Egzamin dla Aktuariuszy z 11 października 2004 r.

Prawdopodobieństwo i Statystyka

Zadanie 1

 8

,

0 1

,

0 09

,

0 09

,

0 0 

1





 0

9

,

0

0

1

,

0







0

0

9

,

0

1

,

0





 1

0

0

0 

rozkład stacjonarny:



Π

81

,

0

1 + Π 4 = Π1



Π

19

,

0

4 =

Π

 ,

0

Π

09

9

,

0

1 +

Π2 = Π2



1



→ Π

9

,

0

2 =

Π1

 ,

0

Π

09

9

,

0

1 +

Π3 = Π3

Π

9

,

0

3 =

Π



1

 ,

0

Π

01

1

,

0

1

,

0

1 +

Π2 + Π3 = Π4

Π + 9

,

0 Π + 9

,

0 Π + 1

,

0 9Π = 1

1

1

1

1

9

,

2 9Π = 1

1

100

Π =

1

299

90

Π = Π =

2

3

299

19

Π =

4

299

100

90

190

ODP = Π + Π =

+

=

≈ ,

0 635

1

3

299

299

299

Zadanie 2

E( NZ

E NZ

N

k P N

k

N )

∞

= ∑ ( N = ) ( = )

k =1

P(min ≤ t) = 1 − P(min > t) = 1 − e− αnx ≅ wykl( αn) 1

E( NZ N =

=

=

N

) k

k

αk

α

E(

k

r

NZ

p

p

p

N )

∞

= ∑ 1  + − 

1

r

k

1

r





1

( − ) =

(1− )

α k

α

k =1





E( 2 2

N Z

E N Z

N

k P N

k

N )

∞

= ∑ ( 2 2 N = ) ( = )

k =1

E( 2 2

N Z

N = k =

= =

=

N

) 2 kE( 2 Z N k

N

) 2 2

2

k

2

2

2

α k

α

E(

2

k

r 1

2

2

N Z

p

p

p

N )

∞

= ∑

 + −  r

k

2

1

(

)

1

r

2 





−

= 2 ( − )

α

k

k =1





α

−

−

−

r

p r

r

p r

r

r

p r

var( NZ

=

−

−

−

=

− −

=

+

=

N )

2

1

1

1

1

1

p

1

p

2

1

p

1

p

2 (

) (

)2

2

2

[ (

)]

(

)

2

2

2

α

α

α

α

α

Zadanie 3

 Z = X + Y

2



 W = X

0 < W < Z < X + Z = ∆

 X = W

dX

dX



0

1

1



dU

dV

U − W

= 1

1 = − ≠

(

0

u, v) ∈ ∆

dY

dY

−

 Y =

2



2

2

2

dU

dV

w:=x



z − x  1

1 − x

f  x;



= e 0<x<z<x+2



2  2

2

czyli odpowiedź (D) prawidłowa

Zadanie 4

1: 0b

Nc

2: 1b

(N-1)c

3: 2b

(N-2)c

...

...

N: (N-1)b

1c

N+1: Nb

0c

P(

N

B 2 B )

+1

1 = ∑ P( B 2 B , 1 k ) P( k B ) 1

k =1

P( k B ) P( 1

B k ) P( k)

1 =

P( B )

1

P(

k −

1

B k )

1

1

=

, P( k) =

N

N + 1

N +1 k

N

N

P( B )

1 = ∑ −1 1

1

1 +

+1



1

 2 +

2 

=

( N + )

1 − ( N + )

1

=

( N +

1



)

1

−  =





N

N

N N

N N

k =

1

(

)

1

2

(

)

1

2

2

2

1

+

+ 



+





P( k B )

2( k − )

1

1 =

N ( N + )

1

P(

k −

B 2 B k )

2

,

1

=

d

la k

= 1 →= 0

N −1

Z tego:

= ⋅

1 2+ ⋅

2 +

3 .. +

. ( N − )

1 N

6

4

4 7

4

4 8

N +1

N +1

ODP = ∑ ( k − 2)2( k − ) 1

=

2

∑

2

( N − )

1 N ( N +

( k − 2)( k − )

