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MSc in Finance and Investment
Presessional Maths and Stats.
Stats Exercises Solutions (Tutorials)
1. The discrete Poisson distribution has one parameter λ. Its density function is given
by
f(y) = !y
ey y = 0, 1, 2,.............where y! = 1x2x3....xy.
(a) Given that this density function is valid, what properties must it satisfy?
0i i
y
!y
ei
= 1 and !y
e
i
yi ≥ 0 i = 0,1,2,........
You might care to initiate a discussion about whether these hold in this case.
There is a well known proof of the first which students do not have to know, the
second is fairly obvious. Also E(y) = λ and Var(y) = λ for the Poisson.
(b) Assuming λ = 2, calculate the following;
(i) Prob{y = 2} = !2
e2 22
= 0.27
(ii) Prob{y < 2} = !0
e2 20
+ !1
e2 21
= 0.135 + 0.27 = 0.405
(iii) Prob{y > 2) = 1 - (0.27 + 0.405) = 0.325
(c) Sketch out the shape of the density function if λ = 2. Is it skewed or symmetric? Explain.
f(0) = 0.135, f(1) = 0.27, f(2) = 0.27, f(3) = 0.18
Its skewed to the right.
3. (a) Given the density function:-
f(x) = x 0 x ½
= (1 – x) ½ x 1
(i) Find
As the density function has a kink at x = ½, we have to integrate in two steps. The
first step is from 0 to ½. The second from ½ to 1. Thus must solve the following
equation;
2/1
0
xdx +
1
2/1
dx)x1( = 1
We find each of these terms in turn;
2/1
0
xdx = 2/1
02 ]2/x[ =
8
1
2/1
dx)x1( = 12/1
2 ]2/xx[ = )8/2/(2/ = 8
So 8
+ 8
= 1 and 14/ or 4
Thus f(x) = 4x 0 x ½
= 4(1 – x) ½ x 1
(ii) Derive the cumulative distribution function F(s)
F(s) = s
0
sds4 = 2s2 0 s ½
F(s) = F(1/2) +
s
2/1
ds)s1(4 = ½ + s 2/1
2s2s4 =
½ + 4s – 2s2 – 2 + ½ = 2s(2 – s) – 1 1/2 s 1
(iii) Sketch f(x) and by inspection what is the mean.
The distribution is symmetric about the mode x = ½. Thus the mean is 1/2
(iv) Check that the value of the mean in part (iii) is the same as E(x).
dxx4 2 = 2/1
03 ]3/x4[ =
6
1
dx)x1(x4 = 12/1
32 ]3/x4x2[ = )24/42/1(3/42 = 3
1
Thus E(X) = ½
(v) Find Var(x)
Var(x) = E(2x ) – 2)x(E
E(2x ) =
2/1
0
3dxx4 +
1
2/1
2 dx)x1(x4
= 2/10
4 ]x[ + 12/1
3 ]3/x4[ – 12/1
4 ]x[
= 1/16 + 4/3 – 4/24 – 1 + 1/16 = 1/8 + 1/3 – 1/6 = 7/24
Var (x) = 7/24 – 2)]x(E[ = 1/24
4. If 12XE and 16XVar , find E(Y) and Var(Y) where Y is the
following :
(i) 34X5Y (ii) 9
4
XY (iii)
4
17X3Y
(iv)
4
12XY
.
(i) E(Y) = 3460 Var(Y) = 400 (ii) E(Y) = 12 Var(Y) = 1 (iii) E(Y) =
4
116 Var(Y) = 9
(iv) E(Y) = 0 Var(Y) = 1
5. A continuous random variable x has a distribution with probability density function
(pdf) given by
)x4(16
3)x(f 2 0 2 x
= 0 otherwise
(a) Check that f(x) is a proper density function.
