5. Let Y = µ + σX, then. MY (t) = Mµ+σX(t). = eµtMX(σt). = eµte σ2t2. 2. = eµt+σ2t2.
2. , which is same as the formula provided in the Casella and Berger.
1
STAT542 HW6 SOLUTION 3.17 EX
∞
Z
1 1 α−1 −x/β x = x e dx = Γ(α)β α Γ(α)β α 0 ν+α ν Γ(ν + α)β β Γ(ν + α) = = Γ(α)β α Γ(α)
ν
ν
Z
∞
x(ν+α)−1 e−x/β dx
0
Note that this formula is valid for all ν > −α. The expectation does not exist for ν ≤ −α.
3.20 a. EX = = EX 2 = V arX =
∞
Z ∞ 2 −x2 /2 2 x2 2 √ e−x /2 d( ) dx = x√ e 2 2π 2π 0 0 r x2 2 2 2 √ (−e− 2 )|∞ [0 − (−1)] = 0 = √ π 2π Z ∞ Z 2π ∞ −x2 −x2 1 2 x2 √ e 2 dx = 1 (X ∼ Normal(0, 1)) x2 √ e 2 dx = 2π 2π 0 0 2 2 EX 2 − (EX)2 = 1 − ( √ )2 = 1 − π 2π Z
√ b. Let Y = g(X) = X 2 . Thus X = g −1 (Y ) = Y ,where X, Y ≥ 0. d −1 2 1 1 −1 fY (y) = fX (g (y)) g (y) = √ e−y/2 · √ = √ y −1/2 e−y/2 = Γ(1/2, 2) dy 2 y 2π 2π where α = 1/2, β = 2.
3.24 a. fX (x) = β1 e−x/β , x > 0. For Y = X 1/γ ,fY (y) = βγ e−y R∞ R ∞ γ −yγ /β γ−1 γ/β f (y)dy = y dy = −e−y |∞ e Y 0 = 1. 0 0 β
γ /β
y γ−1 , y > 0.
Using the transformation z = y γ /β, we calculate γ EY = β n
Z
∞
y 0
γ+n+1 −y γ /β
e
dy = β
n/γ
Z
∞
z n/γ e−z dz = β n/γ Γ(
0
Thus EY = β 1/γ Γ( γ1 + 1) and V arY = β 2/γ [Γ( γ2 + 1) − Γ2 ( γ1 + 1)].
n + 1) γ
Make the transformation Y = 1/X to get fY (y) = R∞ −∞
fY (y)dy =
1 1 ( 1 )a+1 e− by dy Γ(a)ba y
R∞ 0
2
1 xa+1 e−x/b . Γ(a)ba
b. The gamma(a,b) density is fX (x) =
=
R∞ 0
1 1 ( 1 )a+1 e− by . Γ(a)ba y
1 xa+1 e−x/b dx Γ(a)ba
= 1.
The first two moments are Z ∞ 1 a − by1 Γ(a − 1)ba−1 a 1 ( ) e dy = = EY = a a Γ(a)b 0 y Γ(a)b (a − 1)b Z ∞ a−2 1 1 Γ(a − 2)b 1 1 EY 2 = ( )a−1 e− by dy = = a a Γ(a)b 0 y Γ(a)b (a − 1)(a − 2)b2 1 V arY = 2 (a − 1) (a − 2)b2 c. fX (x) = e−x , x > 0. For Y = α − γlogX, fY (y) = e−e R∞
f (y)dy = −∞ Y
R∞ 0
e−e
α−y γ
e
α−y γ
α−y γ
1 dy γ
e
=
α−y γ
R∞ 0
1 , −∞ γ
< y < ∞.
e−x dx = 1.
Calculation of EY and EY 2 cannot be done in closed form. If we define Z ∞ logxe−x dx I1 = Z0 ∞ I2 = (logx)2 e−x dx 0
then EY = E(α − γlogX) = α − γI1 , and EY 2 = E(α − γlogX)2 = α2 − 2αγI1 + γ 2 I2 , then V arY = γ 2 (I2 − I12 ). The constant I1 = .5772 is called Euler’s constant.
