2001 03 24 prawdopodobie stwo i statystykaid 21605

background image

Egzamin dla Aktuariuszy z 24 marca 2001 r.

Prawdopodobieństwo i Statystyka

Zadanie 1

)

5

,

0

(

2

1

,

1

)

2

(

2

2

2

wykl

χ

σ

S

Γ

(

)

63212

,

0

1

2

2

1

2

2

2

2

=



=

e

σ

S

P

σ

S

P


Zadanie 2

1

X

- numer pierwszej kuli

=

=

wpp

0

2

1

2

2

X

X

X

Y

=

=

=

wpp

0

3

2

3

1

3

3

X

X

X

X

X

Y

=

=

=

=

wpp

0

4

3

4

2

4

1

4

4

X

X

X

X

X

X

X

Y

=

=

wpp

0

1

2

2

2

X

X

X

Z

=

=

=

wpp

0

lub

2

3

1

3

3

3

X

X

X

X

X

Z

=

=

=

=

wpp

0

3

4

2

4

1

4

4

4

X

X

X

X

X

X

X

Z

i

i

i

X

Z

Y

=

+

(

)

(

)

=

=

=

=

=

=

=

=

10

1

10

1

1

1

2

2

20

11

100

10

2

11

10

1

10

k

k

k

k

X

P

k

X

Z

E

EZ

(

)

(

)

∑∑

∑∑

= =

= =

=

=

+

=

=

=

=

=

=

10

1

10

1

2

1

10

1

10

1

2

1

3

3

10

11

2

10

2

11

100

1

100

1

10

10

,

,

k

j

k

j

j

k

j

X

k

X

P

j

X

k

X

Z

E

EZ

(

)

∑∑∑

=

=

=

=

=

20

33

(..)

,

,

3

2

1

4

4

P

m

X

j

X

k

X

Z

E

EZ

20

11

5

,

5

2

2

2

=

=

EZ

EX

EY

20

33

5

,

5

,

10

11

5

,

5

4

3

=

=

EY

EY

(

)

7

,

18

9

,

18

20

33

10

11

20

11

4

5

,

5

4

3

2

1

=

+

+

+

=

Y

Y

Y

X

E

ODP

background image

Zadanie 3

(

)

( )

( )

=

+

+

=

=

>

x

x

x

x

x

t

f

t

Y

E

e

x

x

e

t

f

t

Y

E

x

X

Y

E

)

(

)

2

(

2

)

(

( )

( )

x

x

Y

E

e

xe

e

e

x

Y

E

x

x

x

x

+

=

=

1

2

(

)

( )

(

)

(

)

3

2

1

2

=

+

=

+

=

=

=

X

X

E

X

Y

XE

E

X

XY

EE

EXY

2

1

1

)

1

(

=

+

=

+

=

X

E

EY

COV(X,Y)=3-2=1
Nie można obliczyć vary
Odpowiedź (E) jest prawidłowa

Zadanie 4

V

nzl

X

10

1

,

0

2

σ

N

X

)

9

(

10

2

2

χ

σ

V

)

9

(

3

10

3

10

)

9

(

10

9

1

10

2

2

t

V

X

V

σ

σ

X

t

σ

V

σ

X

=

6021

,

0

8331

,

1

3

05

,

0

3

0

=

=





>

c

c

c

V

X

P


Zadanie 5

(

) (

)

αβ

α

β

β

β

α

X

X

X

P

X

X

X

P

+

=

=

=

=

=

=

2

)

1

(

3

,

2

,

1

1

,

1

,

1

1

2

3

1

2

3

(

)

=

+

+

=

)

2

,

3

,

2

(

)

3

,

2

,

3

(

)

3

,

2

,

1

(

1

,

1

1

2

P

P

P

X

X

P

αβ

α

β

β

α

αβ

α

β

αβ

α

αβ

αβ

α

α

β

βα

αβ

α

β

β

α

β

α

β

βα

+

=

+

+

+

+

=

+

+

+

=

2

)

1

(

2

2

2

2

)

1

(

)

1

(

)

1

(

2

2

2

2

)

1

(

2

2

2

)

1

(

β

β

α

αβ

α

β

αβ

α

β

β

αβ

ODP

=

+

+

=


Zadanie 6

Γ

)

,

( θ

n

X

i

(

)

=





=

=

θ

n

θ

θ

X

θ

n

θ

θ

P

θ

θ

θ

P

θ

θ

θ

θ

P

1

99

100

1

101

100

99

100

ˆ

1

101

100

01

,

1

ˆ

99

,

0

40000

1

99

100

96

,

1

95

,

0

1

99

100

1

101

100

=

=





=

n

n

n

X

n

P

background image

Zadanie 7

Atom punkt

θ

x

=

θ

e

θ

X

P

=

=

1

)

