Hw 6 (due Nov 15)
4.19
[Sol]
a. Since X1 and X2 are independent N (0, 1) random
variables, it follows that (X1 − X2 ) is
√
N (0, 2) random variable. Let Y = (X1 − X2 )/ 2, Y is a N (0, 1) random variable. Thus,
1
1/2−1 −y/2
Y 2 = (X1 − X2 )2 /2 ∼ χ21 and the density function of Y is fY (y) = Γ(1/2)2
e
where
1/2 y
0 < y < ∞.
1
b. Let U = φ1 (X1 , X2 ) = X1X+X
and V = φ2 (X1 , X2 ) = X1 + X2 . Thus, X1 = φ−1
1 (U, V ) =
2
−1
1
U V , X2 = φ2 (U, V ) = V (1 − U ) and | J |= V . So, fU,V (u, v) = Γ(α1 )Γ(α2 ) (uv)α1 −1 (v(1 −
Γ(α1 +α2 ) α1 −1
u))α2 −1 e−v v = Γ(α
u
(1 − u)α2 −1 Γ(α11+α2 ) v α1 +α2 −1 e−v = f (u)f (v). This implies that
1 )Γ(α2 )
X1
2
1
∼ Beta(α1 , α2 ). In addition, X1X+X
= 1 − X1X+X
∼ Beta(α2 , α1 ).
X1 +X2
2
2
4.20
[Sol]
a. Since the transformation is not one-to-one, we partition the original support of X1 and
X2 into proper unions of sets, say R2 = A0 ∪ A1 ∪ A2 , where A0 = {(x1 , x2 ) : x1 ∈ R, x2 = 0},
A1 = {(x1 , x2 ) : x1 ∈ R, x2 > 0} and A2 = {(x1 , x2 ) : x1 ∈ R, x2 < 0}. Then conditional on
each Ai , this transformation is one-to-one.
When (X1 , X2 ) ∈ A1 , Y1 = φ11 (X1 , X2 ) = X12 + X22 and Y2 = φ21 (X1 , X2 ) = √ X21 2 . Thus,
X1 +X2
�
√
2 )−1/2
(1−y
2
X1 = φ−1
Y1 Y2 , X2 = φ−1
Y1 (1 − Y22 ) and | J1 |=
. There11 (Y1 , Y2 ) =
21 (Y1 , Y2 ) =
2
y1
− 2 1
−1
−1
1
2
−1/2
fore, f (φ11 (Y1 , Y2 ), φ12 (Y1 , Y2 )) | J1 |= 2πσ2 e 2σ 2 (1 − y2 )
.
When (X1 , X2 ) ∈ A2 , Y1 = φ12 (X1 , X2 ) = X12 + X22 and Y2 = φ22 (X1 , X2 ) = √ X21 2 .
X1 +X2
�
√
2 −1/2
(1−y
−1
−1
2)
Thus, X1 = φ12 (Y1 , Y2 ) = Y1 Y2 , X2 = φ22 (Y1 , Y2 ) = − Y1 (1 − Y22 ) and | J2 |=
.
2
y
− 12 1
−1
−1
1
2
−1/2
Therefore, f (φ12 (Y1 , Y2 ), φ22 (Y1 , Y2 )) | J2 |= 2πσ2 e 2σ 2 (1 − y2 )
.
�
The distribution of Y1 and Y2 is f (y1 , y2 ) is f (y1 , y2 ) = 2i=1 f (φ1i (X1 , X2 ), φ2i (X1 , X2 )) |
y1
Ji |= 2σ1 2 e− 2σ2 π1 (1 − y22 )−1/2 = f (y1 )f (y2 ) for 0 < y1 < ∞ and −1 < y2 < 1.
b. Since f (y1 , y2 ) = f (y1 )f (y2 ) for all y1 and y2 , Y1 and Y2 are independent random
variables. For any points (X1 , X2 ) ∈ R2 , it can be written in terms of a polar coordinate
system, (r cos θ, r sin θ). In this case, Y1 is the squared distance between points (X1 , X2 ) and
the origin, namely r2 , whereas Y2 is cos θ.
4.21
[Sol]
18
Denote random variable T = R2 ∼ χ22 , then by independence assumption, the joint density
1
of (T, θ) can be written as f (t, θ) = 12 e−t/2 2π
where 0 < t < ∞ and 0 < θ < 2π. Define an
√
√
one-to-one transformation as follow: X = φ1 (T, θ) = T cos θ and Y = φ2 (T, θ) = T sin θ.
