251var2 4/19/06 (Open in page layout)
Roger Even Bove
FORMULAS FOR FUNCTIONS OF RANDOM VARIABLES
I.
Basic Computational Formulas for Descriptive Statistics
Consider the following set of observations:
You can easily verify that
y
3
6
9
x
7
15
2
Observation number
1
2
3
x = 8 and y = 6. The formula for the sample variance is
s x2
x

2
 nx 2
n 1
So that
s x2
x

2
 nx 2
n 1

7
2

 15 2  2 2  382 278  192

 43
3 1
2
s x  43 .
and
Similarly
s 2y 
y
2
 ny 2
n 1

3
2

 6 2  9 2  362 126  108

 9 and s y  3
3 1
2
The formula for the sample covariance is s xy 
 xy  nx y 
n 1
1
n 1
 xy  nx y 
So that , for the numbers above
1
(7)(3)  (15)(6)  (2)(9)   3(8)(6)  1 129  144   7.5
s xy 
3 1
2
pg. 58
The only thing that we can usually learn from a covariance is whether the variables x and y move
together or in opposite directions. If a covariance is positive the two variables tend to move together. If the
covariance is negative the two variables tend to move in the same direction. To find out about the strength
of the relationship we compute the correlation. The correlation can only have values between 1.0 and -1.0.
The sign of the correlation means the same thing as the sign of the covariance. A correlation close to 1.0 is
referred to as a strong positive correlation. A correlation close to -1.0 is a strong negative correlation. If the
correlation is 1.0, x and y will tend to move proportionally, that is when x rises, y will rise and when
x falls y will fall. When x takes a big jump, y will take a big jump. When x takes a small jump, y
takes a small jump. If the correlation is -1 we have the same proportionality, but now if x jumps, y will
jump in the opposite direction. If the correlation is zero or close to zero, it is weak, which means there is
not much tendency of y to do anything in particular if x moves.
s xy
The formula for sample correlation is rxy 
and we know that s xy  7.5 , s x  43 and
sx s y
s y  3 , so that, for the numbers above rxy 
7.5
 0.381 .
43 3
The negative covariance tells us that x and y have a tendency to move in opposite directions. The
negative correlation tells us the same thing, but the fact that it is closer to zero than 1 leads us to feel that
the correlation is weak. Actually statisticians tend to measure strength on a zero to one scale by squaring
the correlation. In this case rxy2   0.3812  0.145 , which appears quite weak, though far from
nonexistent. The sample covariance is regarded as an estimate of the true or population covariance, just as
the sample correlation is regarded as an estimate of the population correlation. Formulas for computing
these from a population all of whose points are known are not given here. The next section will deal with
computing population covariances and correlations when probabilities are known.
pg. 59
2
II.
FORMULAS FROM PROBABILITY
Let the following table describe the joint probabilities of x and y :
x
7
y
15
2
sum
px 
3
6
9
p y 
0.1
0.1
0.0
0.2
0.2
0.0
0.1
0.3
0.0
0.3
0.2
0.5
0.3
0.4
0.3
1.0
yp y 
1.4
4.5
1.0
y 2 p y 
9.8
67.5
2.0
x
3
6
9
px 
xpx 
xpx 
0.3
0.4
0.3
1.0
0.9
2.9
2.7
6.0
2.7
14.4
24.3
41.4
 E x 
 
