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COMP 245 Statistics Solutions 4 - Continuous Random Variables −x Z 1. Begin with F(−x) + F(x) = Z x f (u)du + u=−∞ v=−∞ Z x for the first part leads to F(−x) + F(x) = − Z x Z ∞ f (v)dv = f (v)dv = 1. u=−∞ 2. f (v)dv. Taking the change of variable v = −u Z x Z ∞ f (−v)dv + f (v)dv = f (v)dv + v=∞ v=−∞ v=x v=−∞ (a) Since a particle is equally likely to hit anywhere on the plate, for 0 < r < 1 the probability πr2 = r2 . Hence that it will strike inside a circle of radius r is π12 0, x ≤ 0 2 x , 0<x<1 F(x) = 1, x ≥ 1. (b) P(r < X < s) = P(X < s) − P(X < r) = s2 − r2 . (c) From 2a the cdf of X is given by F(x) = x2 for 0 ≤ x ≤ 1. So the pdf, f (x) = F0 (x) = 2x for 0 ≤ x ≤ 1, and 0 everywhere else. (d) The expected distance from the origin is ∞ Z E(X) = 1 Z x f (x)dx = −∞ 0 2x3 x.2xdx = 3 " #1 0 2 = . 3 3. A random variable X ∼ Exp(λ) has density f (x) = λe−λx for x ≥ 0. So Z ∞ Z ∞ Z ∞ h i −λx −λx ∞ E(X) = x f (x)dx = xλe dx = −xe + e−λx dx 0 −∞ 0 " −λx #0∞ −e 1 1 = [0 − 0] − = 0− − = . λ 0 λ λ E X 2 Z ∞ = ∞ Z x f (x)dx = x λe 2 −∞ 2 −λx ∞ Z h i 2 −λx ∞ dx = −x e + 0 0 2xe−λx dx 0 2 2 E(X) = 2 λ λ 2 1 1 ⇒ Var(X) = E X2 − {E (X)}2 = 2 − 2 = 2 . λ λ λ = [0 − 0] + ( 4. X has range [0, 1] and pdf fX (x) = 1, 0 < x < 1 0, otherwise. Thus the cdf for Y = eX is given by Z FY (y) = PY (Y ≤ y) = P(e ≤ y) = P(X ≤ log(y)) = log(y) X −∞ log(y) Z fX (x)dx = 0 dx = log(y), for 0 < log(y) < 1; that is, for 1 < y < e. For the pdf, differentiating FY (y) wrt y gives 1 y, ( fY (y) = 0, 1<y<e otherwise. x−µ , so Y = g(X). First we note that g is clearly a continuous, monotonically σ increasing function of x. Therefore we have 5. Let g(x) = 0 fY (y) = fX {g−1 (y)}|g−1 (y)|. 0 Well g−1 (y) = σy + µ, so g−1 (y) = σ. Since X ∼ N(µ, σ2 ), ) ( (x − µ)2 1 fX (x) = √ exp − 2σ2 σ 2π and hence ) ( ( 2) (σy + µ − µ)2 y 1 1 fY (y) = fX (σy + µ)σ = √ exp − σ = √ exp − =⇒ Y ∼ N(0, 1). 2 2 2σ σ 2π 2π 6. (a) ∀a , 0, FY (y) = PY (Y ≤ y) = P(aX + b ≤ y) = P(aX ≤ y − b). If a > 0, ! ! y−b y−b FY (y) = P X ≤ = FX , a a whereas if a < 0, ! ! y−b y−b FY (y) = P X ≥ = 1 − FX . a a (b) The pdf of a continuous random variable is the derivative of the cdf, fY (y) = F0y (y). So if a > 0, ! ! ! y−b y−b d 1 0 y−b 1 fY (y) = FX = FX = fX , dy a a a a a whereas if a < 0, ( !) ! ! y−b y−b d 1 0 y−b 1 fY (y) = 1 − FX = − FX = − fX , dy a a a a a So either way, we have ! y−b 1 fY (y) = fX . |a| a 7. (a) z = 1.16 or z = -1.16. (b) z = 1.09. (c) z = -1.35 or z = -1.69. 8. (a) 0.3849 (c) 0.6636 (e) 0.8997 (b) 0.2517 (d) 0.1828