1 =

)

1 = 2

N

N N

N

N N

N

N N

k = (

)

1

(

)

1

(

)

1

(

)

1

(

)

1

(

)

1

3

3

2

−

+

−

+ k=2

−

+

Zadanie 5

2

2

2

2 1

1

P(

t

− x

t

− w

− t

X 2 < t ) = P( X ≤ t ) = ∫

x = w

θ

xe

dx =

= ∫

 

θ

e

= 1− θ

e

≅ wykl 

θ

2 xdx = dw

θ 2

 θ 

0

0

 n 

Y

n ≅ wykl



 θ 

θ

n

EaY = a

= θ → a = θ = n n

θ

T = nY

n

n

P

1

1

θ ( nYn − θ > ε ) =

− Pθ (− ε < nYn − θ < ε)

 θ − ε

ε + θ 

= − P 

θ

< Y <

 =

 n

n 

 θ− ε n

ε+ θ n 



ε

ε



−

−

−

= zal : θ > ε = 1−  e n θ − e n θ  = −  −1

1

e e θ −

θ

−1

e

e  =



















 ε 

 ε 

= 1− exp(− )

1 exp  − exp



 −  → ( B)



 θ 

 θ 

Zadanie 6

∑( Xi− µ 2)2

−

1

2

(

2

e

σ

2Π σ )100



2

2 

 µ

µ

µ

µ

2

2

1

1



= exp

X

X

2

∑ i −

−

∑ i +

100

∑( Xi− 1 µ)

 2

2

2

2 

−

 σ

2 σ

σ

2 σ 



1







2 2

e

σ

 2Π σ 

1

µ 2 > µ 1

2 ∑ X i ( µ 2 − µ 1 )

= C ⋅ σ

e

→ STAT = ∑ X

i

∑ X

i ≅ N (

2

100 µ 1

; 00 σ

0

)



t −100 µ 

t −100 µ

P ∑ X > t = P X >

 = ,

0 05 →

= ,164 → t = 1 ,

6 4 σ + 100 µ

0 (

i

)

0

0

0



10 σ



10 σ

cov( X , X =

i

j )

1

2

σ

10





var(∑ X =

+ 



=

+

=

+

⋅

=

i )

100

2

1

2

2

10 !

0

1

2

2

2

2

100 σ

2

σ

100 σ

2

σ

100 σ

99 10 σ

1090 σ

 2  10

2 ⋅ 9 !

8 10

P(∑ X

i >

,

6

1 4 σ + 100 µ

X

N 100 µ 1

; 090 σ

0 ) = ∑

i ≅

(

2

0

) =

≈0,5







8

7

6





1 ,

6 4 σ + 100 µ −100 µ 

1 ,

6 4

0

0





= P X >

 = P X >

≈ 3

,

0 1



1090 σ





1090 









Zadanie 7

m

E ˆ = ( a + a + a + a 1

2

3

4 ) m

2

σ ( ˆ

m) = ( 2 2

2

2

2

2

2

2

a m + 2 a m + 3 a m + 4 a m

= a + 2 a + 3 a + 4 a m 1

2

3

4

) ( 2 2 2 2

1

2

3

4 ) 2

2

ˆ

m

E

= ( 2

2

2

2

a + 2 a + 3 a + 4 a m + a + a + a + a m

1

2

3

4 ) 2

( 1 2 3 4)2 2

E( ˆ

m − m)2 = ( 2

a

2 a

3 a

4 a m

a

a

a

a

m

2 m a

a

a

a

m

1 +

2

2 +

2

3 +

2

4 ) 2 + ( 1 +

2 +

3 +

4 )2

2 −

2 ( 1 + 2 + 3 + )+ 2

4

=

2

= m [2 2

a + 3 2

a + 4 2

a + 5 2

a + 2 a a + 2 a a + 2 a a + 2 a a + 2 a a + 2 a a − 2 a − 2 a − 2 a − 2 a +

1

2

3

4

1 2

1 3

1 4

2 3

2

4

3 4

1

2

3

4

]1

= f ( ˆ a)