1 −(x2 +y 2 )/2
Then T = X 2 + Y 2 and θ = tan−1 (Y /X) and | J |= 2. Thus, f (x, y) = 4π
e
2=
2 /2 1
2 /2
1
−x
−y
√ e
√ e
, 0 < x, y < ∞.
2π
2π
4.22
[Sol]
U −b
Let U = φ1 (X, Y ) = aX + b and V = φ2 (X, Y ) = cY + d. Then X = φ−1
,
1 (U, V ) =
a
−1
−1
−1
V −d
1
Y =� φ2 (U, V� ) = c and | J |= ac . Thus, fU,V (u, v) = fX,Y (φ1 (U, V ), φ2 (U, V )) | J |=
1
f u−b
, v−d
.
ac
a
c
4.23
[Sol]
a. Let U = XY and V = Y . Then X = U/V , Y = V and | det(J) |= 1/V . Since
Γ(α+β) Γ(α+β+γ)
X and Y are independent, fU,V (u, v) = fX,Y (x, y) | J |= Γ(α)Γ(β)
(u/v)α−1 (1 −
Γ(α+β)Γ(γ)
u/v)β−1 v α+β−1 (1 − v)γ−1 (1/v) =
Γ(α+β+γ)
uα−1 v β−1 (1
Γ(α)Γ(β)Γ(γ)
�
− v)γ−1 (1 − u/v)β−1 where 0 <
�1
Γ(α+β+γ)
fU,V (u, v)dv = Γ(α)Γ(β)Γ(γ)
uα−1 u v β−1 (1 −
v < 1 and 0 < u < v. Thus, fU (u) =
�1
Γ(α+β+γ)
Γ(α+β+γ)
v)γ−1 (1 − u/v)β−1 dv = Γ(α)Γ(β)Γ(γ)
uα−1 u (1 − v)γ−1 (v − u)β−1 dv = Γ(α)Γ(β)Γ(γ)
uα−1 (1 −
�
1 1−v γ−1 v−u β−1
1−v
1
u)β+γ−1 u ( 1−u
) ( 1−u ) dv. Let y = v−u
. Then 1 − y = 1−u
and dy = dv 1−u
where
1−u
0 < y < 1.
�1
Γ(α+β+γ)
Γ(α+β+γ)
fU (u) = Γ(α)Γ(β)Γ(γ)
uα−1 (1−u)β+γ−1 0 (1−y)γ−1 y β−1 dy = Γ(α)Γ(β)Γ(γ)
uα−1 (1−u)β+γ−1 Γ(β)Γ(γ)
=
Γ(β+γ)
Γ(α+β+γ) α−1
u (1
Γ(α)Γ(β+γ)
− u)β+γ−1 where 0 < u < 1. Therefore, U ∼ Beta(α, β + γ)
4.27
[Sol]
Let U = X + Y and V = X − Y . Then X =
U +V
2
U −V
2�
and | det(J) |= −1/2. Since X
�2
� u+v 2 u−v 2 �
(
−µ) +( 2 −γ)
U −V
1
√ 1
and Y are independent, fU,V = fX,Y ( U +V
,
)
|
det(J)
|=
exp
− 2
=
2
2
2
2σ 2
2
2πσ
�
�
�
�
2
2
√ 1
√ 1
exp − (u−(µ+γ))
exp − (v−(µ−γ))
= fU (u)fV (v). Therefore, U and V are in2·2σ 2
2·2σ 2
2π2σ 2
2π2σ 2
,Y =
dependent, U ∼ N (µ + γ, 2σ 2 ) and V ∼ N (µ − γ, 2σ 2 ).
4.30
[Sol] Y | X = x ∼ N (x, x2 ) and X ∼ U (0, 1).
a. E(Y ) = E(E(Y | X)) = E(X) = 1/2, V ar(Y ) = E(V ar(Y | X)) + V ar(E(Y |
X)) = E(X 2 ) + V ar(X) = V ar(X) + E(X)2 + V ar(X) = 1/12 + 1/4 + 1/1/2 = 5/12,
Cov(X, Y ) = E(XY ) − E(X)E(Y ) = E(E(XY | X)) − 1/2 · 1/2 = E(XE(Y | X)) − 1/4 =
19
E(X 2 ) − 1/4 = V ar(X) + E(X)2 − 1/4 = 1/12.
b. Let U = Y /X and V = X. Then X = V , Y = U V and | det(J) |= V where
2
2
1
0 < V < 1 and −∞ < U < ∞. Since fX,Y (x, y) = fY |X fX = √2πx
e−(y−x) /(2x ) ,
2
2
fU,V (u, v) = fX,Y (v, uv)v = √12π e−(u−1) /2 . Since fU,V (u, v) can be factored into a function of u and a function of v, thus, U and V are independent.
