Download Chapter 02 - Cambridge University Press

Introductory Econometrics for Finance
Chris Brooks
Solutions to Review Questions - Chapter 2
1. This question simply involves plugging the appropriate values of x into the
formulae.
(a) x=0, f(0) = 3(02) – 40 + 2 = 2.
x=2, f(2) = 3(22) – 42 + 2 = 6.
x=–1, f(–1) = 3(–1)2 – 4(–1) + 2 = 9.
(b) x=0, f(0) =4(02) + (20) – 3 = –3.
x=3, f(3) =4(32) + (23) – 3 = 39.
x=a, f(a) =4(a2) + (2a) – 3 = 4a2+2a–3.
x=(3+a), f(3+a) =4((3+a)2) + (2(3+a)) – 3 = 4(9+6a+a2)+6+2a–
3=4a2+26a+39.
(c) In general, the answer is no. We can see from the final example compared
with the sum of the previous two in part (b) above that the rule does not
follow when f is a quadratic function – in other words, f(3) + f(a)  f(3+a).
2. To answer this question, we simply need to use the rules for manipulating powers
so that we group like terms together.
(a) 4x5  6x3 = 24x8.
(b) 3x2  4y2  8x4  –2y4 = –192x6y6.
(c) (4p2q3)3 =43p23q33 = 64p6q9.
(d) 6x5÷3x2 = (6/3)x5–2 = 2x3.
(e) 7y2÷2y5 = (7/2)y2–5 = 7/(2y3).
© Chris Brooks 2014
1
Introductory Econometrics for Finance by Chris Brooks
(f)
3( xy) 3  6( xz) 4 9 x 3 y 3 x 4 z 4 9 x 7 y 3 z 4


 9 x 2 yz 4 .
2( xy) 2 x 3
x 2 y 2 x3
x5 y 2
(g) (xy)3 ÷ x3y3 = x3y3 ÷ x3y3 = 1.
(h) (xy)3 – x3y3 = x3y3 – x3y3 = 0.
3. Again, here we use the rules for manipulating powers, but now the powers are
not integers and thus effectively involve taking roots.
(a) 1251/3 = 3 125  5 .
(b) 641/3 =
3
64  4 .
(c) 161/4 = 4 16  2 .
(d) 93/2 =
(e) 92/3 =
2
3
93  2 729  27 .
92  3 81  4.33 (to 2 decimal places).
(f) 811/2 + 641/2 + 641/3 =
81  64  3 64  9  8  4  21 .
4. This question involves doing the reverse of the steps in the previous one.
(a) 9 = 32.
(b) 625 = 54.
(c) 125–1 = 5–3.
5. Answering this question involves simply rearranging the expression so that the
terms in x are gathered together on one side of the equation and all of the number
terms are on the other side.
(a) 3x–6 = 6x –12, 3x = 6, x = 2.
(b) 2x–304x + 8 = x + 9–3x+4, –302x = –2x+5, 300x = –5, x = –5/300 = –0.017
(to 3 decimal places).
© Chris Brooks 2014
2
Introductory Econometrics for Finance by Chris Brooks
(c)
x  3 2x  6

, 3(x+3) = 2(2x–6), 3x+9 = 4x–12, x–21 = 0, x = 21.
2
3
6. This question involves using the rules of sigma and pi notation to list and simplify
all of the terms in the expressions.
(a)

(b)
 j
3
j 1
j = 1+2+3 = 6.
5
j 2
2
 
(c)

(d)

x (with x = 2) =
j 1
(e)

i  1 2  3  6 .
n
i 1
 
x (with n = 4 and x = 3) =
3
3
i 1
 
 

 j  3  2 2  2  3  32  3  3  4 2  4  3  52  5  3  80 .