 E x2
6.9
 E y 
79.3
 
 E y2
 xpx  6.0
 E  y    yp y  = 6.9
Note that  x  E x  
and similarly  y
 
 E y   
and  x2  E x 2   x2 = 41.4  6 2  5.4
and 
2
y
2
2
y
= 79 .3  6.9 2  31.69
This implies that  x  5.4  2.32 and that  y  31.69  5.63 .
The formula for the covariance is  xy  E xy    x  y 
 xy px, y   
xy
.
We call this a population covariance, since the probabilities presumably refer to all values of x and y. The
most difficult part of this formula is the evaluation of the expected value of xy , E xy  . The idea here is to
multiply each possible pair of values of x and y by the joint probability of the pair. One way to do this is
to take the joint probability table and to add the values of x and y to it. Notice that in the table below the
probabilities (like 0.1) are in exactly the same place as in the joint probability table above and are followed
by the corresponding x and y.
E xy  
 xy  px, y 
 0.1(3)( 7)  0.2(3)(15)  0.0(3)( 2)
 0.1(6)( 7)  0.0(6)(15)  0.3(6)( 2)
0.0(9)( 7)  0.1(9)(15)  0.2(9)( 2)  36 .0
So,  xy  Exy   x  y  36.0  (6.0)(6.9)  5.4 .
Once again we find a negative covariance, indicating a tendency of x and y to move in opposite
directions.
To measure the strength of the relationship, we must compute the correlation. As with the sample
correlation, the population correlation is computed by dividing the covariance by the standard deviations of
x and y . This time the formula for the correlation reads:  xy 
 xy
 x y
. From above, we know that
5.4
 0.41 .
(2.32 )(5.63)
As with the sample correlation this can only take values between negative and positive one. Since, if we
square -0.41 we get 0.17, this too is a weak correlation.
 xy  5.4, x  2.32 and  y  5.63 . Thus  xy 
pg. 60
3
In many situations, especially with population correlations, we are likely to need the covariance
and know the correlation. The formula for population correlation can be rewritten as  xy   xy x y .
Thus, if we know  xy  0.41,  x  2.32 and  y  5.63 , we can compute the covariance.
 xy  0.41(2.32)(5.63)  5.4
The corresponding formula for the sample covariance is s xy  rxy s x s y .
pg. 61
4
III
FUNCTIONS OF RANDOM VARIABLES
A. Functions of a Single Random Variable.
1. The Mean.
If we know the mean of the distribution of a random variable, we can easily find the mean of a
linear function of the same random variable. For example if we know the mean of x we can find the mean
of 5x  7 . In the following let a and b be constants that either multiply x or are added to x . Of course,
 x  E x  
xp(x) , but these formulas apply to x , the sample mean, as well.

a)
If b is a constant, then E b   b . For example E7  7 .
If a is a constant, then E ax  aEx  . For example, E 5x   5E ( x)  5 x , so
that if the mean of x is  x  3 , then the mean of 5 x will be
b)
E5x   5Ex   5 x  53  15.
c) If b is a constant, then E x  b  E x   b . For example,
E x  7  E x   7 , so that if the mean of x is  x  3 , then the mean of x  7 will be
Ex  7  Ex   7   x  7  3  7  10 .
d) If a and b are both constants, then E ax  b  aEx   b . For example,
E 5x  7  5E x   7 , so that if the mean of x is  x  3 , then the mean of 5x  7 will
E 5x  7  5E x   7  5 x  7  53  7  22 .
be
Note that rules a), b), and c) are really special cases of rule d). Rule a) is rule d) with a set equal
to zero. Rule b) is role d) with b set equal to zero. Rule c) is rule d) with a set equal to one.
2. The Variance.
If we know the variance of a random variable, we can find the variance of a linear function of the
same variable. For example, if we know the variance of x , we can find the variance of 5x  7 . These
formulas are stated in terms of the population variance,

  
 x2  Var( x)  E x   2  E x 2   x2 
s2 
x
2
 nx
n 1
 x
2

px   x2 , but can also be used for the sample variance
2
.
a) If b is a constant, then Var b   0 . For example , Var 7   0. This makes perfect
sense. A constant does not vary, so its variance is zero.
b) If a is a constant, then Varax  a 2 Varx . For example,
Var5x  52 Varx  25Varx , so that if the variance is Varx    x2  20 ,
then the variance of 5 x will be Var 5x   25Var x   2520   500 .
pg. 62
5
c) If b is a constant, then Var x  b  Var x  . For example, Var ( x  7)  Var x 
so that if the variance of x is  x2  20 , then the variance of x  7 is
Varx  7  Varx   x2  20 . Again this is something like common sense. Adding a
constant to x doesn't affect how much it varies, so it doesn't affect its variance.
d) If a and b are both constants, then Varax  b  a 2 Varx . For example,
Var5x  7  5 2 Varx , so that if the variance of x is  x2  20 , then
Var5x  7  5 2 Varx  2520  500 .
We can summarize this in the table below:
b
ax
E y  
b
aEx 
xb
E x   b
If y 
ax  b
a E x   b
Var(y) 
0
a 2Varx
Varx 
a 2Varx
B. The Mean and Variance of Sums of Random Variables.
There are two important rules about sums of random variables. The first one seems to be
intuitively obvious, the second one much less so.
1. The Mean.
If x1 , x2 , x3 , xn are random variables, then
E x1  x 2  x3   x n   Ex1   Ex 2   Ex3    Ex n  .
For example, if E ( x1 )  4,E ( x2 )  7 and E x3   9 , then
E x1  x 2  x3   4  7  9  20 .
2. The Variance.
If x1 , x 2 , x3 ,  x n are independent random variables, then
Varx1  x 2  x3   x n   Varx1   Varx 2   Varx3   Varx n  .
For example, if Var( x1 )  4,Var( x2 )  9 and Ex3   16 and these variables are
independent, then Varx1  x2  x3   4  9  16  29 . This means, of course,
that  xx  x2  x3  29 , and that you cannot add standard deviations..
pg. 63
6
C. Functions of two Random Variables
Since these rules work for sample variances and covariances as well as population variances and
covariances, we will use Covx, y  in place of  xy or sxy , and Corr x, y  in place of  xy or
rxy .
Let us assume that we have two variables, x and y , with the following properties
Ex   1.2,E y   3.7,Varx   4.0,Var y   9.0,Covx, y   3.0 , so that
Corr x, y  
Covx, y 
Var x 
Var  y 

