06-Joint PDF’s
Joint PDFs
 So far: univariate PDFs
 For a continuous PDF, E[r] is the best
forecast for the return, given that we only
know the PDF
 How do variables interact with each
other?
 Can we find relations between past and
future random variables?
 Can we find variables that predict returns?
Joint Discrete PDF
 The numbers in the box are “joint
probabilities”.
 Example: The probability that the return on
IBM is 18% and the return on the market is
-5% is 0.25.
 Notation: f(IBM=.18,M=-0.05)=0.25
Marginal PDF
 MPDF: An individual PDF derived from a joint
PDF
 This is sometimes referred to as an
unconditional probability, because we’re not
conditioning on the outcome of another random
variable.
Marginal Probability Function
 Example:
 Only two possible outcomes for IBM
 0.18 or -0.10
 What is probability IBM will take on these
two outcomes regardless of market
outcome, fIBM(.18) and fIBM(-.10)?
Marginal Probability Function
 Probability IBM will be 18% regardless of
market: 0.40+0.25=0.65
fIBM(.18)=0.65
 Probability IBM will be -10% regardless of
market: 0.05+0.30=0.35
fIBM(-.10)=0.35
Unconditional Expectation
 Unconditional Expected Return for IBM
 E[rIBM]=.65*(.18)+.35*(-.10)=8.2%
Conditional PDF
 Does knowing the outcome for the market change
our view of the probabilities for the outcomes for
IBM?
 Notation: given the market returns 12%, the
conditional PDF for IBM is f(IBM|M=12%)
 “|” means “given”
f ( IBM , M  .12)
f ( IBM | M  0.12) 
f M (.12)
Venn Diagrams
X
Y
f (Y , X )
f (Y | X ) 
f (X )
Conditional Probability Example
 What is probability that IBM = .18 given
that the market is .12?
 fM(.12)=45%
 f(IBM=.18|M=.12)=.40/.45=88.89%
 What is probability that IBM = -.10 given
that the market is .12?
 f(IBM=-.10|M=.12)=.05/.45=11.11%
Conditional Expectation
 Conditional Expected Return for IBM
 E[rIBM|rM=.12]=
.8889*(.18)+.1111*(-.10)=14.9%
 Recall: E[rIBM]=8.2% (unconditional)
 Knowing the market return is 12% changed
our view of the expected return for IBM.
 Knowing that the market return is 12% causes us to
put greater probability on the event that IBM
returns 18%.
Independent Random Variables
 X and Y are independent random
variables if knowing one doesn’t tell
you anything about the other.
 Knowing one variable does not change
our view of probabilities.
 That is:
f (X  X | Y  Y )  fX (X )
*
*
*
Independent Random Variables
 If X and Y are independent
f (X |Y)  fX (X )
But, by definition of conditiona l probability :
f ( X ,Y )
fX (X |Y) 
fY (Y )
which implies
f X ( X ) fY (Y )  f ( X , Y )
Example
 Find fGM(GM=-.15) (one number)
 Find f(GM=-.15|M=.13)
f(GM=-.15|M=.08)
f(GM=-.15|M=-.05)
Example
 Find fGM(GM=-.15)
 fGM(-.15)=10%
Example
 fM(-.05)=20%
 fM(0.08)=20%
 fM(0.13)=60%
 f(GM=-.15|M=-.05)=.02/.20=10%
 f(GM=-.15|M=.08)=.02/.20=10%
 f(GM=-.15|M=.13)=.06/.60=10%
Example
 Note f(GM,M)=fGM(GM)*fM(M)
Perfect Linear Dependence
 If two variables are perfectly linear
dependent, then the outcome for one
random variable can be written as a
linear function of the outcome for the
other random variable.
Perfect Linear Dependence
rA
rI
12
-5
18
0.877
0
-10
0
0.123
30 28
rA    rI
17 17
Perfect Linear Dependence
18  a  12b
18  a  12b
 10  a  5b
10  a  5b
28  17b
b  28 / 17
28
18  a  12 
17
28
30
a  18  12 

17
17