1. SASF CFA Quant. Review
Investment Tools – Probability

2. Probability
A random variable is a quantity whose outcome is uncertain.
Two defining properties of Probability.
Probability of any event E is a number between 0 and 1, p(E).
Sum of the probabilities of any list of mutually exclusive and exhaustive events equals 1.
Mutually exclusive = one and only one event can occur at any time.
Exhaustive = one of the events must occur, jointly cover all possible outcomes.
Empirical probability - probability of an event occurring is estimated from data, usually in the form of a relative frequency.
A priori probability - probability of an event is deduced by reasoning about the structure of the problem itself.
Subjective probability - probability of an event is based on a personal assessment without reference to any particular data.

3. Visualizing Sample Space
1. Listing
S = {Head, Tail}
2. Contingency Table
3. Decision Tree Diagram

4. Contingency Table Experiment: Toss 2 Coins. Note Faces. 2 Coin 1 Coinnd 2
Coin st 1
Coin
Head
Tail
Total
Outcome (Count, Total % Shown Usually)
Simple Event (Head on 1st Coin)
Head HH HT
HH, HT
Tail TH TT
TH, TT
S = {HH, HT, TH, TT}
Sample Space
Total
HH, TH
HT, TT S
To be consistent with the Berenson & Levine text, a simple event is shown. Typically, this is not considered an event since it is not an outcome of the experiment.

5. Tree Diagram H HH H T HT H TH T T TT
Experiment: Toss 2 Coins. Note Faces. H HH H T HT
Outcome H TH T T TT
S = {HH, HT, TH, TT}
Sample Space

6. Probabilities for Two Dice
Second Die
First
Die

7. Forming Compound Events
1. Intersection
Outcomes in Both Events A and B
‘AND’ Statement
 Symbol (i.e., A  B)
2. Union
Outcomes in Either Events A or B or Both
‘OR’ Statement
 Symbol (i.e., A  B)

8. P(A) = P(A B1) + P(A  B2) + P(AB3) +…+ P(A  B6) = 6/36 = 1/6
Joint Probabilities for Two Dice
What is the Probability of
throwing a Seven
with the two Dice?
Second Die
First
Die
Answer: Count the combinations that yield 7 and add the individual probabilities
to find the probability of the event, A, “Rolling a seven”
P(A) = P(A B1) + P(A  B2) + P(AB3) +…+ P(A  B6) = 6/36 = 1/6

9. Addition Rule
1. Used to Get Compound Probabilities for Union of Events:
P(A OR B) = P(A  B) = P(A) + P(B) - P(A  B)
2. For Mutually Exclusive Events:
P(A OR B) = P(A  B) = P(A) + P(B)

10. Addition Rule for Probabilities
What is the Probability of
throwing at least one die
showing 1 with the two Dice?
Second Die
Answer: Count the
combinations that
have one die with 1
and add the individual
probabilities OR
use the Addition Rule
for probabilities.
P(AB) =
P(A) + P(B) - P(A  B)
= 1/6 + 1/6 -1/36
= 11/36
First
Die

11. Addition Rule - Again Second Die First Die What is the Probability of
throwing a four or higher on
either the first or second die?
Second Die
Answer: Count the
combinations that have
one die with 4 or greater
& add the individual
probabilities OR
use the Addition Rule
for probabilities.
P(AB) =
P(A) + P(B) - P(A  B)
= 1/2 + 1/2 -1/4
= 3/4
First
Die

12. Conditional Probability
Probability that one Event will occur Given that Another Event has already occurred.
To calculate a Conditional probability –
Revise Original Sample Space to Account for New Information
Eliminates Certain Outcomes
Result:
P(A | B) = P(A and B) P(B)

13. Conditional Probability
What is the Probability of throwing
at least an 8 if the first die is a 4?
Second Die
Answer: Note we
know we are in the
row associated with
a 4 for the first die.
There are
only 3 ways of
the possible six
outcomes for the
second die that
yield a sum  8.
Hence P(A|B) =
= P(AB)/P(B)
= (3/36)/(6/36) =.5
First
Die