3
j 1

4
i 1
3  3  3  3  3  12 .
2  2 2 2  8.
7. Recall that the equation for a straight line is y = a + bx where a is the intercept
and b is the gradient. In the first three cases, the answers arise naturally by plugging
the appropriate values directly into this formula, but the others require more
thought.
(a) y = –1 + 3x.
(b) y = 4 – 2x.
(c) y = 3 + ½x.
(d) Care is needed here. If the line crosses the x–axis at 3, this is not the
intercept. To get the intercept, we need to solve y = a + (1/2)x for a with y =
0 and x = 3, which gives 0 = a + (1/2)3; so a = –3/2 and thus the line is y = –
3/2 + (1/2)x.
(e) We first need to find the gradient by solving y = a + bx for 2, x = 3 and y = 1;
so 1 = 2 + 3b, b = –1/3 and thus the line is y = 2 –(1/3)x.
(f) We first need to find the intercept by solving y = a + bx for a with b = 4, x = –
2, y = –2; so –2 = a +4–2, a = 6 and thus the line is y = 6 + 4x.
© Chris Brooks 2014
3
Introductory Econometrics for Finance by Chris Brooks
(g) First we need to calculate the gradient as being equal to the change in y
divided by the change in x = y/x = (6–2)/(–2–4) = 4/–6 = –2/3. We can then
use either of the two points together with the intercept to obtain the
gradient. Hence find the gradient by solving y = a + bx for a with b =–(2/3), x
= 4, y = 2; so 2 = a + –(2/3)  4, a = (14/3) and the line is y = (14/3) –(2/3)x.
8. (a) y = 6x,
dy
d2y
 6,
0.
dx
dx 2
dy
d2y
 6x ,
6.
dx
dx 2
(b) y = 3x2 + 2,
(c) y = 4x3 + 10,
(d) y 
dy
d2y
 12x 2 ,
 24 x .
dx
dx 2
1
dy
1 d 2 y
2
 x 1 ,   x 2  2 ,
 2 x 3  3 .
2
x
dx
x
dx
x
(e) y = x,
dy
d2y
 1,
0.
dx
dx 2
(f) y = 7,
dy
d2y
 0,
0.
dx
dx 2
(g) y  6 x 3 
y = 3ln x,
6
 36 d 2 y
144
3 dy
4

12
x
,


36
x

,
 144 x 5  5 .
3
4
2
dx
x
x
dx
x
dy 3 d 2 y
3
 ,
 3x 2  2 .
2
dx x dx
x
(h) y = ln(3x2),
dy f ' ( x) 6 x 2 d 2 y
2
 2 x 2  2 .

 2  ,
2
dx f ( x) 3x
x dx
x
3x 4  6 x 2  x  4
 6 1  4
(j) y 
 3x 
 2  3  3x  6 x 1  x 2  4 x 3 ,
3
x x
x
x
dy
6
2 12
 3  6 x 2  2 x 3  12 x 4  3  2  3  4 ,
dx
x
x
x
d2y
 12  6  48
 12 x 3  6 x 4  48 x 5  3  4  5 .
2
dx
x
x
x
© Chris Brooks 2014
4
Introductory Econometrics for Finance by Chris Brooks
9. All we do here is to apply the standard rule for differentiation for x (treating all
the terms in y as if they were constants) and then separately for y (treating all the
terms in x as if they were constants).
(a) z = 10x3 + 6y2 – 7y,
(b) z = 10xy2 – 6,
(c) z = 6x,
(d) z = 4,
10.
z
z
 30x 2 ;
 12 y  7 .
x
y
z
z
 10 y 2 ;
 20 xy .
x
y
z
z
 6;
 0.
x
y
z
z
 0;
 0.
x
y
(a) x2 – 7x – 8 = (x – 8)(x+1).
(b) 5x – 2x2 = x(5 – 2x).
(c) 2x2 – x – 3 = (2x – 3)(x + 1).
(d) 6 + 5x – 4x2 = (2 – x)(3 + 4x).
(e) 54 – 15x – 25x2 = (9 + 5x)(6 – 5x).
11.
(a) 53 = 125, log5 125 = 3.
(b) 112 = 121, log11 121 = 2.
(c) 64 = 1296, log6 1296 = 4.
12.
(a) log10 10000 = 4.
(b) log2 16 = 4.
(c) log10 0.01 = –2.
(d) log5 125 = 3.
(e) loge e2 = 2.
13.
(a) log5 3125 = 5, 55 = 3125.
© Chris Brooks 2014
5
Introductory Econometrics for Finance by Chris Brooks
(b) log49 7 = ½, 491/2 = 7.
(c) log0.5 8 = –3, 0.5–3 = 8.
14.
(a) log 60 = log(435) = log(2235) = 2log 2 + log 3 + log 5.
(b) log 300 = log(32255) = log(32252) = log 3 + 2log 2 + 2log 5.
15.
(a) log 27 – log 9 + log 81 = log 33 – log 32 + log 34 = 3log 3 – 2log 3 + 4log3 =
5log 3.
(b) log 8 – log 4 + log 32 = log 23 – log 22 + log 25 = 3log 2 – 2log 2 + 5log 2 =
6log 2.
16.
4
 x4 
 5x   x   5x 
(a) log x4 – log x3 = log 5x – log 2x, log  3   log   ,  3     , x =
 2x   x   2x 
x 
5/2.
(b) log (x–1) + log (x+1) = 2log (x+2), log(( x  1)( x  1))  log( x  2) 2 , (x+1)(x–
1) = (x+2)2, x2 – 1 = x2 + 4x + 4, 4x = –5, x = –5/4.
(c) log10 x = 4, x = 104 = 10000.
17.
(a) log 16 = log (82) = log 8 + log 2 = log 8 + log 81/3 = 4/3 log 8 = 4/3  2.1 =
2.9 (to one decimal place)
(b) log 64 = log (88) = log 8 + log 8 = 2log 8 = 4.2 (to one decimal place).
(c) log 4 = log (8/2) = log 8 – log 2 = log 8 – log 81/3 = log 8 – 1/3 log 8 = 2/3
log 8 = 2/3  2.1 = 1.5 (to one decimal place).
18.
(a) 4x = 6, x = log4 6 = 1.3 (to one decimal place).
(b) 42x = 3, 2x = log4 3 = 0.79, x = 0.79/2 = 0.4 (to one decimal place).
(c) 32x–1 = 8, (2x–1) = log3 8 = 1.9, x = 1.4 (to one decimal place).
19. To find the minimum of a function, we need to differentiate the function and set
this first derivative to zero. We would then find the second derivative and
ensure that this is above zero to verify that the turning point is indeed a
© Chris Brooks 2014
6
Introductory Econometrics for Finance by Chris Brooks
minimum. To find the value of the function at the minimum, we need to plug
the value of x obtained back into the original formula for y.
dy
d2y
 12 x  10  0 , x = 5/6.
 12  0 . If x = 5/6,
dx
dx 2
y = –12.2 (to one decimal place).
(a) y = 6x2 – 10x – 8,
(b) There are two ways that we could solve this problem. The first would be to
expand the parentheses so that we have standard terms in x (including x4, x2
and so on) or we could use the rule for differentiating a power of a function,
which is
d
f ( x ) n  nf ( x ) n 1 f ' ( x ) . So for y = f(x) = (6x2 – 8)2,
dx
dy
 2(6 x 2  8)12 x  144 x 3  192 x  0 .
dx
If we solve this, we get three solutions: x=0, x=4/3 and x=–4/3. The second
d2y
 432 x 2  192 . We need to plug each of the roots x=0, x=4/3
dx 2
and x=–4/3 into this equation sequentially to identify which corresponds to a
derivative is
minimum. When x=0,
d2y
d2y
;
when
x=4/3,


