background image

Egzamin dla Aktuariuszy z 12 stycznia 2002 r. 
 
Prawdopodobieństwo i Statystyka 
 
Zadanie 1 
 

J(0,1)

X

 

  

)

1

,

0

(

  

1

=

+

iloczyn

X

J

X

X

X

X

n

n

n

 

Z tego: 

( )

n

n

n

n

n

n

n

n

X

EX

EX

EX

2

2

2

2

1

3

1

var

3

1

2

1

=

=

=

=

       

3

1

4

1

12

1

2

=

+

=

EX

 

1

3

4

2

1

1

3

4

2

1

2

1

4

1

3

1

2

=



=

=

n

n

n

n

n

n

n

ODP

 

 
Zadanie 2 
 
Metodą eliminacji dla 

2

=

A

N

 

(

)

(

)

(

)

(

)

=

=

=

+

=

=

+

=

=

+

=

=

2

2

1

2

2

1

1

1

6

2

6

2

1

1

A

A

A

A

N

A

P

N

A

P

N

A

E

N

N

E

 

3

1

7

66

28

2

4

8

6

2

12

8

8

2

8

2

2

12

8

8

1

4

1

8

1

6

=

+

+

=













+

















+

=

 i pasuje tylko odpowiedź (E) 

moŜna policzyć 

A

EN

EN  

i

 

1

 i sprawdzić czy 

(

)

1

1

EN

N

N

EE

A

=

 

 
Zadanie 3 
 

( )

4

var

2

X

X

Y

 

( )

(

)

=

+

=

+

X

Y

E

X

X

E

X

var

2

var

)

(

4

1

var

2

1

2

 

( )

(

)

X

Y

E

X

E

X

var

)

(

4

1

var

4

1

2

=

+

 

( )

(

)

=



<

=

2

2

2

4

1

4

var

4

1

EX

X

E

X

Y

E

EX

jesli

sprzeczność 

 

background image

Z tego 

( )

4

var

2

X

X

Y

=

 a skoro tak to: 

(

)

(

)




=

=

=

=

2

1

2

1

0

X

X

Y

P

X

Y

P

 

(

)

=

=

=

=

=

2

1

)

(

2

1

)

(

)

(

x

f

x

f

x

x

Y

P

x

Y

P

 

 
Zadanie 4 
 

 

A

 

- nie przetną się 

(

)

(

)

=

Π

Π

=

=

)

,

0

(

),

,

0

(

 

lub

 

)

2

,

(

),

2

,

(

4

3

4

3

2

x

P

x

P

x

P

x

P

P

x

P

A

P

 

2

2

2

2

2

2

2

2

2

0

2

2

2

1

)

2

(

4

)

2

(

2

2

2

2

1

2

1

Π

+

Π

=

Π

+

+

Π

Π

=

Π

+

Π

Π

=



Π

+



Π

=

Π

x

x

x

x

x

x

x

x

x

 

Π

=

+

=

Π

Π

+

Π

=

2

0

2

2

3

2

3

2

1

1

2

2

1

)

(

x

x

A

P

 

3

1

3

2

1

=

=

ODP

 

 
Zadanie 5 
 

(

)

(

)

(

)

=

+

+

=

+

+

=

2

0

1

2

0

1

0

1

2

)

1

(

)

1

(

)

1

(

ˆ

X

p

z

X

z

E

pX

X

X

z

zX

E

W

W

E

 

[

]

=

+

+

+

+

=

2

0

2

1

0

2

1

2

)

1

(

)

1

)(

1

(

2

)

1

(

X

p

z

X

X

p

z

z

X

z

E

 

(

)

(

)

(

)

(

)

=

+

+

+

+

+

+

+

=

2

0

2

2

1

0

2

2

1

2

2

2

2

1

2

2

2

2

4

1

2

EX

p

pz

p

z

z

X

X

E

p

zp

z

z

EX

z

z

 

(

)

(

)

(

)

(

)

c

pEX

EX

X

X

pE

X

X

E

EX

z

EX

X

X

E

EX

z

+

+

+

+

+

=

>

2

0

2

0

1

0

1

0

2

1

0

2

0

1

0

2

1

2

2

2

2

4

2

2

4

4

4

4

8

4

4

4

4

7

6

(

)

(

)

2

0

1

0

2

1

1

0

2

1

2

0

*

2

4

2

)

2

(

2

)

1

(

2

2

2

EX

X

X

E

EX

p

X

X

E

p

EX

EX

a

b

z

+

+

+

+

=

=

 

(

)

(

)

p

X

X

X

X

E

EX

EX

=

=

=

=

1

0

1

0

2

1

2

0

,

cov

1

1

 

background image

2

1

)

1

(

4

)

1

)(

2

(

2

)

1

(

4

)

2

(

2

)

2

(

2

)

1

(

4

2

4

2

2

2

2

4

2

)

2

(

2

)

1

(

2

2

2

*

p

p

p

p

p

p

p

p

p

p

p

p

p

p

p

z

+

=

+

=

+

+

=

+

+

=

+

+

+

+

=

 
Zadanie 6 
 

 

 

2

1

 

 

2

2

1

3

 

.. 

 

.. 

 

 

 

ORR   

OROO  

ROROO 

 

1

2

1

2

2

1

=

k

k

 

 

ROO   

RORR  

ORORR 

 

=

+

+

=

=

=

2

2

1

...

