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Chapter 7 Problems 2. This is simply because F (x) − F (x−) = P{X = x}. 3. Note that 1 Z EX = xf (x) dx 0 Z 1 1 xa+1−1 (1 − x)b−1 dx B(a , b) 0 B(a + 1 , b) = . B(a , b) = Similarly, Z 2 E(X ) = 1 x2 f (x) dx 0 Z 1 1 xa+2−1 (1 − x)b−1 dx B(a , b) 0 B(a + 2 , b) . = B(a , b) = Therefore, Var(X) = B(a + 2 , b) − B(a , b) B(a + 1 , b) B(a , b) 2 . 6. We compute directly: For all real numbers a, Z a F (a) = exp −x − e−x dx −∞ Z ∞ = exp (y − ey ) dy [y = −x] −a Z ∞ dz [z = ey ; y = ln z; dy = dz/z] = exp (ln z − z) z −a e Z ∞ = e−z dz [since exp(ln z) = z] −a e = exp −e−a . 7. Here, f (x) = λe−λx if x ≥ 0; else, f (x) = 0. Let us first compute the density of Y = eaX in the case that a > 0. In that case, for all y > 1, FY (y) = P eaX ≤ y 1 1 = P X ≤ ln y = FX ln y . a a 1 Therefore, whenever y > 1, 1 1 λ λ λ −λ/a λ fY (y) = fX ln y = exp − ln y = y = y −(λ/a)−1 . ay a ay a ay a Else, fY (y) = 0 if y ≤ 1. Consequently, Z ∞ EY = 1 λ λ y y −(λ/a)−1 dy = a a ∞ Z y λ/a 1 λ/a dy = 1 + (λ/a) ∞ 1 if λ > a, otherwise. If, on the other hand, a < 0, then aX ≥ 0, and this means that 0 ≤ Y ≤ 1 with probability one. That is, for all 0 < y < 1, FY (y) = P eaX ≤ y 1 1 ln y . = P X ≥ ln y = 1 − FX a a Therefore, λ −(λ/a)−1 − y fY (y) = a 0 if 0 < y < 1, otherwise. In this case, λ EY = a Z 1 y −λ/a 0 λ/a dy = 1 − (λ/a) ∞ if 0 < λ < a, otherwise. 15. For all c, h i 2 E (X − c) = E(X 2 ) − 2cEX + c2 , provided that E(X 2 ) < ∞. Call this expression f (c) and note that f 0 (c) = −2EX + 2c f 00 (c) = 2. This implies that f is minimized at cmin = EX. Also, h i 2 f (cmin ) = E (X − EX) = Var(X). Chapter 8 Problems 4. We know that for all a ≥ 0, Z a Z FX+Y (a) = a−y g(x + y) dx dy. 0 0 2 Note that for all y fixed, the inner integral is equal to: Z a Z a−y g(z) dz. [z = x + y] g(x + y) dx = y 0 Therefore, Z a a Z Z a Z 0 y Z dy g(z) dz = g(z) dz dy = FX+Y (a) = z 0 0 a zg(z) dz. 0 by reversing the order of the two integrals. Consequently, ( zg(z) if z ≥ 0, fX+Y (z) = 0 otherwise. 5(b). We know from our previous assignment that c = 1/(2π), and so f (x , y) = 1 3/2 2π (1 + x2 + y 2 ) . Integrate [dy] to find that Z ∞ 1 1 fX (x) = dy 2π −∞ (1 + x2 + y 2 )3/2 Z ∞ p 1 1 = 1 + x2 dz 3/2 2π −∞ (1 + x2 + (1 + x2 )z 2 ) Z ∞ dz C 1 = , = 2π(1 + x2 ) −∞ (1 + z 2 )3/2 1 + x2 [z = y/ p 1 + x2 ] for some C which does not depend on x. It follows then that X is Cauchy, and C = 1/π (why?). 6. First, note that w . 1 + z2 Here, we are interested over the range x, y ≥ 0. Therefore, r r w w x= and y = xz = z . 1 + z2 1 + z2 z 2 x2 = w − x2 ⇒ x2 = Now, 1 1 ∂x = = , ∂w ∂w/∂x 2x Also, ∂x 1 x2 = =− . ∂z ∂z/∂x y 3 Similarly, ∂y 1 1 = = , ∂w ∂w/∂y 2y and ∂y 1 1 = = . ∂z ∂z/∂y x Therefore, ∂x ∂y ∂x ∂y − ∂w ∂z ∂z ∂w 2 1 1 x 1 = × − − × 2x x y 2y 2 2 2 x x +y 1 . = 2+ 2 = 2x 2y 2x2 y 2 J(w , z) = Solve for (w , z) instead of (x , y): x2 + y 2 = w2 ; and x2 y 2 = w z2w wz 2 × = . 1 + z2 1 + z2 (1 + z 2 )2 Therefore, J(w , z) = w2 1 z2 = w3 . 2 2 2 2(1 + z ) /wz 2 (1 + z 2 )2 Now plug to find that z2 1 3 w fX,Y (x , y) 2 (1 + z 2 )2 1 z2 1 = w3 2 (1 + z 2 )2 2π (1 + x2 + y 2 )3/2 fW,Z (w , z) = = z2 w3 1 . 4π (1 + z 2 )2 (1 + w2 )3/2 In particular, W and Z are independent, since fW,Z (w , z) is the product of a function of w and a function of z; why does this do the job? 10. Solve for x and y to obtain: x= u+v , 2 y= u−v . 2 Therefore, J(u , v) = ∂x ∂y ∂x ∂y 1 − =− , ∂u ∂v ∂v ∂u 2 and therefore, 1 fU,V (u , v) = fX,Y 2 4 u+v u−v , 2 2 . If X = N (0 , σ 2 ) and Y = N (0 , τ 2 ), then 2 2 y 1 x − . fX,Y (x , y) = exp − 2πστ 2σ 2 2τ 2 And hence, fU,V (u , v) = 1 (u − v)2 (u + v)2 − . exp − 4πστ 8σ 2 8τ 2 If τ 6= σ, then this cannot be written as a function of u times a function of v. Therefore, U and V are not independent. Else, if τ = σ, then 1 1 2 2 fU,V (u , v) = exp − 2 (u + v) + (u − v) 4πσ 2 8σ 1 −u2 /4σ2 −v2 /4σ2 e e . = 4πσ 2 This has the correct product form, so U and V are independent in this case. 5