14. Statistical Independence
1. Event Occurrence Does Not Affect Probability of Another Event
Toss first die, what does its outcome tell you about the likely results in tossing the second die?
Answer: Nothing. The two die are independent
2. Causality Not Implied
3. Tests For Statistical Independence
P(A | B) = P(A)
P(A and B) = P(A)*P(B)

15. Multiplication Rule
1. Used to Get Compound Probabilities for Intersection of Events
Called Joint Events
2. P(A and B) = P(A  B) = P(A)*P(B|A) = P(B)*P(A|B)
3. For Independent Events: P(A and B) = P(A  B) = P(A)*P(B)

16. Multiplication Rule with Independence
What is the Probability of throwing
“snake eyes” i.e. a 1 on each die?
Answer: The result
on the first die is independent
of the second die. We use
The multiplication
rule to find prob.
of A and B
Hence P(A  B) =
= P(A)*P(B)
= (1/6)*(1/6)
= (1/36)
Second Die
First
Die

17. Discrete R.V.’s - Summary Measures
Expected Value
Mean of Probability Distribution
Weighted Average of All Possible Values
X = E(X) = Xi P(Xi)
= p(X1) X1+ p(X2) X2+ … + p(Xn) Xn
Variance
Weighted Average Squared Deviation about Mean
X2 = E[ (Xi-X(Xi-XP(Xi)
= p(X1)(X1- X)2 + p(X2 )(X2- X)2 + … + p(Xn) )(Xn- X)2
Standard deviation
Square root of the variance. Measure of risk shows dispersion of possible outcomes around expected level of outcomes.
Population notation is used since all values are specified.

18. Mean/Variance Calculation Table2 X
P(X ) X
P(X ) X -  (X -  ) ( X -  ) 2 P( X ) i i i i i i i i
Total  X
P(X )  ( X -  2 ) P( X ) i i i i

19. Asset Expected Return & Risk
Calculate the Expected Returns of alternative assets.
Rki= Return to Asset k if Event i occurs
P(Rki) = Probability Event i occurs.

20. Calculating ER and Risk
Can use the previous table to calculate ER and variance (risk) for each asset.
Show calculations for Asset A below.
ERA = 9.1%
A = 3.56%
ERA/A = 2.55

21. Covariance
Covariance between two random variables X and Y is defined as
A negative covariance between X and Y means that when X is above its mean its is likely that Y is below its mean value.
If the covariance of the two random variables is zero then on average the values of the two variables are unrelated.
A positive covariance between X and Y means that when X is above its mean its is likely that Y is above its mean value.
Covariance of a random variable with itself, its own covariance, is equal to its variance.

22. Correlation
Correlation between two random variables X and Y measured as:
Correlation takes on values between –1 and +1. Correlation is a standardized measure of how two random variables move together, i.e. correlation has no units associated with it.
A correlation of 0 means there is no straight-line (linear) relationship between the two variables.
Increasingly positive (negative) correlations indicate an increasingly strong positive (negative) linear relationship between the variables.
When the correlation equals 1 (-1) there is a perfect positive (negative) linear relationship between the two variables.

23. Portfolio Probability Calculations
Portfolio consisting of two assets A and B, wA invested in A. Asset A has expected return rA and variance s2A. Asset B has expected return rB and variance s2B. The correlation between the two returns is rAB.
Portfolio Expected Return:
E(rp) = wA rA + (1-wA )rB
Portfolio Variance: or

24. Exam Questions on Probability
28. The probability that two or more events will happen concurrently is:
A. joint probability.
B. multiple probability.
C. concurrent probability.
D. conditional probability.

25. Exam Question on Expected Values
24. An analyst developed the following probability distribution of the rate of return for a common stock:
Scenario Probability Rate of return
The standard deviation of the rate of return is closest to: A B C D
To calculate standard deviation must first calculate expected value, mean, of the rate of return.
= [.25 x x x .16]
= 0.12
Calculate variance as probability-weighted squared deviations of values from expected value.
s2 = .25[ ]2 + .5[ ] [ ]2 =
Finally calculate standard deviation as square root of variance.
s = SQRT(0.0008) =

26. Common Probability Distributions

27. Probability Distributions
Probability distribution specifies the probabilities of the possible outcomes of a random variable.
Discrete random variable can take on at most a countable number of possible values, such as coin flip or rolling dice.
Continuous random variable can take on an uncountable (infinite) number of possible values, such as asset returns or temperatures.
Probability function specifies the probability that the random variable takes on a specific value: P(X = x).
Two Key Properties of a Probability Function.
0 ≤ p(x) ≤ 1 because a probability lies between 0 and 1.
The sum of probabilities p(x) over all values of X equals 1.

