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Probabilistyka Wykład 1w12.pwr.wroc.pl/lmg/wp-content/uploads/KM/Probabilistyka_W3.pdf · 𝑃 = t1 = l𝑃 = t1 = m 𝑖 l+ m=1 m=1− l = t1∙ l+ t2∙ m= t1∙ l+ t2∙(1−

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  • 3

  • lim𝑥→−∞

    𝐹 𝑥 = 0 lim𝑥→∞

    𝐹 𝑥 = 1

    𝐹: ℝ → [0,1]

    𝐹 𝑥 = 𝑃𝑟 𝑋 ≤ 𝑥 , 𝑥 ∈ ℝ

  • 1

    x

    F(x) 𝑓 𝑥 =𝑑𝐹(𝑥)

    𝑑𝑥

    x𝐹 𝑥 =

    −∞

    𝑥

    𝑓 𝑥 𝑑𝑥

  • 1) 𝑓(𝑥) ≥ 0

    2)

    −∞

    +∞

    𝑓 𝑥 𝑑𝑥 = 1

    𝑃𝑟 𝑎 < 𝑋 < 𝑏 =

    𝑎

    𝑏

    𝑓 𝑥 𝑑𝑥

    𝑃𝑟 𝑋 < 𝑏 =

    −∞

    𝑏

    𝑓 𝑥 𝑑𝑥

  • 𝑚𝑘 = 𝐸𝑋𝑘 =

    −∞

    +∞

    𝑥𝑘𝑓 𝑥 𝑑𝑥

    𝑚𝑘 = 𝐸𝑋𝑘 =

    𝑖=1

    𝑛

    𝑥𝑖𝑘Pr(𝑋 = 𝑥𝑖)

    x

    𝜇𝑘 = 𝐸(𝑋 − 𝐸𝑋)𝑘

    𝜇2 = 𝐷2𝑋 = 𝐸𝑋2 − (𝐸𝑋)2

    𝜎 = 𝜇2

    m

    s

  • 𝐽𝑒ż𝑒𝑙𝑖 𝑌 = 𝑎𝑋 + 𝑏, 𝑡𝑜

    𝐽𝑒ż𝑒𝑙𝑖 𝑋, 𝑌 𝑠ą 𝑛𝑖𝑒𝑧𝑎𝑙𝑒ż𝑛𝑒, 𝑡𝑜

    𝐽𝑒ż𝑒𝑙𝑖 𝑍 = 𝑎𝑋 + 𝑏𝑌, 𝑡𝑜

    𝐸𝑌 = 𝑎𝐸𝑋 + 𝑏

    𝐸(𝑋𝑌) = 𝐸𝑋 ∙ 𝐸𝑌

    𝐸𝑍 = 𝑎𝐸𝑋 + 𝑏𝐸𝑌

    𝐷2𝑌 = 𝑎2𝐷2𝑋

    𝐷2𝑌 = 𝑎2𝐷2𝑋 + 𝑏2𝐷2𝑋

    𝐽𝑒ż𝑒𝑙𝑖 𝑌 = 𝜑 𝑋 , 𝑡𝑜 𝐸𝑌 = 𝐸 𝜑 𝑋 =

    −∞

    +∞

    𝜑 𝑥 𝑓 𝑥 𝑑𝑥

  • x

    1 2 3

  • x

    ϵ

    𝑜𝑟𝑎𝑧Pr(𝑋 ≤ 𝑥𝑝) ≥ 𝑝 Pr 𝑋 ≥ 𝑥𝑝 ≥ 1 − 𝑝

    𝐹 𝑥𝑝 = 𝑝 1

    1 xp 6 x

    F(x)

    p

    xp

    −∞

    𝑥𝑝

    𝑓 𝑥 𝑑𝑥 = 𝑝

  • xX1/2

  • 𝑃𝑟 𝑋 = 𝑥1 = 𝑝 𝑃𝑟 𝑋 = 𝑥1 = 𝑞 𝑔𝑑𝑧𝑖𝑒 𝑝 + 𝑞 = 1

    𝑞 = 1 − 𝑝

    𝐸𝑋 = 𝑥1 ∙ 𝑝 + 𝑥2 ∙ 𝑞 = 𝑥1 ∙ 𝑝 + 𝑥2 ∙ (1 − 𝑝)

