Solutions for Homework 5 Part II

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By the “memoryless” property (on page 101 of Casella & Berger) of the exponential distribution: P(X>s|X>t) = P(X>s − t),. P(X − 4 ≤ x|X ≥ 4) = P(X ≤ 4 + x|X ...
Autumn 08 Stat 620

Solutions for Homework 5 Part II 4.17 (a) P (Y = i + 1) =

R i+1 i

e−x dx = (−e−x )|i+1 = e−i (1 − e−1 ), which is geometric with p = 1 − e−1 . i

(b) Since Y = i + 1 if and only if i ≤ X < i + 1, Y ≥ 5 ⇔ X ≥ 4.

P (X − 4 ≤ x|Y ≥ 5) = P (X − 4 ≤ x|X ≥ 4)

By the “memoryless” property (on page 101 of Casella & Berger) of the exponential distribution: P (X > s|X > t) = P (X > s − t),

P (X − 4 ≤ x|X ≥ 4) = P (X ≤ 4 + x|X ≥ 4) = 1 − P (X > 4 + x|X ≥ 4) = 1 − P (X > 4 + x − 4) = 1 − P (X > x) = P (X ≤ x) = 1 − e−x , x ∈ {0, 1, 2, . . .}

4.20 (a) This transformation is not one-to-one because you cannot determine the sign of X2 from Y1 0 = {−∞ < x < ∞, x = 0}, S 1 = and Y2 . So partition the support of (X1 , X2 ) into SX 1 2 X 2 = {−∞ < x < ∞, x < 0}. The support of (Y , Y ) is {−∞ < x1 < ∞, x2 > 0} and SX 1 2 1 2 0 } = 0. The inverse SY = {0 < y1 < ∞, −1 < y2 < 1}. It is easy to see that P {X ∈ SX p 1 is x = y √y and x = transformation from SY to SX y1 − y1 y22 with Jacobian 1 2 1 2

J1 (y1 , y2 ) =

1 √y2 2 y1 √ 2 1−y2 1 √ 2 y1

√ y2 √

y1



y1

1−y22

1 = p 2 1 − y22

p 2 is x = y √y and x = − y − y y 2 with J = −J . The inverse transformation from SY to SX 1 2 1 1 2 2 1 1 2 Then 1 −y1 /(2σ2 ) 1 1 −y1 /(2σ2 ) −1 1 −y1 /(2σ2 ) 1 p p p fY1 ,Y2 (y1 , y2 ) = e + e e = 2 2 2 2 2 2πσ 2πσ 2πσ 2 1 − y2 2 1 − y2 1 − y22 (b) We can see in the above expression that the joint pdf factors into a function of y1 and a function of y2 . So Y1 and Y2 are independent. Y1 is the square of the distance from (X1 , X2 ) to the origin. Y2 is the cosine of the angle between the positive x1 -axis and the line from (X1 , X2 ) to the origin. So 1

Autumn 08 Stat 620 independence says that the distance from the origin is independent of the orientation (as measured by the angle). 4.23 (a) X ∼ beta(α, β), Y ∼ beta(α + β, γ). X and Y are independent. The inverse transformation is y = v, x = u/v with Jacobian J(u, v) = fU,V (u, v) =

∂x ∂u

∂x ∂v

∂y ∂u

∂y ∂v

1 v = 0

−u 2 1 v = 1 v

Γ(α + β) Γ(α + β + γ) u α−1 u 1 ( ) (1 − )β−1 v α+β−1 (1 − v)γ−1 , 0 < u < v < 1 Γ(α)Γ(β) Γ(α + β)Γ(γ) v v v

So

fU (u) = = = =

Z Γ(α + β + γ) α−1 1 β−1 v − u β−1 u ) dv v (1 − v)γ−1 ( Γ(α)Γ(β)Γ(γ) v u   Z 1 dv Γ(α + β + γ) α−1 v−u β+γ−1 β−1 γ−1 u (1 − u) , dy = y (1 − y) dy y = Γ(α)Γ(β)Γ(γ) 1−u 1−u 0 Γ(β)Γ(γ) Γ(α + β + γ) α−1 u (1 − u)β+γ−1 Γ(α)Γ(β)Γ(γ) Γ(β + γ) Γ(α + β + γ) α−1 u (1 − u)β+γ−1 , 0 < u < 1. Γ(α)Γ(β + γ)

Thus, U ∼ beta(α, β + γ). 4.24 Let z1 = x + y, z2 =

x x+y ,

then x = z1 z2 , y = z1 (1 − z2 ) and

|J(z1 , z2 )| =

∂x ∂z1

∂x ∂z2

∂y ∂z1

∂y ∂z2

z2 z 1 = = z1 1 − z −z 2 1

The support SXY = {x > 0, y > 0} is mapped onto the set SZ = {z1 > 0, 0 < z2 < 1}. So 1 1 (z1 z2 )r−1 e−z1 z2 · (z1 − z1 z2 )s−1 e−z1 +z1 z2 · z1 Γ(r) Γ(s) Γ(r + s) r−1 1 = z r+s−1 e−z1 · z (1 − z2 )s−1 , Γ(r + s) 1 Γ(r)Γ(s) 2

fZ1 ,Z2 (z1 , z2 ) =

2

0 < z1 < +∞, 0 < z2 < 1.