dx)x4(16
3 2
2
0
=
2
0
3
16
x
16
x12
= 1.5 − 0.5 = 1 it is “proper”
(b) Find the following probabilities:
(i) P ( X > 1)
dx)x4(16
3 2
2
1
=
2
1
3
16
x
16
x12
= 1 − 11/16 = 5/16
(ii) P ( 0.5 X 1.7)
dx)x4(16
3 2
7.1
5.0
=
7.1
5.0
3
16
x
16
x12
= 1.275 − 0.3071 − 0.375 + 0.0078 = 0.5851
(b) Find E(x)
dx)x4(16
x3 2
2
0
=
2
0
42
64
x3
64
x24
= 96/64 − 48/64 = 48/64 = 0.75
(c) Find the variance of x
)x(E 2 = dx)x4(
16
x3 2
2
0
2
=
2
0
53
80
x3
64
x16
= 2 − 96/80 = 0.8
Variance = 0.8 − 0.5625 = 0.2375
6. Suppose there is a sample of data, ix i = 1,2,3,....N
The sample mean is defined as N
xx
i
The sample variance is defined as N
)xx( 2
i
Derive a simpler expression for the sample variance which can be used in
calculations.
Sample variance = N
)xx( 2
i = ])x(xx2x[N
1 2i
2i
= N
)x( 2
i −
N
)x(x2 i +
N
)x( 2
N
)x( 2
i −
2)x(2 + 2)x( =
N
)x( 2
i −
2)x(
7. If z = ax + by where E(x) = 1 ,E(y) = 2 , var(x) = 2
1 and var(y) = 2
2
and cov(x,y) = 2
12 , a and b are constants, show that
the variance of z = E[z – E(z)2] =
2
1
2a + 2
2
2b + 12ab2 .
The long proof is;
Var(z) = E[(ax + by – (a 1 + b 2 )2) ] = E(
2
2
22
1
22222 baybxa
+ 212
2
121
2 abybyabxab2xa2abxy2 ) =
2
2
22
1
22222 ba)y(Eb)x(Ea + 2abE(xy)
21
2
2
2
2112
2
1
2 ab2b2ab2ab2a2
= ))x(E(a 2
1
22 + ))y(E(b 2
2
22 + 2ab( 21)xy(E )
= )xvar(a 2 + )bvar(b2
+ 2abcov(x,y)
A shorter proof starts from
Var(z) = E[(ax + by – (a 1 + b 2 )2) ] = E[(a(x – 1 )+b(y – 2 )
2) ] =
2
12 )x(Ea +
22
2 )y(Eb + 2 )y)(x(abE 21 =
)xvar(a 2 + )bvar(b2
+ 2abcov(xy)
10.
X
f(y)
0 1 2
0 0.05 0.10 0.03 0.18
Y 1 0.21 0.11 0.19 0.51
2 0.08 0.15 0.08 0.31
f(x)
0.34
0.36
0.30
1.00
(a) Compute the following three probabilities:
P(Y < 2) = P(Y = 0) + P(Y = 1) = 0.18 + 0.51 = 0.69.
P( Y < 2 , X > 0) = P( Y = 0 , X = 1) + P( Y = 0 , X = 2) + P( Y = 1 , X =1) + P( Y = 1 , X = 2) = 0.10 + 0.03 + 0.11 + 0.19 = 0.43
P(Y = 1 and X 1) = P(Y = 1 , X = 1) + P(Y = 1 and X = 2) = 0.11 + 0.19 = 0.30.
(b) Find the marginal distributions of X and Y Given above
(c) Calculate E(X), E(Y), Var(X), Var(Y), Cov Y,X and )YX(E 2.
x
ii )x(fxXE 0(0.34) + 1(0.36) + 2(0.30) = 0.96
y
ii )y(fyYE 0(0.18) + 1(0.51) + 2(0.31) = 1.13.
22 XEXEXVar where
x
i
2
i
2 xfxXE 20 (0.34) + 21 (0.36) +
22 (0.30) = 1.56
So 296.056.1 XVar = 0.6384
22 YEYEYVar where
y
i
2
i
2 yfyYE 20 (0.18) + 21 (0.51) +
22 (0.31) = 1.75
So 213.175.1 YVar = 0.4731
YEXEXYEYXCov , 13.196.0XYE
where yxpxyXYEy x
, .
See the table below – values in the brackets are the values of xy.
X
0 1 2
0 0.05
(0)
0.10
(0)
0.03
(0)
Y 1 0.21
(0)
0.11
(1)
0.19
(2)
2 0.08
(0)
0.15
(2)
0.08
(4)
)08.0)(4()15.0)(2()19.0)(2()11.0)(1( XYE = 1.11
So 0252.00848.111.1, YXCov .
(d) Find the conditional distribution of Y given xX .
Need to find xXyYP for x = 0, 1 or 2 and y = 0, 1 or 2 where
)(
,
xXP
yYxXPxXyYP
Consider the conditional distribution of Y given X = 0.
00 XYP
0
0,0
XP
YXP
34
5
34.0
05.0 using the joint table and (b).
01 XYP
34.0
1,0 YXP
34
21
34.0
21.0 “ “ “
02 XYP
34.0
2,0 YXP
34
8
34.0
08.0 “ “ “
So the conditional distribution of Y given X = 0 is given by
Y 0 1 2 Total
0Xyf
or 0yf
34
5
34
21
34
8
1.0
Consider the conditional distribution of Y given X = 1.
10 XYP
10,1
XP
YXP
18
5
36.0
10.0 using the joint table and (b).
11 XYP
36.0
1,1 YXP
36
11
36.0
11.0 “ “ “
12 XYP
36.0
2,1 YXP
12
5
36.0
15.0 “ “ “
So the conditional distribution of Y given X = 1 is given by
Y 0 1 2 Total
1Xyf
or 1yf
18
5
36
11
12
5
1.0
Consider the conditional distribution of Y given X = 2.
20 XYP
2
0,2
XP
YXP
10
1
30.0
03.0 using the joint table and (b).
21 XYP
30.0
1,2 YXP
30
19
30.0
19.0 “ “ “
22 XYP
30.0
2,2 YXP
15
4
30.0
08.0 “ “ “
So the conditional distribution of Y given X = 2 is given by
Y 0 1 2 Total
2Xyf
or 2yf
10
1
30
19
15
4
1.0
You could put all 3 conditional distributions together in one table to give the conditional distribution of Y given X = x as follows :
X
0 1 2
0 34
5 18
5 10
1
Y 1 34
21 36
11 30
19
2 34
8 36
15 15
4
What is the conditional mean 1XYE and the conditional variance 1XYVar ?
1XYE = the mean of the conditional distribution of Y given X = 1
= y
1Xyfy
= 0
18
5 + 1
36
11 + 2
36
15=
36
41 or 1.1389 to 4 dec. places.
1XYVar = the variance of the conditional distribution of Y given X = 1
= 2
2 11 XYEXYE
where y
XypyXYE 11 22
18
502
+
36
1112
+
36
1522
=
36
71.So
6752.036
41
36
711
2
XYVar to 4 dec. places.
11. Grosvenor Motors Ltd. is developing a marketing plan to better target advertising and
sales promotion to subgroups. As part of the market research they have prepared the
table given below which indicates the probabilities for subgroups defined by age of
car and owner age group. For example, X = 1 indicates that the car owners are aged
16 – 25 years and Y = 2 indicates that the car is between 2 and 4 years old.
Age of car (Y)
1
(16 – 25 yrs)
Age Group (X)
2
(26 – 45 yrs )
3
(46 – 65 yrs)
1 ( 2 yrs) 0.05 0.17 0.06
2 ( 2 – 4 yrs) 0.15 0.20 0.07
3 ( 5 yrs ) 0.10 0.08 0.12
(a) What age group would you concentrate your advertising and sales promotion
on if you were attempting to sell cars that are over 5 years old?
As Y is the age of the car, when Y = 3 the car is at least 5 years old. When Y = 3, the X value with the highest associated probability is X = 3 since 0.12 > 0.10 > 0.08, so the age-group you would concentrate on is 46 – 65 years.
(b) Calculate the following probabilities : P( X = 2, Y = 3), P( X = 3, Y =
1), 32 YXP , 32 XYP , 2,2 YXP and define
each probability in words.
(i) P( X = 2, Y = 3) = 0.08 = the probability that the car owner is aged 26 – 45 and the car is at least 5 years old.
(ii) P( X = 3, Y = 1) = 0.06 = the probability that the car owner is aged 46 – 65 and the car is at most 2 years old.
(iii) 15
4
30.0
08.0
)3(
3,232
YP
YXPYXP = the probability that the car owner
is aged 26 – 45 years old if his/her car is at least 5 years old.
(iv) 25
7
25.0
07.0
)3(
2,332
XP
YXPXYP = the probability that the car is 2
– 4 years old if the car owner is aged 46 – 65.
(v) 2,2 YXP 21,21 orYorXP = P( X = 1, Y = 1) + P( X = 1, Y = 2) +
P( X = 2, Y = 1) + P( X = 2, Y = 2) = 0.05 + 0.15 + 0.17 + 0.20 = 0.57 = the probability that the car owner is no more that 45 years old and his/her car is at most 4 years old.
(c) Find the marginal distributions of X and Y and hence calculate 2XP and
2YP . See the joint table above with row and column totals.
The marginal distribution of X is given by
x 1 2 3 Total
xpX 0.30 0.45 0.25 1.00
The marginal distribution of Y is given by
y 1 2 3 Total
ypY 0.28 0.42 0.30 1.00
2XP = 0.30 + 0.45 = 0.75 or 2XP = 1 – P(X = 3) = 1 – 0.25 = 0.75.
2YP = 0.28 + 0.42 = 0.70 or 2YP = 1 – P(Y = 3) = 1 – 0.30 = 0.70.
(d) Calculate E(X), E(Y) and cov( X , Y).
x
X xxpXE 1(0.30) + 2(0.45) + 3(0.25) = 1.95
y
Y yypYE 1(0.28) + 2(0.42) + 3(0.30) = 2.02
YEXEXYEYXCov , 02.295.1XYE
where yxpxyXYEy x
, .
See the table below – values in the brackets are the values of xy.
X
1 2 3
1 0.05
(1)
0.17
(2)
0.06
(3)
Y 2 0.15
(2)
0.20
(4)
0.07
(6)
3 0.10
(3)
0.08
(6)
0.12
(9)
XYE 1(0.05) + 2(0.17) + 3(0.06) + 2(0.15) + 4(0.20) + 6(0.07) + 3(0.10) +6(0.08) +
9(0.12) = 3.95
So 011.0)02.2(95.195.3, YXCov
(e) Explain why the random variables X and Y are not independent.
If X and Y were independent, then YXCov , would be 0 so as it is not 0, it means
that X and Y cannot be independent.
12. Let f(x,y) = )yx(k 22 , 0 x 1, 0 y 1 be the joint probability density
function of x and y where k is a constant. Find the following;
(a) value of k.
1
0
1
0
22 dxdy)yx(k =
1
0
1
0
32 dx)
3
kyykx(
= dx)
3
kkx(
1
0
2
=
1
0
3
3
kx
3
kx
= 2k/3 = 1 Thus k = 3/2
(b) the marginal density function of x.
f(x) =
1
0
22
2
dy)yx(3 =
1
0
32
2
y
2
yx3
=
2
)13( 2 x
(c) the marginal expectation of x.
E(x) = dx2
x
2
x31
0
3
=
1
0
24
4
x
8
x3
= 5/8
(c) the marginal variance of x.
)( 2xE = dx2
x
2
x31
0
24
=
1
0
35
6
x
10
x3
= 3/10 + 1/6
= 14/30 Var(x) = 14/30 - 25/64 = 0.467 – 0.391 = 0.076
(d) the conditional distribution of y given x. [6 marks]
f(y | X) = )x(f
)y,x(f =
)1X3(2
)yX(62
22
13. Suppose 1̂ and 2̂ are independent unbiased estimators of the same unknown
population parameter . It is known that var( 1̂ ) = 2
1 and var( 2̂ ) = 2
2 . A
statistician proposes a new estimator
3̂ = a 1̂ + (1− a) 2̂
where a is a constant whose value is chosen by the statistician.
(a) Show that 3̂ is unbiased.
E( 3̂ ) = aE( 1̂ ) + (1− a)E ( 2̂ ) = a + (1− a) =
(b) Derive the variance of 3̂ .
Var( 3̂ ) = var( 1̂ ) + var( 2̂ ) + 2a(1-a)cov( 1̂ 2̂ )
Since the covariance term is zero,
Var( 3̂ ) = 2
1 + 2
2
(c) Find the value of a that minimises the variance of 3̂ (5 marks)
First order condition gives 2a2
1 - 2(1-a) 2
2 = 0 or
Thus optimal a = 2
2
2
1
2
2
16 A random sample ( )x,....,x,x n21 of size n is drawn from a population of known mean
and unknown variance 2 . The following estimator of
2 is proposed
2̂ = 2)ix(
n
1
(a) Is 2̂ biased or unbiased? Give details.
Short proof; since E( 2
i )x(n
1) = ))x((E
n
1 2
i =
2
n
1 =
2 thus unbiased
Slightly longer proof, E[ 2
i )x(n
1]
= n
)x(E 2i
– n
)x(E2 i +
2 = )x(E 2i
– 2 =
)x(Var i = 2
(b) The sample variance is defined as
2s = 2
i )xx(n
1
where x is the sample mean. Explain why 2s is a biased estimator of
2 .
2s = 2
i )xx(n
1 = 2
ixn
1 − ix
n
x2 +
2x
= 2ix
n
1 −
2x
E(2s ) = )x(E
n
1 2i − E(
2x ) = E(2ix ) − E(
2x )
= var( ix ) + 2
i )]x(E[ − (var( x ) + 2)]x(E[ )
= var( ix ) − var( x )
Thus because 2s contains x and not , the E(
2s ) depends on the variance of x
as well as the var( ix ) . The variance of x unlike the variance of is non zero.
17. Construct confidence intervals for the population mean when the sample mean is 11.5
and unbiased estimate of the population variance is 100.
(a) Use a level of confidence of 95 % and a sample size of 25
critical value of t(24) is 2.064. Standard deviation of the sample mean is
25/100 = 2 Thus 95% confidence interval is given by 11.5 (2.064x2)
7.372 15.628 where is the population mean
(iii) a sample size of 1,000,000
critical value of t(999,999) is 1.96 approx. Standard deviation of the sample
mean is 000,000,1/100 = 0.01 Thus 95% confidence interval is given by
11.5 (1.96x0.01)
11.48 11.52 where is the population mean
(b) Use a level of confidence of 90 %
(i) a sample size of 25 critical value of t(24) is 1.711. Standard deviation of the sample mean is
25/100 = 2 Thus 90% confidence interval is given by 11.5 (1.711x2)
8.078 14.922 where is the population mean
(iii) a sample size of 1,000,000
critical value of t(999,999) is 1.645 approx. Standard deviation of the
sample mean is 000,000,1/100 = 0.01 Thus 90% confidence
interval is given by 11.5 (1.645x0.01)
11.48 11.52 where is the population mean
18. Suppose X is a normally distributed random variable with mean = 200 and standard
deviation = 20.
(a) Calculate P(180X220)
P(180X220) = P(
220X
180 ) = P(
20
200220X
20
200180
)
= P(-1 X 1) = 1 – 0.1587 – 0.1587
= 0.6826
(b) Calculate P(160X240)
P(160X240) = P(
240X
160 ) = P(
20
200240X
20
200160
) =
P(–2 X 2) = 1 – 0.0228 – 0.0228
= 0.9544
20 Explain what is meant by a Type 1 and a Type 2 error in hypothesis testing. (a) Consider the following decisions and indicate whether you think a type 1 or a type
2 error is more important
i. A doctor tests a patient for a serious infectious disease. The null hypothesis is that the patient has the disease.
ii. A bank is deciding whether or not to make a large loan. The null hypothesis is that the customer will default on the loan.
iii. The Bank of England is deciding on interest rate policy. The null
hypothesis is that the rate of inflation will be greater that the target level set by the government.
(b) Discuss whether you think a type 1 or a type 2 error is more important in general.
Type 1 error is rejecting a true null, type 2 is accepting a false one. The answer to (a) depends on the investigator’s loss function which I have not discussed except by implication. This is an opportunity for a discussion, There is no right or wrong answer. (b) In general there is no reason to favour either type of error. It all depends on the context. 21Test the null hypothesis that the population mean is 10 against a two sided alternative if the sample mean is 11.5 and an unbiased estimator of the population variance is 100 (as in 1 above).
(a) Set the probability of making a type 1 error is 5 per cent.
(i) if N = 25
test statistic is 5/10
)105.11( = 0.75
critical value of t(24) is 2.064, thus do not reject null.
(ii) if N = 100
test statistic is 10/10
)105.11( = 1.5
critical value of t(99) is 1.99, thus do not reject null.
(iii) if N = 1,000,000
test statistic is 1000/10
)105.11( = 150.0
critical value of t(999,999) is 1.96, thus reject null.
(b) Calculate approximate p values for each of these tests.
(i) t table value corresponding to 0.75 (24 degrees of freedom) is 0.23 approx. Thus p value is 0.23x2 = 0.46
(ii) t table value corresponding to 1.5 (99 degrees of freedom) is 0.16 approx. Thus p value is 0.16x2 = 0.32
(iii) t table value corresponding to 150 (999,999 degrees of freedom) is 0.000 approx. Thus p value is 0.000.
(c) Would your results in (a) be any different if the population variance was known to be 100. Explain in detail.
It might make a difference for the two cases N = 4 and 25. Note in all cases
where the sample size is small (below 30 approx) the population must be
assumed to be normally distributed.
Critical values would then be 1.96, but the decision not to reject null would be
the same.
22. Suppose you wish to test a hypothesis about the mean growth rate of a certain kind
of companies. A random sample of 51 companies give you an estimator of the mean
growth rate of 8.7 per cent. The estimated standard error of this estimator is 0.9. You
know that the estimator has a normal distribution.
(a)Test the hypothesis that the mean growth rate is 10 per cent using a 5 per cent
level of significance.
0H population mean = 8.7
1H population mean 8.7
t = (8.7−10)/0.9 = − 1.444 this has a t distribution with 50 degrees of freedom.
Critical value is 2.009. Thus do not reject null.
(b)Test the null that the mean rate of growth is 6 per cent using a 5 per cent level of
significance
0H population mean = 6
1H population mean 6
t = (8.7−6)/0.9 = 3 this has a t distribution with 50 degrees of freedom.
Critical value is 2.009. Thus reject null.
(c) A new larger sample becomes of 200 companies. The estimator of the mean
growth is 8.5 and standard error is 0.3. Test the null that the mean growth rate is
10 per cent
0H population mean = 10
1H population mean 10
z = (8.5 −10)/0.3 = − 5 thus reject the null at 5 % and at 1%
23 (a) The mean of the annual average rate of return (in per cent) for the stock of
the public companies in the smallest decile (by size) in the US for 33 years
(1970-2002) is 13.30. The standard deviation of the sample is 24.91. If you were
to test the hypothesis that the mean annual rate of return is at least 10 per cent,
write down the null and alternative hypotheses and the test statistic.
0H population mean = 10
1H population mean > 10
Test statistic is 33/91.24
103.13 = 0.7611. This has a t(32) if the null is true. 5%
critical value for a one sided test is 1.629 approx. Thus do not reject.
(b) The mean annual average rate of return for the largest decile for the same
period is 11.87 per cent. The sample standard deviation is 18.23. Test the
null hypothesis that the annual average rate of return is 11 per cent. Use a
significance level of 5 %.
0H population mean = 11
1H population mean 11
Test statistic is 33/23.18
1187.11 = 0.274 do not reject but note relevant critical
value for a two sided test is 2.035 (approx)
Note: the standard deviations given are the square root of an unbiased
estimator of the population variance.