3.26 a. fT (t) = β1 e−t/β , t ≥ 0,and FT (t) =
Rt
1 −x/β e dx 0 β
= −e−x/β |t0 = 1 − e−t/β .Thus,
fT (t) (1/β)e−t/β 1 hT (t) = = = 1 − FT (t) 1 − (1 − e−t/β ) β γ
b. fT (t) = βγ tγ−1 e−t /β , t ≥ 0, Rt R tγ/β γ γ/β γ and FT (t) = 0 βγ xγ−1 e−x /β dx = 0 e−u du = −e−u |t0 = 1 − e−t /β ,where u = xγ/β . Thus, γ /β
fT (t) (γ/β)tγ−1 e−t hT (t) = = 1 − FT (t) e−tγ /β c. FT (t) = Thus,
1 1+e−(t−µ)/β
and fT (t) =
1 −(t−µ)/β e β (1+e−(t−µ)/β )2
1
=
γ γ−1 t . β
.
e−(t−µ)/β
fT (t) 1 β (1+e−(t−µ)/β )2 hT (t) = = = FT (t). 1 1 − FT (t) β 1 − 1+e−(t−µ)/β
3
4.1
Since the distribution is uniform, the easiest way to calculate these probailities is as the ratio of areas, the total areal being 4. a. The circle x2 + y 2 ≤ 1 has area π, so P (X 2 + Y 2 ≤ 1) = π4 . b. The area below the line y=2x is half of the area of the square, so P (2X − Y > 0) = 12 . c. Clearly P (|X + Y | < 2) = 1.
4.5 a. P (X >
√
Y)=
R1R1
√
0
b. P (X 2 < Y < X) =
y
(x + y)dxdy =
R 1 R √y 0
y
2xdxdy =
7 20 1 6
4.40 a. Z
Z Z
1
1−x
Z
f (x, y)dxdy = R2
0
Z
0 1
= C Z0 1 = C
Cxa−1 y b−1 (1 − x − y)c−1 dydx Z 1−x y b−1 y c−1 y a−1 b+c−1 x (1 − x) ( ) (1 − ) d( )dx 1−x 1−x 1−x 0 xa−1 (1 − x)b+c−1 B(b, c)dx
0
= C B(b, c) B(a, b + c) ≡ 1. Hence C =
Γ(a+b+c) . Γ(a)Γ(b)Γ(c)
b. Z
Z
1−x
Cxa−1 y b−1 (1 − x − y)c−1 dy 0 Z 1−x a−1 = Cx y b−1 (1 − x − y)c−1 dy
f (x, y)dy = R
0
= Cxa−1 (1 − x)b+c−1 B(b, c) = B(a, b + c)xa−1 (1 − x)b+c−1 , 0 ≤ x ≤ 1. Hence X is Beta(a, b + c). Similarly, Y is Beta(b, a + c). Z Z 1−y f (x, y)dx = Cxa−1 y b−1 (1 − x − y)c−1 dx R 0 Z 1−y b−1 = Cy xa−1 (1 − x − y)c−1 dy 0
= Cy b−1 (1 − y)a+c−1 B(a, c) = B(b, a + c)y b−1 (1 − y)a+c−1 , 0 ≤ y ≤ 1.
y b−1 y c−1 1 1 = B(b,c) ( 1−x ) (1 − 1−x ) 1−x , 0 ≤ y ≤ 1. Let Z = c. fY |X=x (y) = ffX(x,y) (x) 1 b−1 fZ (z) = fY |X=x ((1 − x)z)|1 − x| = B(b,c) z (1 − z)c−1 , 0 ≤ z ≤ 1. Hence Z is Beta(b,c).
Y , 1−x
d. Z Z EXY
=
xyf (x, y)dxdy Z
R2 1 Z 1−x
xyCxa−1 y b−1 (1 − x − y)c−1 dydx 0 Z 10 Z 1−x xa y b (1 − x − y)c−1 dydx = C 0 Z0 1 = C xa (1 − x)b+c B(b + 1, c)dx
=
0
= CB(b + 1, c)B(a + 1, b + c + 1) ab = . (a + b + c + 1)(a + b + c) COV (X, Y ) = EXY − EXEY ab a b = − (a + b + c + 1)(a + b + c) a + b + c a + b + c ab = − (a + b + c)2 (a + b + c + 1)
4.43 COV (X1 + X2 , X2 + X3 ) = = COV (X1 + X2 , X1 − X2 ) = = =
E(X1 + X2 )(X2 + X3 ) − E(X1 + X2 )E(X2 + X3 ) (4µ2 + σ 2 ) − 4µ2 = σ 2 E(X1 + X2 )(X1 − X2 ) − E(X1 + X2 )E(X1 − X2 ) EX22 − EX12 0
Additional Problem Let X be standard normal N(0,1), Z
∞
x2 1 etX √ e− 2 dx 2π Z∞∞ (x−t)2 t2 1 √ e− 2 + 2 dx = 2π −∞ Z ∞ 2 (x−t)2 t 1 √ e− 2 dx = e2 2π −∞
MX (t) =
t2
= e2.
4
5
Let Y = µ + σX, then MY (t) = Mµ+σX (t) = eµt MX (σt) = eµt e
σ 2 t2 2
= eµt+
σ 2 t2 2
,
which is same as the formula provided in the Casella and Berger.