    BB: 𝑃𝑟 𝑋 = 𝑥1 ∙ 𝑃𝑟 𝑋 = 𝑥1 = 𝑝2

    CC: 𝑃𝑟 𝑋 = 𝑥2 ∙ 𝑃𝑟 𝑋 = 𝑥2 = (1 − 𝑝)2

    BC: 𝑃𝑟 𝑋 = 𝑥1 ∙ 𝑃𝑟 𝑋 = 𝑥2 = 𝑝(1 − 𝑝)

    CB: 𝑃𝑟 𝑋 = 𝑥2 ∙ 𝑃𝑟 𝑋 = 𝑥1 = 1 − 𝑝 𝑝

  • BB: 𝑃𝑟 𝑋 = 𝑥1 ∙ 𝑃𝑟 𝑋 = 𝑥1 = 𝑝2

    CC: 𝑃𝑟 𝑋 = 𝑥2 ∙ 𝑃𝑟 𝑋 = 𝑥2 = (1 − 𝑝)2

    BC: 𝑃𝑟 𝑋 = 𝑥1 ∙ 𝑃𝑟 𝑋 = 𝑥2 = 𝑝(1 − 𝑝)

    CB: 𝑃𝑟 𝑋 = 𝑥2 ∙ 𝑃𝑟 𝑋 = 𝑥1 = 1 − 𝑝 𝑝

    𝑝2 + 1 − 2𝑝 + 𝑝2 + 𝑝 − 𝑝2 + 𝑝 − 𝑝2 = 1

  • 𝑃𝑟 𝑋 = 𝑘 =𝑛

    𝑘𝑝𝑘(1 − 𝑝)𝑛−𝑘

  • 𝐸 𝑥𝑖 = 1 ∙ 𝑝 + 0 ∙ 1 − 𝑝 = 𝑝

    𝑥 =

    𝑖

    𝑥𝑖

    𝐸 𝑥 =

    𝑖=1

    𝑛

    𝐸(𝑥𝑖) = 𝑖=1

    𝑛

    𝑝 = 𝑛𝑝

  • 𝐷2 𝑥𝑖 = 𝐸(𝑥𝑖)2 − (𝐸 𝑥𝑖 )

    2

    𝐸(𝑥𝑖)2 = 12 ∙ 𝑝 + 02 ∙ (1 − 𝑝) = 𝑝

    𝐷2 𝑥𝑖 = 𝑝 − 𝑝2 = 𝑝(1 − 𝑝)

    𝐷2 𝑥 = 𝑛𝑝(1 − 𝑝) = 𝑛𝑝𝑞

    𝑘=1

    𝑛

    𝐷2(𝑥𝑘) =𝑛𝐷2(𝑥𝑖) =

  • 𝑃𝑟 𝑋 = 𝑘 =𝜆𝑘𝑒−𝜆

    𝑘!

    𝐸𝑋 = 𝜆 𝐷2𝑋 = 𝜆 𝑀𝑒 ≈ 𝜆 +1

    3+

    0,02

    𝜆

  • 1

    𝑏 − 𝑎

    0 a bx

    f(x)

    −∞

    +∞

    𝑓 𝑥 𝑑𝑥 = 1

    𝑎

    𝑏

    𝑓 𝑥 𝑑𝑥 = 1

    𝑓 𝑥

    𝑎

    𝑏

    𝑑𝑥 = 𝑓 𝑥 𝑏 − 𝑎 = 1

    𝑓 𝑥 =

    1

    𝑏 − 𝑎𝑑𝑙𝑎 𝑥 ∈ [𝑎, 𝑏]

    0 𝑑𝑙𝑎 𝑝𝑜𝑧𝑜𝑠𝑡𝑎ł𝑦𝑐ℎ 𝑥

  • 𝐹 𝑥 =

    −∞

    𝑥

    𝑓 𝑥 𝑑𝑥

    𝑎

    𝑥1

    𝑏 − 𝑎𝑑𝑥 =

    1

    𝑏 − 𝑎

    𝑎

    𝑥

    𝑑𝑥 = 𝑥

    𝑏 − 𝑎

    𝑥𝑎

    =𝑥 − 𝑎

    𝑏 − 𝑎

    𝐹 𝑥 =

    𝑑𝑙𝑎 𝑥 ∈ (−∞, 𝑎)

