Логика. Множества. Вероятность. Лексаченко В.А. - 108 стр.

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Sravniva
f
ξ
(
x)
i
1
2
πσ
e
(
x m
)
2
2
σ
2
,
poluqim sistemu uravneni$i
(
1
/(2σ
2
) =9
/
2
m/(
σ
2
) =9
/
2
otkuda sleduet
σ = 1/
3, m = 1/
2, k
=
3
e
9/
8
2
π
.
P
ξ
([0; 1.
0]) = Φ(1
,
5) Φ(
1
, 5) = 2Φ(1,
5)
1 = 0,
866. J
15 .
Opredelit~ parametry
m
ξ
, m
η
, σ
ξ
, σ
η
, r
ξ
i postonnu
k
sovmestno$i normal~no$i plotnosti
1)
f
ξη
(x, y
) =
k
exp
8(x
+ 3)
2
/3
4(
x + 3)(y
1)
6(
y
1)
2
.
2)
f
ξη
(x, y
) =
k exp
(
x
+ 2)
2
/
4 + (
x
+ 2)(
y
+ 4)
/
3
(
y + 4)
2
.
3) f
ξη
(
x, y
) =
k
exp
3(
x + 1)
2
/8 +
2(x
+ 1)y/
4 y
2
/4
.
4) f
ξη
(
x, y
) =
k
exp
2(x 1)
2
+ (
x
1)(
y + 2)/3 (
y + 2)
2
/
18
.
5) f
ξη
(
x, y) =
k exp
2(
x 2)
2
+ 2(x
2)(y + 1)
(
y + 1)
2
.
6)
f
ξη
(x, y) =
k exp
9(x + 1)
2
/2 2(x
+ 1)(
y
3)/
5
(
y
3)
2
/
10
.
7) f
ξη
(
x, y) = k
exp
4x
2
4x(y
4)
2(
y
4)
2
.
8)
f
ξη
(x, y
) =
k
exp
9(x
2)
2
+ 3(x
2)(y
2)
/
2
9(y 2)
2
/8
.
9)
f
ξη
(
x, y) =
k exp
2(
x
5)
2
/
3 + 2(
x
5)(
y + 1)
6(y + 1)
2
.
10) f
ξη
(x, y
) =
k exp
2(
x
3)
2
+ 4(
x
3)(y 2)
4(y
2)
2
.
R e x e n i e dl
f
ξη
(x, y
) = k exp
3(x2)
2
+
6(
x
2)(y+4)
3(y+4)
2
.
Sravniva
f
ξη
(x, y)
s dvumerno$i normal~no$i plotnost~
1
2
πσ
1
σ
2
p
1
ρ
2
exp
(
x
m
1
)
2
2
σ
2
1
(1
ρ
2
)
+
ρ(x
m
1
)(y
m
2
)
σ
1
σ
2
(1
ρ
2
)
(y m
2
)
2
2
σ
2
2
(1
ρ
2
)
,
poluqim m
1
= 2, m
2
=
4, a σ
1
, σ
2
, ρ nahodts iz uravneni$i
1
/
(2σ
2
1
(1
ρ
2
)) =3
ρ/
(σ
1
σ
2
(1
ρ
2
)) =
6
1/(2σ
2
2
(1
ρ
2
)) =3
.
Iz pervogo i tret~ego uravneni poluqim ravenstvo σ
1
= σ
2
, a
zatem, razdeliv vtoroe uravnenie na pervoe, poluqim ρ = 1/
6.
Podstavl ρ
v pervoe uravnenie, poluqim σ
1
=
σ
2
= 1/
5
i
zatem
k
=
30
/(2π
).
J
16 .
Ispol~zu primer 6.1, dokazat~ formulu e
itm
(
)
2
/2
dl
harakteristiqesko$i funkcii normal~no$i sluqa$ino$i veliqiny s
matematiqeskim oidaniem m
i dispersie$i
σ
2
.
17 . Dokazat~, qto summa nezavisimyh normal~nyh sluqa$inyh
veliqin
ξ
1
,
. . . , ξ
n
s matematiqeskimi oidanimi
m
1
, . . . , m
n
i
dispersimi σ
2
1
, . . . , σ
2
n
vlets normal~no$i sluqa$ino$i veliqi-
no$i s
m =
n
X
k
=1
m
k
i σ
2
=
n
X
k
=1
σ
2
k
.
108