192

0
 576  0 ; and when
dx 2
dx 2
d2y
x=–4/3,
 576  0 . So the function has two minima when x=4/3 and when
dx 2
x=–4/3 and one maximum when x=0. When x=4/3, y=64/9 and when x=–4/3,
y=64/9.
20. There is clearly no shortage of possible examples, but suppose we picked:
 3 0
4 1 
A
,B  

.

1
6
2
7




The first thing to note is that the matrices are conformable for multiplication as A is
22 and B is 22 so AB will also be 22.
 3 0 4 1 12 3 
12 3 
AB  

, AB 1  





 1 6 2 7  8 41
 8 41
We also have B 1 
© Chris Brooks 2014
1

1  41  3

.
468  8 12 
1  7  1
1 6 0
1
 2 4  , A  1 3 , so that
26 
18 


7
Introductory Econometrics for Finance by Chris Brooks
B 1 A1 
1 1  7  1 6 0
1  41  3
.





26 18  2 4  1 3 468  8 12 
Hence this example has illustrated that (AB)–1 =B–1A–1.
21.
(a) For a pair of matrices to be capable of being multiplied together, they
must be conformable so that the number of columns of the first matrix is
equal to the number of rows of the second. On this basis, the following
multiplications are permissible: AB, BA, AC, DA, BC, DB, and CD.
To multiply matrices by a scalar (including scalars less than one), we simply
multiply each element of that matrix by the scalar. So
 1 6  2 12
 3  8  9  24
,
2 A  2

, 3B  3



4   18 12 
  2 4   4 8 
6
1 
6  2  3
1
1


D  0  1  0  0.5 .
 

2
2
0 
3 0  1.5
(c) Recall that the trace is the sum of the elements on the leading diagonal.
So Tr(A) = 1 + 4 = 5, Tr(B) = –3+4 = 1.
 1 6  3  8  2  2
A B  

and Tr(A+B) = –2 + 8 = 6.

4   4
8 
  2 4  6
Tr(A) + Tr(B) = 5 + 1 = 6 = Tr(A+B).
(d) The rank is the number of linearly independent rows or columns in a
square matrix. We can see in the case of the matrix A that all the columns
and rows are linearly independent of one another and hence the matrix is of
full rank, 2.
(e) To find the eigenvalues, , of the matrix A+B, we need to solve
( A  B)  I  0 , where I is a 22 identity matrix. From part (c) above,
  2  2
.
A B  
8 
4
  2  2
1 0
 
4

  0 , so
8

0 1
(4–2) = 0. So we have –16 –
© Chris Brooks 2014
 2    2 
 0 and thus (–2–)(8–) –
 4
8   