3.0

4 9
3
23
 0.3 .
1. Linear Functions of Two Random Variables.
Let us introduce two new variables, w and v , so that w  ax  b , and v  cy  d , where
a, b, c, and d are constants. From the earlier part of this section we know the following:
 w2  Varw  a 2Varx  a 2 x2
 w  Ew  Eax  b  aEx   b
 v2  Var v   c 2Var  y   c 2 y2
 v  E v   E cy  d   cE y   d
To this we now add a new rule: Covw, v   wv  acCovx, y   ac xy
To find the correlation between w and v , recall that  wv 
 w2  a 2 x2
 wv 
 wv
. But since
 w v
and  v2  c 2 y2 , then
ac xy
a 2 x2 c 2 y2

ac xy
ac  x y

ac  xy
 signac xy .
ac  x y
Note that, because the ac in the numerator cancels the ac in the denominator, the only
thing that ac contributes to the result is its sign. If the product of a and c is negative we
reverse the sign of  xy . Signac thus takes the values 1 or 1 .
For example, let w  5 x  1 and v  3x  2 so that a  5 , b  1 , c  3 and d  2 .
Then  wv  ac xy  53 xy . But we already know that  xy  3 , so that
 xy  5 33  45 . Now remember that  x2  4.0 and  y2  9.0 , so that
 w2  a 2 x2  5 2 4.0  100 and  v2  c 2 y2   32 9.0  81 . Thus, the correlation
between w and v can be found in two ways.

