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