4.31
[Sol] Y | X ∼ Bin(n, X) and X ∼ U (0, 1).
a. E(Y ) = E(E(Y | X)) = E(nX) = n/2, V ar(Y ) = E(V ar(Y | X)) + V ar(E(Y | X)) =
E(nX(1 − X)) + V ar(nX) = n(E(X) − E(X 2 )) + n2 /12 = n(1/2 − (1/12 + 1/4)) + n2 /12 =
n2 /12 + n/6.
� �
b. fX,Y (x, y) = fY |X (y | x)fX (x) = ny xy (1 − x)n−y where y = 0, 1, . . . , n and 0 ≤ x ≤ 1.
� �
�1
�1� �
c. fY (y) = 0 fX,Y (x, y)dx = 0 ny xy (1−x)n−y dx = ny Γ(y+1)Γ(n−y+1)
where y = 0, 1, . . . , n.
Γ(n+2)
4.34
[Sol]
� 1 1−p Γ(α+β) α−1
Γ(α+β) � 1 α−2
β−1
b. E(X) = E(E(X | P )) = E( r(1−p)
)
=
r
p
(1−p)
dp
=
r
p (1−
p
Γ(α)Γ(β) 0
0 p Γ(α)Γ(β)
Γ(α+β) Γ(α−1)Γ(β+1) � 1
Γ(α+β)
rβ
β
α−1−1
β+1−1
p) dp = r Γ(α)Γ(β) Γ(α+β)
p
(1 − p)
dp = α−1 .
0 Γ(α−1)Γ(β+1)
4.39
[Sol]
�
a. Define a set B1 = {xk : k �= i, j : k�=i,j xk = m − xi − xj }. Then the joint pmf of (Xi , Xj )
20
is given by
f (xi , xj ) =
�
{xk ∈B,k�=i,j}
=
px1 1
=
m!
x
px1 1 · · · pxi i · · · pj j · · · pxnn
x1 ! · · · xi ! · · · xj ! · · · xn !
1 xi xj
p p
xi !xj ! i j
�
m!
x1 ! · · · xi−1 !xi+1 ! · · · xj−1 !xj+1 ! · · · · · · xn !
{xk ∈B,k�=i,j}
xj−1 xj+1
xi−1 xi+1
· · · pi−1 pi+1 · · · pj−1
pj+1
· · · pxnn
m!
x
pxi p j (1 − pi − pj )m−xi −xj
xi !xj !(m − xi − xj )! i j
�
m!
x1 ! · · · xi−1 !xi+1 ! · · · xj−1 !xj+1 ! · · · · · · xn !
�
�x1
�
�xi−1 �
�xi+1
�
�xj−1
p1
pi−1
pi+1
pj−1
···
···
1 − pi − pj
1 − pi − pj
1 − pi − pj
1 − pi − pj
�xj+1
�
�xn
�
pj+1
pn
···
1 − pi − pj
1 − pi − pj
m!
x
=
pxi p j (1 − pi − pj )m−xi −xj .
xi !xj !(m − xi − xj )! i j
{xk ∈B,k�=i,j}
Thus, the joint distribution of (Xi , Xj ) ∼ multinomial(m, pi , pj ). Note that the fourth
equality holds by Theorem 4.6.4. By using the same technique, we can find marginally xj
will follow a binomial(m, pj ) (see class notes).