 45
 wv  wv 
 0.5
 w v
100 81
or  wv  signac xy  10.5  0.5 .
Again, remember that these rules hold for sample data too. That is
s
s w2  a 2 s x2 , s v2  c 2 s 2y , swv  ac s xy , and rwv  wv  signacrxy .
s w sv
pg. 64
7
2. Sums of Random Variables.
The question now is what happens if we add together two random variables. We learned in
Section B above that you can add means. This means that if u  x  y the mean of u will be
the sum of the mean of x and the mean of y .
Formally Eu   Ex  y   Ex  E y    x   y , or, since this also applies to
samples, u  x  y . For example, if  x  1.2 and  y  3.7 , then  u  1.2  3.7  4.9 .
But the situation with variances is not so simple. We learned in Section B that we can add
variances only if the variables are independent. If the variables are independent their
covariance and correlation will both be zero. But if they are not zero, we must take the value
of the covariance into account. Often the covariance will not be available and we must
compute  xy   xy x y . Then, if u  x  y , we can use the formula:
 u2  Var u   Var x  y   Var x   2Covx, y   Var  y    x2  2 xy   y2
Or, if we are working with sample data: s u2  s x2  2s xy  s 2y . For example let us assume
that  x2  4,  y2  9 and  xy  0.5 . Then  x  2,  y  3 and
 xy   xy  x y  0.523  3 .
Thus Var x  y    x2  2 xy   x2  4  23  9  19 .
3. Sums of Functions of Random Variables
By combining the information from the last two sections, we can look at the situation that
occurs when we deal with a sum of two random variables. To keep things simple, let w  ax ,
u  cy and v  w  u , so that v  ax  cy . As far as the mean is concerned, we can say
that Ev   E w  u   Eax  cy   E w  E u  . But E w  E ax  aEx 
E u   E cy   cE y  , so that E v   aEx   cE y  . Alternately,  v  a x  c y .
For example, If w  3x and u  5 y so that v  3x  5 y , then E v   3E x   5E  y 
and if E x   1.2 and E  y   3.7 , Ev   31.2  53.7  22.1 .
Also Varv   Varw  u   Varw  2Covw, u   Var u  . But since w  ax and u  cy ,
Varw  a 2Varx , Varu   c 2Var y  and Covw, u   acCovx, y  , we can write
Varv  a 2Varx  2acCovx, y   c 2Var y  or  v2  a 2 x2  2ac xy  c 2 y2 .
To summarize then, Varax  cy  a 2Varx  2acCovx, y   c 2Var y  .
For example, Let v  3x  5 y . Then Var 3x+5 y   3 2  x2  235 xy  5 2  y2 .
So if  x  2 ,  y  3 , and  xy  3 , Var3x+5 y   3 2 2 2  2353  5 2 32
 36 + 90 + 225 = 351 .
pg. 65
8
IV. APPLICATION TO PORTFOLIO ANALYSIS
We can now use the formulas above to find the mean and variance of a portfolio of stocks, To
keep things simple, assume that we are offered only two stocks, and that the return of the first stock is R1 ,
while the return of the second stock is R 2 . Let us assume that each stock sells for $1.00 a share, and that
we have exactly $1 to invest. Since we can buy fractional shares, we divide our dollar into two parts,
P1 and P2 , where P1+P2=1 . Our total return is R=P1 R1+P2 R2 . For example, if P1=.60 ,
P2=.40 , R1=.08 and R2=.06 , our total return is R=.60 .08   .40 .06   .072
.
A. Mean Return
We know that E ax+cy  aEx   cE y  , so that
E P1 R1+P2 R2   P1 E R1 +P2 E R2  . For example, if we split our money equally
between two stocks P1 and P2 both equal .50 . Then the expected return is
E R   E .50 1 R+.50 R2   .50 E R1   .50 E R2  . In particular, if E R1   0.20
and E R2   0.24 , E R   .50.20   .50.24   0.22 .
B. Variance of the Return
We know that Varax+cy=a 2Varx+c 2Var y +2acCovx,y , so that
VarR=VarP1 R1+P2 R2 =P12VarR1 +P22VarR2 +2P1 P2 CovR1 ,R2 
is the variance of the return. Thus if P1 and P2 are both .50 , we can say
VarR =.25VarR1 +.25VarR2 +.50CovR1 R2  .
For example, assume that  R1  0.20 ,  R2  0.30 , but  R1R 2 is unknown. Then
CovR1 ,R2    R1R2  R1  R2 = R1R2 .20.30=.06 R1R2 . If we use the formula for VarR 
immediately above, VarR  .25.202 +.25.302 +.50.06 R1R2
= .0100 + .0225 + .0300 R1R 2  .0325 + .0300 R1R 2 . Now we can see the effect various
values of R1R 2 will have on VarR  and  R  Var R  .
If  R1R2  1, VarR=.0325+.03001 =.0625 and  R  .2500.
If  R1R2  0, VarR=.0325+.03000 =.0325 and  R  .1803.
If  R1R2  1,VarR=.0325+.0300-1=.0025 and  R  .0500 .
pg. 66
9
C. Variance Minimization.
The purpose of this section is to show how to find the minimum value for Var R  . Since
variance is a measure of risk, minimizing variance minimizes risk, though actually, the best
measure of risk is probably the coefficient of variation, the standard deviation divided by the
mean, in this case C 
R
E R 
.
Remember that VarR= R2 =P12VarR1 +P22VarR2 +2P1 P2 CovR1 R2  .
Also recall that, since P1 and P2 are shares of $1.00, P1+P2  1 , then P2  1  P1 .
Remember too, that CovR1 ,R2    R1R2  R1  R2 . If we put all this together,
VarR  P12VarR1   1-P1 2 VarR2   2P1 1-P1  R1R2  R1 R2 .
Now let us assume some values for the standard deviations and the correlation. Let
 R1  0.4 soVarR1   0.16 ,  R2  0.3 soVarR2   0.09 and
CorrR1 ,R2    R1R2  0.5 . Then
VarR   P12 0.16   1-P1 2 0.09   2P1 1-P1 0.50.40.6.
 0.16 P12  0.09 1-P1 2  20.06 P1 1  P1 



 0.16P12  0.09 1  2P1  P12  0.12 P1-P12
 0.16P12

 0.09  0.18P12
 0.09P12  0.12P1  0.12P12
If we collect terms in P1 and P12 , we get
Var R  0.16  0.09  0.12 P12   0.18  0.12 P1  0.09
or Var R  0.13P12  0.06P1  0.09 .
  
 



In order to minimize risk we pick our value of P1 to give us a minimum variance. The way
That we find this minimum variance is by taking the first derivative of VarR  with respect to
P1 and setting it equal to zero.
Since
d
Var R   0.26 P1  0.06 , if we set the variance equal to zero we get
dP1
0.06
 0.2308 . Now since
0.26
P1  P2  1 , we set P2  1  P1  0.7692 . That is, to minimize risk, we put about 23% of
our money in stock 1 and 77% in stock 2.
0.26 P1  0.06  0 , which implies that P1 
pg. 67
10