2

1

3

2

1

2

2

1

k

k

k

ODP

X

 

...

2

1

3

2

1

2

2

1

3

2

+

+

=

X

 

2

3

2

4

1

1

2

1

1

2

1

1

...

2

1

2

1

2

1

2

2

1

1

2

3

2

=

+

=

+

=

+

+

+

=

X

 

3

2

2

3

=

=

X

 

 
Zadanie 7 
 



n

σ

µ

N

X

2

,

 



+

=

n

σ

N

Y

µ

Y

X

2

,

0

,

 

(

)

4

2

2

2

4

4

3

2

2

2

2

3

4

4

6

3

4

2

4

4

µ

n

σ

µ

n

σ

µ

Y

µ

µ

Y

Y

µ

Y

µ

Y

E

X

E

+

+

=

+

+

+

+

+

=

 

(

)

2

2

2

2

1

σ

ES

n

σ

X

X

E

i

=

=

 



+

=



=

2

4

2

2

4

2

2

2

4

2

n

S

n

X

S

X

E

n

S

X

E

µ

E

 

(

)



+

=

=

2

2

2

2

2

2

2

µ

n

σ

σ

X

E

ES

X

S

E

 

 

background image

1

)

1

(

)

1

(

2

)

1

(

)

1

(

)

1

(

2

2

4

2

4

2

2

=

+

=



n

n

n

σ

n

S

E

n

χ

σ

n

S

 

1

)

1

(

)

1

)(

1

(

)

1

(

4

2

4

4

+

=

+

=

n

n

σ

n

n

n

σ

ES

 

=



+

+



+

+

+

=

4

4

2

2

2

4

4

2

2

2

4

1

)

1

(

1

2

6

3

var

µ

n

n

σ

n

µ

σ

n

σ

n

µ

n

σ

µ

n

σ

 

=

+

+

+

=

+

+

+

=

)

1

(

)

1

(

4

)

1

(

)

1

(

2

2

6

3

2

4

2

2

2

4

2

4

2

2

2

4

2

2

2

4

n

n

n

σ

n

σ

µ

n

σ

n

n

n

σ

n

µ

σ

n

σ

n

σ

µ

n

σ

 

2

2

4

2

2

2

4

2

2

2

4

4

4

)

1

(

2

4

)

1

(

2

4

)

1

(

)

1

(

)

1

(

σ

µ

n

σ

n

n

σ

µ

n

n

n

σ

n

σ

µ

n

n

n

n

σ

σ

n

+

=

+

=

+

+

+

=

 

 
Zadanie 8 
 
Innymi słowami: znaleźć n takie by test najmocniejszy istniał 

(

)

(

)

05

,

0

1

,

0

005

,

1

ln

2

1

2

1

0

2

10

2

1

,

10

0

2

2

=

+

>

=

>

Π

Π

n

t

X

P

t

e

e

P

i

X

n

X

n

i

i

 

n

n

n

t

n

n

n

t

n

n

X

P

i

10

64

,

1

1

,

0

005

,

1

ln

64

,

1

05

,

0

10

1

,

0

005

,

1

ln

10

0

+

=

+

→=

=

+

>

=

 

moc: 

0

1

,

10

10

64

,

1

5

,

0

1

,

10

10

64

,

1

1

,

10

+



+

>

n

n

n

n

n

n

n

n

X

P

i

 

271

96

,

268

4

,

16

64

,

1

1

,

0

n

n

n

 

 
Zadanie 9 
 
S: c-kwantyl rzędu 0,01 z chi(5), z tego c=0,554 

=

+

+

=

=

=

3

1

2

3

2

1

2

123

2

456

456

2

123

123

3

2

1

,

211

,

0

2

,

211

,

0

2

:

i

i

X

X

X

X

S

S

G

S

G

N

 

=

+

+

=

6

4

2

6

5

4

2

456

3

2

1

i

i

X

X

X

X

S

 

)

2

(

2

       

 

od

 

nzl

 

2

2

123

2

456

2

123

χ

σ

S

S

S

 

background image

5

5

 

bo

 

554

,

0

5

554

,

0

5

2

2

2

2

=



=

=

σ

S

E

σ

S

E

EG

S

 

(

)

(

)

=

=

2

1

wykl

(2)

 

nzl

 

   

,

max

2

211

,

0

2

,

max

211

,

0

2

2

1

2

2

456

2

123

χ

X

X

X

σ

S

S

G

i

N

 

( )

( )

S

S

N

G

E

G

E

σ

σ

EG

58

,

1

5

554

,

0

211

,

0

3

211

,

0

3

2

2

3

211

,

0

2

2

=

=

=

 

 
Zadanie 10 
 

2

var

EX

λ

S

=

 

2

var

X

E

λ

S

=

 

2

2

2

λ

EX

=

 

λ

EX

1

=

 

+

=

=

=

=

=

=

λ

a

a

t

λ

a

λ

a

t

λ

x

λ

λ

e

e

λ

t

e

dt

e

λ

t

t

λ

a

x

dx

e

λ

λ

a

x

X

E

0

0

2

2

2

2

2

2

 

λ

e

S

a

=

2

var

 

0217

,

1

36

,

0

ln

36

,

0

ln

2

36

,

0

2

2

var

=

=

=

=

a

a

λ

λ

e

λ

S

a