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Probabilistyka WykΕad 1w12.pwr.wroc.pl/lmg/wp-content/uploads/KM/Probabilistyka_W3.pdfΒ Β· π =...
Transcript of Probabilistyka WykΕad 1w12.pwr.wroc.pl/lmg/wp-content/uploads/KM/Probabilistyka_W3.pdfΒ Β· π =...
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limπ₯βββ
πΉ π₯ = 0 limπ₯ββ
πΉ π₯ = 1
πΉ: β β [0,1]
πΉ π₯ = ππ π β€ π₯ , π₯ β β
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1
x
F(x) π π₯ =ππΉ(π₯)
ππ₯
xπΉ π₯ =
ββ
π₯
π π₯ ππ₯
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1) π(π₯) β₯ 0
2)
ββ
+β
π π₯ ππ₯ = 1
ππ π < π < π =
π
π
π π₯ ππ₯
ππ π < π =
ββ
π
π π₯ ππ₯
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ππ = πΈππ =
ββ
+β
π₯ππ π₯ ππ₯
ππ = πΈππ =
π=1
π
π₯ππPr(π = π₯π)
x
ππ = πΈ(π β πΈπ)π
π2 = π·2π = πΈπ2 β (πΈπ)2
π = π2
m
s
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π½πΕΌπππ π = ππ + π, π‘π
π½πΕΌπππ π, π π Δ ππππ§πππΕΌππ, π‘π
π½πΕΌπππ π = ππ + ππ, π‘π
πΈπ = ππΈπ + π
πΈ(ππ) = πΈπ β πΈπ
πΈπ = ππΈπ + ππΈπ
π·2π = π2π·2π
π·2π = π2π·2π + π2π·2π
π½πΕΌπππ π = π π , π‘π πΈπ = πΈ π π =
ββ
+β
π π₯ π π₯ ππ₯
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x
1 2 3
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x
Ο΅
ππππ§Pr(π β€ π₯π) β₯ π Pr π β₯ π₯π β₯ 1 β π
πΉ π₯π = π 1
1 xp 6 x
F(x)
p
xp
ββ
π₯π
π π₯ ππ₯ = π
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xX1/2
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ππ π = π₯1 = π ππ π = π₯1 = π πππ§ππ π + π = 1
π = 1 β π
πΈπ = π₯1 β π + π₯2 β π = π₯1 β π + π₯2 β (1 β π)
BB: ππ π = π₯1 β ππ π = π₯1 = π2
CC: ππ π = π₯2 β ππ π = π₯2 = (1 β π)2
BC: ππ π = π₯1 β ππ π = π₯2 = π(1 β π)
CB: ππ π = π₯2 β ππ π = π₯1 = 1 β π π
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BB: ππ π = π₯1 β ππ π = π₯1 = π2
CC: ππ π = π₯2 β ππ π = π₯2 = (1 β π)2
BC: ππ π = π₯1 β ππ π = π₯2 = π(1 β π)
CB: ππ π = π₯2 β ππ π = π₯1 = 1 β π π
π2 + 1 β 2π + π2 + π β π2 + π β π2 = 1
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ππ π = π =π
πππ(1 β π)πβπ
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πΈ π₯π = 1 β π + 0 β 1 β π = π
π₯ =
π
π₯π
πΈ π₯ =
π=1
π
πΈ(π₯π) = π=1
π
π = ππ
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π·2 π₯π = πΈ(π₯π)2 β (πΈ π₯π )
2
πΈ(π₯π)2 = 12 β π + 02 β (1 β π) = π
π·2 π₯π = π β π2 = π(1 β π)
π·2 π₯ = ππ(1 β π) = πππ
π=1
π
π·2(π₯π) =ππ·2(π₯π) =
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ππ π = π =πππβπ
π!
πΈπ = π π·2π = π ππ β π +1
3+
0,02
π
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1
π β π
0 a bx
f(x)
ββ
+β
π π₯ ππ₯ = 1
π
π
π π₯ ππ₯ = 1
π π₯
π
π
ππ₯ = π π₯ π β π = 1
π π₯ =
1
π β ππππ π₯ β [π, π]
0 πππ πππ§ππ π‘πΕπ¦πβ π₯
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πΉ π₯ =
ββ
π₯
π π₯ ππ₯
π
π₯1
π β πππ₯ =
1
π β π
π
π₯
ππ₯ = π₯
π β π
π₯π
=π₯ β π
π β π
πΉ π₯ =
πππ π₯ β (ββ, π)
=
ββ
π₯
0ππ₯ = 0
πππ π₯ β [π, π)
πΉ π₯ =
πππ π₯ β [π, +β)
π
π1
π β πππ₯ +
π
+β
0ππ₯ = π₯
π β πππ
=π β π
π β π= 1
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1
π β ππΉ π₯ =
0 πππ π₯ < ππ₯ β π
π β ππππ π β€ π₯ < π
1 πππ π₯ β₯ π
0 a bx
f(x)
0 a bx
F(x)
1
EX
πΈπ =π + π
2
π·2π =(π β π)2
12
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π π₯ = 0 πππ π₯ < 0
ππβππ₯ πππ π₯ β₯ 0, π > 0
x
f(x)
πΈπ =1
ππ·2π =
1
π2
πΉ π₯ = 0 πππ π₯ < 0
1 β πβππ₯ πππ π₯ β₯ 0, π > 0
F(x)
x
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π π₯ =
0 πππ π₯ < 0
π
π
π₯
π
πβ1
πβ( π₯
π)π
πππ π₯ β₯ 0, π > 0
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πΉ π₯ = 0 πππ π₯ < 0
1 β πβ( π₯
π)π
πππ π₯ β₯ 0, π > 0
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π π₯ =1
π 2ππ
(βπ₯βπ 2
2π2)
( , s)
(0, 1)
π π₯ =1
2ππ(β
π₯2
2 )
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π = π1 + π2 + β― + πππ‘π π ππ πππ§πΕππ π 0,1 ππππ§
πΈπ = πΈπ1 + πΈπ2 + β― + πΈππ
π·2π = π·2π1 + π·2π2 + β― + π·
2ππ
Ξ¦ π₯ =
ββ
π₯1
2ππ(β
π₯2
2 )ππ₯
Ξ¦ π₯ =1
21 + πππ
π§
2
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π π₯ =
0 πππ π₯ < 0ππ
Ξ(π)π₯πβ1π(βππ₯) πππ π₯ β₯ 0, π > 0
Ξ π =
0
β
π₯πβ1πβπ₯ππ₯
πΉ π₯ =ππ
Ξ(π)
0
π₯
π₯πβ1 π(βππ₯)ππ₯
πΈπ =π
ππΈπ2 =
π(π + 1)
π2π·2π =
π
π2
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