Autumn 08 Stat 620 Z1 and Z2 are independent because fZ1 ,Z2 (z1 , z2 ) can be factored into two terms, one depending only on z1 and the other depending only on z2 . Inspection of these kernals shows Z1 ∼ gamma(r + s, 1), Z2 ∼ beta(r, s). 4.26 (a) The only way to attack this problem is to find the joint cdf of (Z, W ). Now W = 0 or 1. For 0 ≤ w < 1,

P (Z ≤ z, W ≤ w) = P (Z ≤ z, W = 0) = P (min(X, Y ) ≤ z, Y ≤ X) = P (Y ≤ z, Y ≤ X) Z zZ ∞ 1 −x/λ 1 −y/µ = e e dxdy λ µ 0 y   λ 1 1 = 1 − exp{−( + )z} λ+µ µ λ and

P (Z ≤ z, W = 1) = P (min(X, Y ) ≤ z, X ≤ Y ) µ = P (X ≤ z, X ≤ Y ) = λ+µ



1 1 1 − exp{−( + )z} µ λ



can be used to give the joint cdf. (b) Z

∞Z ∞

P (W = 0) = P (Y ≤ X) = 0

y

λ 1 −x/λ 1 −y/µ e e dxdy = λ µ µ+λ

P (W = 1) = 1 − P (W = 0) =

µ µ+λ

P (Z ≤ z) = P (Z ≤ z, W = 0) + P (Z ≤ z, W = 1) = 1 − exp{−(

1 1 + )z} µ λ

Therefore, P (Z ≤ z, W = i) = P (Z ≤ z)P (W = i), for i = 0, 1, z > 0 which implies that Z and W are independent. 4.36 (a) EY =

n X i=1

EXi =

n X

E(E(Xi |Pi )) =

i=1

n X i=1

3

E(Pi ) =

n X i=1

α nα = α+β α+β

Autumn 08 Stat 620 (b) Since the trials are independent,

VY =

n X

VXi =

i=1

=

n X

n X

[V(E(Xi |Pi )) + E(V(Xi |Pi ))] =

i=1

=

[V(Pi ) + E(Pi (1 − Pi ))]

i=1

[V(Pi ) − E((Pi )2 ) + E(Pi )] =

i=1 n X

n X

n X

[−(E(Pi ))2 + E(Pi )]

i=1

[−(

i=1

α 2 α ) + ]= α+β α+β

n X i=1

αβ nαβ = (α + β)2 (α + β)2

α (c) We show that Xi ∼ Bernoulli α+β . Then, because Xi0 s are independent, Y =

Pn

i=1 Xi



α ). Let fPi (p) denote the beta(α, β) density for the Pi s and conditioning on Pi gives Binomial(n, α+β 1

Z 1 Γ(α + β) α−1 P (Xi = x|Pi = p)fPi (p) dp = px (1 − p)1−x p (1 − p)β−1 Γ(α + β) 0 0 Z Γ(α + β) 1 α+x−1 = p (1 − p)β−x+1−1 dp Γ(α)Γ(β) 0    β Γ(α + β) Γ(α + x)Γ(β − x + 1)  α+β x = 0 = =  Γ(α)Γ(β) Γ(α + β + 1)   α x=1 α+β Z

P (Xi = x) =

4.47 Because the sample space can be decomposed into the subsets [XY > 0] and [XY < 0] (and a set of (X, Y ) probability zero), we have for z < 0, by definition of Z,

P (Z ≤ z) = P (X ≤ z and XY > 0) + P (−X ≤ z and XY < 0) = P (X ≤ z and Y < 0) + P (X ≥ −z and Y < 0)

(because z < 0)

= P (X ≤ z)P (Y < 0) + P (X ≥ −z)P (Y < 0)

(because X and Y are independent)

= P (X ≤ z)P (Y < 0) + P (X ≤ z)P (Y < 0)

(symmetry of the distributions of X and Y )

= P (X ≤ z)

By a similar argument, for z > 0, we get P (Z > z) = P (X > z), and hence, Z ∼ X ∼ N (0, 1). (b) By definition of Z, Z > 0 ⇔ either (i) X < 0 and Y > 0 or (ii) X > 0 and Y > 0. So Z and Y always have the same sign, hence they cannot be bivariate normal. 4

Autumn 08 Stat 620 4.58 These parts are exercises in the use of conditioning. (a)

Cov(X, Y ) = E(XY ) − E(X)E(Y ) = E[E(XY |X)] − E(X)E[E(Y |X)] = E[XE(Y |X)] − E(X)E[E(Y |X)] = Cov(X, E(Y |X))

(b) Cov(X, Y − E(Y |X)) = Cov(X, Y ) − Cov(X, E(Y |X)) = 0

by (a)

So, X and Y − E(Y |X) are uncorrelated. (c)

V ar(Y − E {Y |X}) = V ar(E {Y − E {Y |X} |X}) + E {V ar [Y − E {Y |X} |X]} = V ar (E {Y |X} − E {Y |X}) + E {V ar (Y − E {Y |X} |X)} = V ar(0) + E {V ar(Y |X)} = E {V ar(Y |X)}

5

E {Y |X} is constant wrt the [Y |X] distribution