28. Discrete Probability Distributions
Discrete Uniform Random Variable: The uniform random variable X takes on a finite number of values, k, and each value has the same probability of occurring, i.e. P(xi) = 1/k for i = 1,2,…,k.
Bernoulli random variable is a binary variable that takes on one of two values, usually 1 for success or 0 for failure.
Think of a single coin flip as an example of a Bernoulli r.v.
Binomial random variable: X ~ B(n, p) is defined is the number of successes in n Bernoulli random trials where p is the probability of success on any one Bernoulli trial.
For 4 = X ~ B(10, 0.5) think –“what’s probability of 4 heads in ten coin flips?”
Probability distribution for a Binomial random variable is given by:

29. Population Distribution
Uniform Distribution
Example of distribution of a uniform random variable.
Summary Measures
Population Distribution  x i N X   1 2 5 . .3 .2 .1 .0 1 2 3 4   x i N X    2 1 12 . [ }
Have students verify these numbers.

30. Binomial Dist’n – Coin Flips
P(X)
N = p = .5
Probability of exactly 4 heads in 10 tosses = Bar Height .3 .2 .1
Probability of 4 heads or fewer in 10 tosses = Sum of Bar Heights .0 X
As the number of vertical bars (n) increases, the errors due to approximating with the normal decrease. 2 4 6 8 10

31. Binomial Probability Dist’n FunctionX x n p ( | , ) !   1
P(X= x | n,p) = probability that X = x given n & p
n = sample size { =3 in previous example }
p = probability of ‘success’ { Pr(Price up)=.75 }
x = number of ‘successes’ in sample (X = 0, 1, 2, ..., n)

32. Binomial Dist’n - Characteristics
n = 5 p = 0.1
Mean
P(X) .6   E ( X )  np .4 x .2 .0 X 1 2 3 4 5
Standard Deviation
n = 5 p = 0.5
Distribution has different shapes.
1st Graph:
If inspecting 5 items & the probability of a defect is 0.1 (10%), the probability of finding 0 defective item is about 0.6 (60%).
If inspecting 5 items & the probability of a defect is 0.1 (10%), the probability of finding 1 defective items is about .35 (35%).
2nd Graph:
If inspecting 5 items & the probability of a defect is 0.5 (50%), the probability of finding 1 defective items is about .18 (18%).
Note:
Could use formula or tables at end of text to get probabilities.   np ( 1  p )
P(X) x .6 .4 .2 .0 X 1 2 3 4 5

33. Evolution of Share Price
Assuming:
S=$130
S=$110
.75
.25
P = .4219
1. Share Price goes
up or down by $10
each week.
S=$120
.75
P = .5625
P = 3(.75)(.75)(.25)
2. Probability that
Share Price goes
down by $10 is
25% each week. i.e.
pdown = .25
S=$110
.75
P = 2(.25)(.75)
P = .75
S=$100
.25
S=$90
.75
.25
P = .4219
S=$100
3. Probability that
Share Price goes
up by $10 is 75% each week.
pup = .75
S=$90
.25
P = .25
.75
P = .375
S=$80
.25
P = .0625
S=$70
.75
.25
P = .1406
P = .0156
4. Probability that
Share Price goes
up this week is NOT affected by what happened before.
Week 1
Week 2
Week 3
Week 4

34. Normal Distribution
Normal distribution is a continuous, symmetric probability distribution that is completely described by two parameters: its mean, μ, and its variance, σ2.
General Normal random variable – X ~ N(μ, σ2)
The normal distribution is said to be bell-shaped with the mean showing its central location and the variance showing its “spread”.
A linear combination of two or more Normal random variables is also normally distributed.
Standard Normal distribution – Z ~ N(0, 1).
is a Normal distribution with mean μ=0, and variance σ2=1.