The conditional distribution of Xi | Xj is given as
f (xi | xj ) =
=
x
m!
pxi p j (1 − pi − pj )m−xi −xj
xi !xj !(m−xi −xj )! i j
x
m!
p j (1 − pj )m−xj
xj !(m−xj )! j
�
��
�xi �
�m−xi −xj
m − xj
pi
pi
xi
1 − pj
1−
1 − pj
pi
Thus, the conditional distribution of Xi | Xj ∼ Binomial(m − xj , 1−p
).
j
�
X1
m!
t1 X1 +···+tn xn
t1 X1 +···+tn xn
Xn
b. MX (t) = E(e
) = X∈X e
p · · · pn
X1 !···Xn ! 1
�
m!
t1
X1
tn
Xn
t1
=
· · · (e pn )
= (p1 e + · · · + pn etn )m . Thus, Cov(Xi , Xj ) =
X∈X X1 !···Xn ! (e p1 )
E(Xi Xj ) − E(Xi )E(Xj ) =
∂ 2 MX (t)
|t=0
∂ti ∂tj
−m2 pi pj = m(m − 1)pi pj − m2 pi pj = −mpi pj .
4.42
[Sol] E(XY · Y ) = E(XY 2 ) = E(X)E(Y 2 ) = µX (µ2Y + σ 2Y ), Cov(XY, Y ) = E(XY 2 ) −
E(XY )E(Y ) = µX µ2Y +µX σ 2Y −µX µY µY = µX σ 2Y and V ar(XY ) = E((XY )2 )−(E(XY ))2 =
21
E(X 2 )E(Y 2 ) − (E(X)E(Y ))2 = (σ 2X + µ2X )(σ 2Y + µ2Y ) − µ2X µ2Y . Then Corr(XY, Y ) =
2
√ µX σ Y 2 .
V ar(XY )σ Y
4.43
[Sol] E(Xi ) = µ, V ar(Xi ) = σ 2 and Cov(Xi , Xj ) = 0 for i �= j. Then i) Cov(X1 + X2 , X2 +
X3 ) = E((X1 + X2 )(X2 + X3 )) − E(X1 + X2 )E(X2 + X3 ) = E(X1 X2 ) − E(X1 )E(X2 ) +
E(X1 X3 ) − E(X1 )E(X3 ) + E(X2 X3 ) − E(X2 )E(X3 ) + E(X22 ) − E(X2 )2 = σ 2 , and ii)
Cov(X1 + X2 , X1 − X2 ) = E((X1 + X2 )(X1 − X2 )) − E(X1 + X2 )E(X1 − X2 ) = 0.
4.45
[Sol] see the bivariate normal distribution handout
4.47
[Sol]
a. By definition of Z, for z < 0,
P (Z ≤ z) = P (X ≤ z, XY > 0) + P (−X ≤ z, XY < 0) = P (X ≤ z, Y < 0) + P (X ≥ −z, Y < 0)
= P (X ≤ z)P (Y < 0) + P (X ≥ −z)P (Y < 0) = P (X ≤ z)P (Y < 0) + P (X ≤ z)P (Y < 0)
= P (X ≤ z)(P (Y < 0) + P (Y < 0)) = P (X ≤ z)
Note that the fourth equality holds due to symmetry of X.
By a similar argument, for z > 0, we get P (Z > z) = P (X > z), and hence, P (Z ≤ z) =
P (X ≤ z). Thus, Z ∼ n(0, 1).
b. By definition of Z, Z > 0 is equivalent to 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.
4.53
[Sol] The equation Ax2 + Bx + C has real root if and only if the discriminant B 2 ≥ 4AC
or equivalently −2 log B ≤ − log 4 − log A − log C. Let X = −2 log B ∼ exponential(2),
Y = − log A − log C ∼ gamma(2, 1), the probability of having real root is P (X ≤ − log 4 +
� ∞ � y−log 4 1 −x/2 −y
�∞
y−log 4
e
ye dxdy = log 4 (1 − e− 2 )ye−y dy =
Y ) = P (log 4 ≤ Y − X) = log 4 0
2
( log2 2 + 1/4) − 2( log6 2 + 1/18) = log6 2 + 5/36.
4.54
[Sol] Since Xi ∼ unif orm(0,
�1) independently, we have − log Xi ∼ Gamma(1, 1), independently. Therefore, T2 = ni=1 − log Xi ∼ Gamma(n, 1). Consider the transformation
T = φ(T2 ). Then T2 = φ−1 (T ) = − log T and | det(J) |= 1/T . The density function of T is
1
1
given by f (t) = Γ(n)
(− log t)n−1 elog t 1t = Γ(n)
(− log t)n−1 , 0 < t < 1.
22
4.63
[Sol] Since g(X) = − log X is a concave function, by Jensen’s inequality, 0 = E(Z) =
E(g(X)) > g(E(X)) = − log E(X). BY taking exponential on both side, we have e0 = 1 <
E(X).
23