(

dalej wykładniczy

tn

e

t

P

t

P

=

>

=

1

)

(min

1

)

(min

tn

ne

t

f

=

)

(

min

n

θ

e

θ

P

θ

P

=

>

=

=

1

)

(min

1

)

(min

2

2

ˆ

2

ˆ

)

(

θ

θ

E

θ

θ

E

θ

R

+

=

(

)

+

+

=

+

=

θ

θ

n

θ

n

n

θ

xn

θ

e

n

e

n

θ

e

θ

ne

x

θ

E

2

2

2

2

2

2

2

1

ˆ

(

)

+

=

+

=

θ

θ

n

n

θ

xn

θ

e

n

e

θ

xne

θ

E

1

1

ˆ

θ

n

θ

n

θ

n

θ

n

e

n

θ

θ

e

n

θ

θ

e

n

e

n

θ

θ

R

=

+

+

+

=

2

2

2

2

2

2

2

2

2

2

)

(


Zadanie 8

(

)

(

)

3

2

1

2

3

2

1

2

X

od

nzl

bo

var

5

var

X

X

X

X

X

X

X

v

+

+

=

+

=

(

)

(

)

(

)

=

=

=

+

=

=

=

=

+

=

10

5

5

5

5

2

1

1

2

2

1

2

!

5

10

)!

5

(

5

!

5

5

5

5

e

e

j

e

j

X

X

P

j

X

j

X

P

X

X

j

X

P

j

j

)

5

,

0

;

5

(

2

1

5

10

!

5

)!

5

(

!

5

5

10

5

10

5

dwum

j

e

j

j

e





=

=

(

)

4

5

2

1

2

1

5

5

var

2

1

2

=

=

=

+

X

X

X

ODP=1,25+5=6,25

Zadanie 9

 

X

E

EX

X

E

>=

<

 

(

)

(

)

∑ ∫

=

+

=

=

+

=

=

=

+

+

=

2

1

3

2

1

1

1

1

)

1

(

1

...

2

k

k

k

k

k

λ

k

λ

k

λ

k

λ

x

λ

x

λ

x

λ

e

ke

e

e

k

e

λ

k

e

λ

e

λ

X

E

...

2

2

+

+

=

λ

λ

e

e

u

...

2

3

2

+

+

=

λ

λ

λ

e

e

ue

(

)

(

)

2

2

1

1

...

1

λ

λ

λ

λ

λ

λ

λ

e

e

u

e

e

e

e

e

u

=

=

+

+

=

 

(

)( )

 

 

1

1

1

1

1

1

2

=

=

=

=

X

E

e

X

E

e

e

e

e

e

e

X

E

λ

λ

λ

λ

λ

λ

λ

background image

)

ln(

)

1

ln(

1

c

c

λ

c

c

e

λ

+

=

+

=

λ

EX

1

=

)

(

ln

)

1

ln(

1

A

c

c

c

ODP

=

+

=


Zadanie 10

9

4

)

2

(

9

4

)

0

(

9

1

)

2

(

:

2

=

=

=

p

p

p

S

81

16

)

4

(

81

32

)

2

(

81

24

)

0

(

81

8

)

2

(

81

1

)

4

(

:

4

=

=

=

=

=

p

p

p

p

p

S

3

3

3

3

3

3

3

6

9

64

)

6

(

9

192

..

64

128

)

4

(

9

240

..

16

128

96

)

2

(

9

160

..

32

96

32

)

0

(

9

60

..

24

32

4

)

2

(

9

12

..

4

8

)

4

(

9

1

)

6

(

:

=

=

+

=

=

+

+

=

=

+

+

=

=

+

+

=

=

+

=

=

p

p

p

p

p

p

p

S

(

) (

)

(

)

=

=

=

=

=

=

=

=

5

,...,

5

,

2

...

5

,...,

5

,

2

5

,...,

5

,

2

6

1

10

8

1

10

9

1

10

S

S

S

P

S

S

S

P

S

S

S

P

r

[

]

k

S

k

k

k

k

odpada

k

=

=

=

=

=

=

2

4

;

4

;

4

;

2

,

0

,

2

,

4

;

6

4

(

)

(

) (

)

=

=

=

=

=

=

=

5

6

6

4

6

10

2

5

,

2

k

k

S

P

k

S

P

S

S

P

(

) (

) (

) (

) (

) (

) (

) (

)

=

=

=

+

=

=

+

=

=

+

=

=

=

2

4

0

2

2

0

4

2

6

4

6

4

6

4

6

4

S

P

S

P

S

P

S

P

S

P

S

P

S

P

S

P

2265

,

0

9

60

16

9

160

32

9

240

24

9

192

81

8

5

5

5

3

+

+

+

=


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