8 + 2 + 2 + 8 = 0, 2 – 6 – 8 = 0. This
8
Introductory Econometrics for Finance by Chris Brooks
equation does not factorise and thus the quadratic formula can be used,
giving the eigenvalues as  = –1.12 and  = 7.12 to two decimal places.
(f) An identity matrix of order 12 will be a square 1212 matrix having one as each
element of the leading diagonal and zero elsewhere. Since the trace is the sum of
the elements on this leading diagonal, it will be 1+1+…+1 = 12.
 2  1  3 0   2  3  1  0  1  1
22. (a) 



.
 7 4   7  4  7  7 4  4  0 0 
0  2  3  1  0  5  1
 2  1  3
(b) 




.
 7 4   7  4   7  7 4  4  14 8 
 3  1 1  2
(c) The inverse of 
 is 2  4

 4 2 
1  1 0.5
.

3 2 1.5 
(d) No, since the determinant is 32 – 32 = 0, the matrix is not of full rank and
therefore its inverse does not exist.
23.
(a) E(ax + by) = aE(x) + bE(y).
(b) E(axy) = aE(x)E(y).
(c) E(axy) = aE(xy).
24.
(a) The pdf (probability density function) describes how likely it is that a
random variable following a given distribution (e.g., the normal) will take on
a value within a given range. The cdf (cumulative distribution function), on
the other hand, is a function giving the probability that the random variable
will take on a value lower than some pre–specified value (call this z). The cdf
gives the area under the pdf from minus infinity up to that point z and thus
we can think of the cdf as the integral of the pdf whereas the pdf is the
derivative of the cdf. To illustrate, a pdf could give the probability that a
share price will increase by between 10% and 20% in a given year, while the
cdf could give the probability that the increase will be less than 5%.
(b) The pdf has a “bell shape”, while the cdf is a sigmoid (S–shape).
25. The central limit theorem (CLT) states that the mean of a sample of data having
any distribution converges upon a normal distribution as the sample size tends to
infinity. This is an important result in statistics since it shows that even if the raw
© Chris Brooks 2014
9
Introductory Econometrics for Finance by Chris Brooks
data are not normally distributed, the distribution of their mean will converge on a
normal as the sample increases. Hence the conventional approach to statistics
involving hypothesis testing using the normal or t–distribution tables can be applied.
Without the CLT, it would not be valid to use these tables to find critical values.
26. All three are measures of central tendency – i.e. they capture the average or
‘typical’ behaviour of a series. The (arithmetic) mean is simply calculated as the sum
of all values in a series divided by the number of values, whereas the mode
measures the most frequently occurring value in a series and the median is the
middle value in a series when the elements are arranged in an ascending order. It is
not possible to say that one of these three measures is better than the other two –
each has its own advantages and disadvantages. The benefit of the mean is that it
fully encapsulates the information from all data points in the series and it is the
most familiar method to most researchers, but can be unduly affected by extreme
values and in such cases, it may not be representative of most of the data. The
mode is arguably the easiest to obtain, but is not suitable for continuous, non–
integer data (e.g., returns or yields) or for distributions that incorporate two or more
peaks (known as bimodal and multi–modal distributions respectively). The median is
often considered to be a useful representation of the ‘typical’ value of a series, but
has the drawback that its calculation is based essentially on one observation. Thus if,
for example, we had a series containing 10 observations and we were to double the
values of the top three data points, the median would be unchanged.
27. Again, it is not possible to state that one approach is always better than the
other – it really depends on what the estimated mean will be used for. Geometric
returns are harder to calculate but give the fixed return on the asset or portfolio
that would have been required to match the actual performance, which is not the
case for the arithmetic mean. Thus, if you assumed that the arithmetic mean return
had been earned on the asset every year, you would not reach the correct value of
the asset or portfolio at the end. But it could be shown that the geometric return is
always less than or equal to the arithmetic return, and so the geometric return is a
downward–biased predictor of future performance. Hence, if the objective is to
summarise historical performance, the geometric mean is more appropriate, but if
we want to forecast future returns, the arithmetic mean is the one to use.
28. Not necessarily. The covariance between two random variables, call them x and
y, will scale with x multiplied by y. So, if we multiplied all of the values in a series x
by 10, the covariance would also increase by a factor of 10 but the variables would
not really be any more closely related than they were before. Thus the numerical
value that the covariance takes does not have a straightforward interpretation and
0.99 may or may not indicate a high degree of association depending on the scaling
© Chris Brooks 2014
10
Introductory Econometrics for Finance by Chris Brooks
of the data. If the correlation figure was 0.99 by contrast, this would be a clear
demonstration that the two series are strongly positively correlated since the
correlation measure is constructed so that it must lie between –1 and +1.
© Chris Brooks 2014
11