∂ = 4 a + 2 a + 2 a + 2 a − 2 = 0

1

2

3

4

∂ a 1

∂ = 6 a + 2 a + 2 a + 2 a − 2 = 0

2

1

3

4

∂ a

4 a − 2 a = 0

2

2

1

∂

=

a = 5

,

0 a

8 a + 2 a + 2 a + 2 a − 2 = 0

2

1

3

1

2

4

∂ a 3

∂ =10 a + 2 a + 2 a + 2 a − 2 = 0

4

1

2

3

∂ a 4

5 a

a

a

1 + 2 3 + 2 4 = 2

3 a

a

a

a = 1 − a − 5

,

2 a

1 + 8 3 + 2 4 = 2

3

4

1

3 a

a

a

1 + 2 3 + 10 4 = 2

3 a

a

a

a

1 +

(81− 4 − 5,

2

1 ) + 2 4 = 2



3 a

a

a

a

1 +

(21− 4 − 5,

2

1 ) + 10 4 = 2

17 a

a

1 + 6 4 = 6



2 a

a

1 − 8 4 = 0

6 − 6 a

6

6

4

a =

=

−

a

1

4

17

17

17

 6

6



2

−

a  − 8 a = 0

4

4

17 17





12 

12

8 +

 a =

4



17 

17

12

17

12

3

a =

=

=

4

17 17 ⋅ 8 + 12

148

37

6

6 3

12

a =

−

=

1

17

17 37

37

3

5 12

4

a = 1 −

−

=

3

37

2 37

37

6

a =

2

37

f ( ˆ a) ≈ 3

,

0 2 →to jest minimum

f (0) = 1

Zadanie 8

ODP = M

M

N (ln

)

1

(

Y

)

M

)

1

(

= E( eY

Y

)=1⋅ P( X ≤ d)+ E( X−

e

d X > d ) P( X > d) =

∞

−

= 1− 2 d

e

+ ∫ x− d −2 x

−2 d

−2 d

−

e

2 e

dx = 1 − e

+ 2 e

= 1+ 2 d

e

d

λ( − )

1

M ( t)

e

N

=

t

e

−

−

3(

2 d

1+ e

)

2 d

1

3 e

ODP =

−

e

= e

Zadanie 9

cov(3 X + ... + X , X + ... + 2 X

= 3 σ + ( n − )

2 σ + 2 σ = ( n + ) 3 σ

1

n

1

n )

2

2

2

2

2

2

2

var U = 9 σ + ( n − ) 1 σ = ( n + )

8 σ

2

2

2

var V = ( n − )

1 σ + 4 σ = ( n + ) 3 σ

( n + )

3

2

σ

n + 3

corr( U , V ) =

=

2

σ

( n + )

3 ( n + )

8

n + 8

Zadanie 10

1 4 4 −2

θ e θ χ

θ

x

f (

4

:

4

4

f x

x θ f θ

θ x ,...,

θ

x

=

=

1

4 )

( ,...,

1

4

)

( ;

0 )

( )

3

f ( x ,..., x

1

4 )

...

∞

∞

f (

4

θ

4

θ

4 1

x

2

x ... x

e

χ ( θ)

x

e

e

e

1

) = ∫ −2

4

= ∫

−2

44

=

( −2 ) −

=

2

3

3

3 2

3

0

x max

4 −2 θ

x

−

f (

e

χ

θ

max

θ x ... x

=

= e θe x χ

θ

1

4 )

( ;

0 )

3

2

2

2 max

( ;

0 )

2

x max

−2 x max

e

3

P( θ >

θ

x

x

1

ln 2

3 x ... x

jezel

i x

3

2

max

max

e

e

dθ

e

e

x

3

1

4 )

∞

=

max <

= ∫ −2 2

2

−

=

6 > → max > −

2

2

3

P( θ > 3 x ...

θ

x

x

x

x

x

e

e

dθ

e

e

1

4 )

∞

= dl

a max > 3 = ∫

−2

2

2

2

1

max

max

−

2

=

max

= 1 > 2

x max

Z tego:



ln 2 

K =  x

> 3

4

:

4

−





2 