    =

    −∞

    𝑥

    0𝑑𝑥 = 0

    𝑑𝑙𝑎 𝑥 ∈ [𝑎, 𝑏)

    𝐹 𝑥 =

    𝑑𝑙𝑎 𝑥 ∈ [𝑏, +∞)

    𝑎

    𝑏1

    𝑏 − 𝑎𝑑𝑥 +

    𝑏

    +∞

    0𝑑𝑥 = 𝑥

    𝑏 − 𝑎𝑏𝑎

    =𝑏 − 𝑎

    𝑏 − 𝑎= 1

  • 1

    𝑏 − 𝑎𝐹 𝑥 =

    0 𝑑𝑙𝑎 𝑥 < 𝑎𝑥 − 𝑎

    𝑏 − 𝑎𝑑𝑙𝑎 𝑎 ≤ 𝑥 < 𝑏

    1 𝑑𝑙𝑎 𝑥 ≥ 𝑏

    0 a bx

    f(x)

    0 a bx

    F(x)

    1

    EX

    𝐸𝑋 =𝑏 + 𝑎

    2

    𝐷2𝑋 =(𝑏 − 𝑎)2

    12

  • 𝑓 𝑥 = 0 𝑑𝑙𝑎 𝑥 < 0

    𝜆𝑒−𝜆𝑥 𝑑𝑙𝑎 𝑥 ≥ 0, 𝜆 > 0

    x

    f(x)

    𝐸𝑋 =1

    𝜆𝐷2𝑋 =

    1

    𝜆2

    𝐹 𝑥 = 0 𝑑𝑙𝑎 𝑥 < 0

    1 − 𝑒−𝜆𝑥 𝑑𝑙𝑎 𝑥 ≥ 0, 𝜆 > 0

    F(x)

    x

  • 𝑓 𝑥 =

    0 𝑑𝑙𝑎 𝑥 < 0

    𝑘

    𝜆

    𝑥

    𝜆

    𝑘−1

    𝑒−( 𝑥

    𝜆)𝑘

    𝑑𝑙𝑎 𝑥 ≥ 0, 𝜆 > 0

  • 𝐹 𝑥 = 0 𝑑𝑙𝑎 𝑥 < 0

    1 − 𝑒−( 𝑥

    𝜆)𝑘

    𝑑𝑙𝑎 𝑥 ≥ 0, 𝜆 > 0

  • 𝜙 𝑥 =1

    𝜎 2𝜋𝑒

    (−𝑥−𝑚 2

    2𝜎2)

    ( , s)

    (0, 1)

    𝜙 𝑥 =1

    2𝜋𝑒(−

    𝑥2

    2 )

  • 𝑌 = 𝑋1 + 𝑋2 + ⋯ + 𝑋𝑛𝑡𝑜 𝑌 𝑚𝑎 𝑟𝑜𝑧𝑘ł𝑎𝑑 𝑁 0,1 𝑜𝑟𝑎𝑧

    𝐸𝑌 = 𝐸𝑋1 + 𝐸𝑋2 + ⋯ + 𝐸𝑋𝑛

    𝐷2𝑌 = 𝐷2𝑋1 + 𝐷2𝑋2 + ⋯ + 𝐷

    2𝑋𝑛

    Φ 𝑥 =

    −∞

    𝑥1

    2𝜋𝑒(−

    𝑥2

    2 )𝑑𝑥

    Φ 𝑥 =1

    21 + 𝑒𝑟𝑓

    𝑧

    2

  • 𝑓 𝑥 =

    0 𝑑𝑙𝑎 𝑥 < 0𝜃𝑘

    Γ(𝑘)𝑥𝑘−1𝑒(−𝜃𝑥) 𝑑𝑙𝑎 𝑥 ≥ 0, 𝑏 > 0

    Γ 𝑘 =

    0

    𝑥𝑘−1𝑒−𝑥𝑑𝑥

    𝐹 𝑥 =𝜃𝑘

    Γ(𝑘)

    0

    𝑥

    𝑥𝑝−1 𝑒(−𝜃𝑥)𝑑𝑥

    𝐸𝑋 =𝑘

    𝜃𝐸𝑋2 =

    𝑘(𝑘 + 1)

    𝜃2𝐷2𝑋 =

    𝑘

    𝜃2

  • 3