 ∩ 1 = 0.25; 2 × 1 = 0.4 × 0.6 = 0.24
• 2 ∩ 1 ≠ 2 × 1 → Not independent.
1 2 3
1 0.15 0.25 0.20
2 0.10 0.15 0.15
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• 40% of ECON5007 students attend Sinan’s lectures.
• For those who attend Sinan’s lectures, 70% of the students get HD.
• For those who do not attend Sinan’s lectures, only 10% of them can get
HD.
• Q1: What is the probability of a student to get HD?
• Q2: Given that a student gets HD in ECON5007, what is the probability
that the student attends Sinan’s lectures?
Example 3 - Probability
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• 40% of ECON5007 students attend Sinan’s lectures.
• For those who attend Sinan’s lectures, 70% of the students get HD.
• For those who do not attend Sinan’s lectures, only 10% of them can get
HD.
• Q1: What is the probability of a student to get HD?
• We can use the law of total probability rule. 0.4 × 0.7 + 0.6 × 0.1 =
0.34
• Q2: Given that a student gets HD in ECON5007, what is the probability
that the student attends Sinan’s lectures?
• We use Bayes’ Theorem,
0.28
0.34
= 0.82
Example 3 - Probability
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The price of stock X is $100 today. It might increase by $5 with a 20%
chance but decrease by $1 with an 80% chance. Will you buy this stock?
The expected price = $105 × 0.2 + $99 × 0.8 = 100.2
Example 4 - Probability
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The following table gives the joint probability density function = , = = (, ) of
two random variables X and Y:
Evaluate () and ()
Example 5 - Probability
Y X 0 1 2
10 0.05 0.3 0.1
20 0.1 0.25 0.2
The expected value of X is given by:
• E = Pr = 0 ∗ 0 + Pr = 1 ∗ 1 + Pr = 2 ∗ 2 = 0.15 ∗ 0 + 0.55 ∗ 1 + 0.3 ∗ 2 =
1.15
• = 1.15
Similarly for Y:
• E = Pr = 10 ∗ 10 + Pr = 20 ∗ 20 = 0.45 ∗ 10 + 0.55 ∗ 20 = 15. 5
• = 15.5
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Find the conditional distribution ( = | = 10) and its mean.
• = 0| = 10 =
=0,=10
=10
=
0.05
0.45
=
1
9
(11. ሶ1%)
• = 1| = 10 =
=1,=10
=10
=
0.3
0.45
=
6
9
66. ሶ6%
• = 2| = 10 =
=2,=10
=10
=
0.1
0.45
=
2
9
(22. ሶ2%)
• The conditional mean of X is therefore:
• = 10 = = 0| = 10 ∗ 0 + = 1| = 10 ∗ 1 + = 2| = 10 ∗ 2
• = 10 =
1
9
∗ 0 +
6
9
∗ 1 +
2
9
∗ 2 =
10
9
Y X 0 1 2
10 0.05 0.3 0.1
20 0.1 0.25 0.2
Example 5 - Probability
The following table gives the joint probability density function = , = =
(, ) of two random variables X and Y:
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Compare () and (| = 10). Are the conditional and unconditional
expectations the same? If no, why are they different?
We know that:
• = 1.15
• = 10 =
10
9
≈ 1.11
• The two expectations are not the same;
• It is also clear that the distribution of X is affected by Y, so we can say that these
two variables are not independent.
Example 5 - Probability
Y X 0 1 2
10 0.05 0.3 0.1
20 0.1 0.25 0.2
The following table gives the joint probability density function = , = =
(, ) of two random variables X and Y: