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For practice

MSc in Finance and Investment

Presessional Maths and Stats.

Stats Exercises Solutions (Tutorials)

1. The discrete Poisson distribution has one parameter . 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) = and Var(y) = for the Poisson.

(b) Assuming = 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 = 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

= (1 x) x 1

(i) Find

As the density function has a kink at x = , we have to integrate in two steps. The

first step is from 0 to . The second from 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 = 2/102 ]2/x[ =

8

1

2/1

dx)x1( = 1 2/12 ]2/xx[ = )8/2/(2/ =

8

So 8

+ 8

= 1 and 14/ or 4

Thus f(x) = 4x 0 x

= 4(1 x) x 1

(ii) Derive the cumulative distribution function F(s)

F(s) = s

0

sds4 = 2s2 0 s

F(s) = F(1/2) + s

2/1

ds)s1(4 = + s 2/12s2s4 =

+ 4s 2s2 2 + = 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 = . Thus the mean is 1/2

(iv) Check that the value of the mean in part (iii) is the same as E(x).

dxx42

= 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) =

(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[ + 1 2/13 ]3/x4[ 1 2/1

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 22

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 22

1

= 2

1

3

16

x

16

x12

= 1 11/16 = 5/16

(ii) P ( 0.5 X 1.7)

dx)x4(16

3 27.1

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 22

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 22

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( 2i

Derive a simpler expression for the sample variance which can be used in

calculations.

Sample variance = N

)xx( 2i = ])x(xx2x[N

1 2i

2i

= N

)x( 2i

N

)x(x2 i +

N

)x( 2

N

)x( 2i

2)x(2 + 2)x( = N

)x( 2i

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] = 212a + 22

2b + 12ab2 .

The long proof is;

Var(z) = E[(ax + by (a 1 + b 2 )2) ] = E( 22

22

1

22222 baybxa

+ 2122

121

2 abybyabxab2xa2abxy2 ) =

2

2

22

1

22222 ba)y(Eb)x(Ea + 2abE(xy)

212

2

2

2112

2

1

2 ab2b2ab2ab2a2

= ))x(E(a 2122 + ))y(E(b 22

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 1

0,1

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.