Probabilità Esercizi

-3 ≤ x ≤ -1

1 ≤ x ≤ 3

x > 3

E[Lx] =

-7

∫(3/8)dx + ∫(x/8)dx + ∫(3/64)x²dx + ∫(x²/64)dx

= [3/8 x + x²/16 + 3/64 x³]ʃ -3 + [x³/64]ʃ -3

= –3

+ (3/64) – 27/64] + ∫(9/64) – (27/64) + 9/64] + ∫(3/8 – 2/8)

= (3/8 x + (x²/16)]ʃ + (3/8 x + x²/16)]ʃ –3 + [8/8 x³ x dx]

= 3/8

x dx dx x dx + x dx +[3/8]ʃ –4 [8 x³/2]ʃ –7 [26/3]

+7 [26/3] = 13/4 + 13/12 + 39 +73/12 + 52/12 + 73/3

σ²_x = EE[x³] – E[x³]² + 13/3 – (–7)² + 13/3 – 7 = 10/3

X₁ ~ N(0,1)

X₂ ~ N(0,1)

cov (x₁, x₂) = 0

Y₁ = X₂ - 3X₁

-z

Y₁ ~ N(μ_y1, σ_y1²)

Y₂ = 2X₂ + X₁

Y₂ ~ N(-2, 10)

Y₂ ~ N(μ_y2, σ_y2²)

μ_y1 = 0

μ_y2 ~ N(0, 5)

EE[Y₁] = 0 - 0 = 0

EE[Y₂] = 0 + 0 = 0

σ_y1² = 9 + 1 - 18⋅cov(x₁, x₂) = 9 + 1 - 18⋅0 = 9 + 1 - 0 = 10

σ_y2² = 1 + 4 + 4⋅cov(x₁, x₂) = 1 + 4 - 4⋅0 = 1 + 4 - 0 = 5

P(Y₂ | Y₁ > 6) = Q ( ^{6 + z} / _√¹⁰ ) = Q ( ⁸ / _3,16 ) = Q ( 2,53) = 0,60570 = 0,57%

ρ_{y1, y2} = cov(Y₁, Y₂) / √(σ_Y1² ⋅ σ_Y2²) = ^-1 / √(10⋅5) = ^-1 / √50 = ^-1 / 5√2 ≈ -^√2 / ₁₀ ≈-0,14

cov(x₁, Y₁, Y₂) = E[X₁ Y₁ Y₂] - E[Y₁]E[Y₂] = E[Y₁ Y₂] - 0 ⋅ E[E[X₂² X₁ X₂ - 6X₁ X₂]₀]

-3X₁² + 4X₁² 2X₁] ⋅ 0 - E[E[X₂² - 3X₁²] ⋅ 0 - E[E[3X₁] ⋅ 0 = 2 - 3 = -1

E[E[X₂ X₁²]]! = 2E[E[X₂²]] → var(X_a) = E[E[X₂²]] - E[E[X₂]²]_11¹]

7 Giugno 2017

F(x) =

• 1, x < -2
• -x - 2, -2 ≤ x < -1
• -1, -1 ≤ x < 1
• x - 1, 1 ≤ x < 2
• 0, x ≥ 2

A = {-2, 2}

∂ =

• x < -2
• -1, -2 ≤ x < -1
• 0, -1 ≤ x < 1
• 1, 1 ≤ x < 2
• 2, x ≥ 2

∂ 1/2 (x+1)² = (x+1)² / 2 + x+2, x ≥ -2

∂ 1/2 [1 + (x-1)²] =

∫[x(x+2)] -2² = ∫[x-1] =

x² + x, -2^x

(1/3 x³ + x²) | -2² = [(3/3 x² + (x²) | -2]

[(3/3 x³ - x²/2)]

+ [2/3³ - 2/2] - [(1/3³) 1/2]

[(1/3³ + 8 - 4) + [(8/3 - 2)]

[3 - 2, 5 - x] 16 - 12 + 8 + 3, 5/6

x =

x²(x+2)&sup> -

[(2(-1)³ - 2(-2)³] 1/

+ [1/3³ - 2/2

+ [2/3³ - 4/2]

- 2/4 (= -1/4(

+ 2/4³ - , ) 1/

1/6 - 4 x 8 + 4/3

3 - 8 + 12 + 64 , 12/4

5 - 3 / 10

6/12 + 2

(-13 - 2)