35. Effect of Varying Parameters (x & x)
f(X)
mA = mB, sA > sB B
mA < mC, sA = sC A C X

36. N(a < X < b| μ, σ2) = N(X < b| μ, σ2) - N( X < a| μ, σ2) .
Normal Distribution
General Normal random variable X ~ N(μ, σ2)
X can be standardized to a Standard Normal random variable.
Resulting variable has mean zero and variance equal to 1.
Calculating probabilities for a normal random variable
X ~ N(μ, σ2) taking on a range of specified values, say a < X < b, directly as the area under the normal curve using the cumulative Normal distribution function as:
N(a < X < b| μ, σ2) = N(X < b| μ, σ2) - N( X < a| μ, σ2) .
You should be able to show what this looks like using a diagram of the Normal distribution.

37.  Confidence Intervals x X X= x ± ZX x x-2.58x x-1.65x
90% level
95% level
99% level

38. Questions on Probability Distributions
30. A normal distribution would least likely be described as:
A. asymptotic.
B. a discrete probability distribution.
C. a symmetrical or bell-shaped distribution.
D. a curve that theoretically extends from negative infinity to positive infinity.
31. An investment strategy has an expected return of 12 percent and a standard deviation of 10 percent. If investment returns are normally distributed, the probability of earning a return less than 2 percent is closest to:
A. 10%.
B. 16%.
C. 32%.
D. 34%.
2% is 1 st. dev. (s=10%) away from the mean of 12%.
We know that for a Normal distribution there is a 68% chance of being within 1 st. dev. of the mean.
We are 1 st.dev below, so there is (1-.68)/2 = 16% chance of this occurring.

39. Question on Confidence Intervals
32. Based on a normal distribution with a mean of 500 and a standard deviation of 150, the Z-value for an observation of 200 is closest to: A B C D
34. If the standard deviation of a population is 100 and a sample size taken from that population is 64, the standard error of the sample means is closest to: A B C D

40. Hypothesis Testing Process
I Believe the
Population
Mean Age is 50.
(Hypothesis)
Population
The Sample
Mean Is 20
Sample Is
REJECT
X = 20
= 50? No!
Hypothesis    X

41. Null Hypothesis 1. States What Is Tested by the Data
2. Has Serious Outcome If Incorrect Decision Made
3. Always Has Equality Sign: , , or 
4. Designated H0
Pronounced H Sub-Zero or H-Oh
5. Example - Stated H0: X  3

42. Alternative Hypothesis
1. Opposite of Null Hypothesis (the Complement of the Null Hypothesis)
2. Always Has Inequality Sign: ,, or 
3. Designated H1
4. Stated H1: X < 3

43. Basic Idea for Hypothesis Test
Sampling Distribution
(Reflects potential error between sample and population)
…it is unlikely that we would get a sample mean of this value ... 20
= 50
If in fact this were the population mean H0
... therefore, we reject the hypothesis that X = 50. 
Sample Mean X

44. Level of Significance 1. Cast in terms of a Probability
Defines maximum unlikely Values of Sample Statistic if Null Hypothesis true.
Called Rejection Region of Sampling Dist’n
4. Designated  (alpha)
Typical Values Are .01, .05, .10
5. Selected by Researcher at start of study

45. Sampling Distribution
Two-Tailed Test
Sampling Distribution
Nonrejection
Region
1 - 
Level of Confidence
1/2 
Rejection
Region
Rejection region does NOT include critical value.
Critical
Value H0
Sample Statistic
Value

46. Z-Test Statistic (X Known)
1. Convert Sample Statistic to Standardized Z Variable
2. Compare to critical Za/2 Values
If Z-Test Statistic falls in Critical Region, Reject H0,
Otherwise Do Not Reject H0 x n X Z    
Rejection region does NOT include critical value.
Reject if:
Z-Test Statistic > Critical Z Value or
Z-Test Statistic < Critical Z Value

47. Errors in Making Decision
1. Type I Error
Reject a Null Hypothesis that is actually true.
Has Serious Consequences
Probability of Type I Error Is 
Called Level of Significance
2. Type II Error
Do Not Reject a Null Hypothesis that is actually false.
Probability of Type II Error Is (Beta)