[ -1/6 , (-19/12)

[ - (47 - 2) /12] = (-13 - 2) /

45 - 1

+ [(45 - 21)/12

40 CARTE

A B C D

10 10 10 10

E₁ = {

E₂ = {

E₃ = {

UNO DEI GIOCATORI HA QUATTRO 10”}

E₄ =

(⁴⁰⁄₁₀) (²⁹⁄₁₀) (¹⁄₇₀) + (³⁹⁄₁₀) (²⁹⁄₁₀) (²⁰⁄₁₀) = t + 3 × 300 × 5015 + 67960 = 2 ⁄ 42 ≈ 24,2 %

E₂ (³⁶⁄₁₀) + (⁴⁄₄₀) (²⁶¹⁄₃₀) + 261 ≈ 0,35 ≈ 35 %

E₃ = (⁴⁄₄₀) (³⁶⁄₁₀) + (⁴⁄₃₀) (²⁶⁄₂₀) + ¹⁴⁄ ³⁶⁄₁₀ = 5 × 481 + 42

5 DADI

3 BONUS

₅C₃ = 10

P(6) = ¹/₆

2 TRUCATI

P(6) = ^u/₆

X = N. LANCI PER FARE CINQUE 6

1. 2. 3. 4.

DF: ¹/₆ ¹/₆ (1-u) ^u/₆

u²/₆ LIFE GEOMETRICA (P, X)

╚→ 2/6

a) E[X] ≥ 300 → ³⁰⁰/_u² < 300

u² > ³⁰⁰/₂₁₆ u² < 300 → u² < ^300u2/216

b) DISPERSIONE: √VAR = √¹/_p

E[X] = ¹/_p √^p/_1-p = 1

√^p/_1-p = ¹/_1-p

DISP MASSIMA PER U MINIMA

X ~ N(1,1)

Y ~ N(z, 4)

Z = X + Y ρ_{x, y} = ¹/₂

Z ~ N(μ_z, σ_z²)

μ_z = E[X] + tE[Y] = 1 - 2t

σ_z² = 1 + 4t² - t cov(X, Y) = 1 + t² - 2t

cov(X, Y) = ρ_{x, y} √σ_x² σ_y² = ¹/₂ (1-2) = ¹/₂ 2 = 1

4t² - 2t + 1 = ^{2 ± √4 - 16, İMP/₈}

t PER COV, 1 + 4t² - 2t = 1 → 4t² - 2t - 1 = 1 → 4t² - 2t= 0 → t(4t + 2) = 0.

t = 0

t = 0

4t - 2t = 0 → 4t = 2

t = 0

t = ^1/₂

t = 0

t = ^1/₂

P_R{ |Z| > 1 } = Q ( ^{1-2(0)/₁) = Q ( 0/₁) = Q(0)}

P_R{ |Z| > 1 } = Q(1) = Q(^{1/₁) = Q ( 1/₁) = Q(1)}

X₁ ~ N(0,1)   X₂ ~ N(0,1)   cov(x₁, x₂) = 0

Y₁ = 2X₁ + X₂

Y₂ = 3X₁ - 2X₂ + 5

a) Y₁ ~ N(μ_Y1, σ²_Y1)   Y₂ ~ N(0,5)

E[Y₁] = 0   E[Y₂] = E[Y2] ~ N(5,13)

σ²_Y2 = 3² var(X₁) + 2² var(X₂) - 2cov(X₁,X₂) = 4 + 1 - 0 = 5

Y₂ ~ N(μ_Y2, σ²_Y2)

E[Y₂] = 5

σ2_Y1 = 3² var(X₁) + 2 var(X₂) - 2cov(X₁,X₂) = 9 + 4 - 0 = 13

b) cov(Y₁, Y₂) = E[Y₁Y₂] - E[Y₁]E[Y₂] = E[(2X₁ + X₂)(3X₁ - 2X₂ + 5)] - 0

= E[6X₁² - 4X₁X₂ + 10X₁ + 3X₂² + 5X₂] - 0 = 30 - 0 + 0 + 0 - 10 + 25 + 5

= 4 E[Y₁²] = var(Y₁) = E[X₁²] - E[Y₁]² = 30

4 E[Y₁ Y₂] = 4 E[Y₁] E[Y₂] = (4 - 0)(4 - 5) = 0