MATH 5010–001 SUMMER 2003
SOLUTIONS TO ASSIGNMENT 9
Problems from pp. 290–295:
51. Let r = u1 (x, y) = (x2 + y 2 )1/2 and θ = u2 (x, y) = tan−1 (y/x), and note that the
Jacobian of this transformation is
!
√ x
√ y
2
2
2
2
x +y
x +y
J = det
1
1
1
− xy2 1+(y/x)
2
x 1+(y/x)2
!
√ 2x 2 √ 2y 2
x +y
x +y
= det
y
x
− x2 +y
2
x2 +y 2
=p
So,
1
x2 + y 2
.
p
r
fR,Θ (r, θ) = fX,Y (x, y) x2 + y 2 = ,
π
0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π.
52. We use the above to get the Jacobian. So
p
fR,Θ (r, θ) = fX,Y (x, y) x2 + y 2 = r,
0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π.
√
√
53. Let x = g1 (u, z) = 2z cos u and y = g2 (u, z) = 2z sin u. Then, the Jacobian of
this transformation is,
!
√
− 2z sin u √12z cos u
J = det √
= −1.
√1 sin u
2z cos u
2z
Therefore,
fX,Y (x, y) = fU,Z (u, z)
1
= fU,Z (u, z).
|J|
Now, by independence,
fU,Z (u, z) = fU (u)fZ (z) =
1 −z
e ,
2π
0 ≤ u ≤ 1, z ≥ 0.
Therefore,
1 −z
e ,
0 ≤ u ≤ 1, z ≥ 0.
2π
It remains to solve for (x, y). Note that x2 + y 2 = 2z, so z = 12 (x2 + y 2 ). Also, (y/x) =
tan(u). Because we are interested in 0 ≤ u ≤ 1, the latter yields u = tan−1 (y/x).
Plug to find,
2
2
1 −(x2 +y2 )/2 e−x /2
e−y /2
fX,Y (x, y) =
e
= √
× √
.
2π
2π
2π
fX,Y (x, y) =
1
All values of x are allowed.
54. (a) Check that the Jacobian of the transformation from (x, y) to (u, v) is J = −2x/y.
Therefore,
1
1
fU,V (u, v) = fX,Y (x, y)
= 3 ,
x, y ≥ 1.
|J|
2x y
√
Nowpwe solve for (u, v). Note that uv = x2 , and u/v = y 2 . So, x = uv and
y = u/v, and we get
fU,V (u, v) =
1
,
2u2 v
u, v ≥ 0, uv ≥ 1, u/v ≥ 1.
The conditions on u and v are: v ≥ (1/u) and v ≤ u. Therefore u must be greater
than one, and v is between 1/u and u.
54 (b) Integrate [du] to get fV and integrate [dv] to get fU , viz., you need to draw a
picture of the region to see that
Z u
1
ln u
dv = 2 ,
u ≥ 1.
fU (u) =
2
u
1/u 2u v
Now for fV , things depend on whether or not v ≤ 1. With this consideration in mind,
we get
(R∞
1
du = 2v1 2 ,
if v ≥ 1,
2
fV (v) = Rv∞ 2u1v
1
du = 2 ,
if 0 ≤ v ≤ 1.
1/v 2u2 v
55. (a) The Jacobian of the map from (x, y) to (u, v) is −(x/y) − (1/y). So
fU,V (u, v) = fX,Y (x, y)
y
.
x+1
Now we solve in terms of u and v. Note that x = yv. Plug this into the first equation
u
uv
to see that u = yv + y = y(1 + v). So y = 1+v
. Therefore, x = yv = 1+v
. Plug to get
fU,V (u, v) =
u/(1 + v)
,
1 + (uv/(1 + v))
0≤
uv
u
≤ 1, 0 ≤
≤ 1.
1+v
1+v
To simplify note that u and v are both positive, so 0 ≤ u ≤ 1 and 0 ≤ uv ≤ 1.
Therefore,
u
fU,V (u, v) =
,
0 ≤ u ≤ 1, 0 ≤ uv ≤ 1.
1 + v + uv
55. (b) Here, J = −x/y 2 . Moreover, y = u/v, so
fU,V (u, v) =
u
v2
0 ≤ u ≤ 1, u ≤ v.
2
55. (c) This time, J = −1/(x + y) = −1/u. So
fU,V (u, v) = u
0 ≤ x, y ≤ 1.
Now, v = x/u, so x = uv and y = u − x = u(1 − v). Therefore,
0 ≤ uv ≤ 1, 0 ≤ u(1 − v) ≤ 1.
fU,V (u, v) = u
Theoretical Problem from p. 298:
22. (a) To ease the notation, we start with the seemingly special case µx = µy = 0 and
σx = σy = 1. At the end, we will reduce to this case. So now we have
2
1
1
2
p
fX,Y (x, y) =
x + y − 2ρxy .
exp −
2(1 − ρ2 )
2π 1 − ρ2
We start with the marginal of Y .
Lemma 1. Y itself is N (0, 1).
Proof: fY is obtained by integrating, in fX,Y , the variable x; i.e.,
fY (y) =
1
2π
p
1 − ρ2
=
=
=
∞
2
1
2 2
2 2
p
dx
x − 2xρy + ρ y − ρ y
exp −
2(1 − ρ2 )
2π 1 − ρ2 −∞
Z ∞
2
2
e−y /2
1
2 2
p
dx
x − 2xρy + ρ y
exp −
2(1 − ρ2 )
2π 1 − ρ2 −∞
Z ∞
2
e−y /2
(x − ρy)2
p
exp −
dx
2(1 − ρ2 )
2π 1 − ρ2 −∞
Z
2
2
e−y /2
e−y /2 ∞ −z2 /2
e
dz = √
,
2π
2π
−∞
−
=
2
1
2
x + y − 2ρxy
exp −
dx
2(1 − ρ2 )
−∞
Z
e
y2
2(1−ρ2 )
Z
∞
p
where z = (x − ρy)/ 1 − ρ2 . Thus, Y is N (0, 1).
Now we prove the assertion in the special case µx = µy = 0 and σx = σy = 1. Namely,
2
y2
fX,Y (x, y)
1
1
2
fX|Y (x|y) =
=p
x + y − 2ρxy +
exp −
fY (y)
2(1 − ρ2 )
2
2π(1 − ρ2 )
2
2
2
y
1
1
ρ
.
= p
exp −
x
−
2ρxy
−
2
2
2(1 − ρ )
2 1 − ρ2
2π(1 − ρ )
3
Now complete the square to see that x2 −2ρxy = x2 −2x(ρy) = x2 −2x(ρy)+(ρy)2 −ρ2 y 2 =
(x − ρy)2 − ρ2 y 2 . Therefore,
1
1
2
fX|Y (x|y) = p
(x − ρy)
exp −
.
2(1 − ρ2 )
2π(1 − ρ2 )
Thas is, given {Y = y}, X is normal with mean ρy and variance (1 − ρ2 ). This is the
desired formula in the special case µx = µy = 0 and σx = σy = 1. More generally, suppose
X and Y are as given in this solution, and consider
U = σu X + µu ,
V = σv Y + µv .
The Jacobian of the transformation that makes (x, y) into (u, v) is J = σu σv . Therefore,
fU,V (u, v) =
1
f (x, y).
σu σv X,Y
That is, (U, V ) now has the bivariate normal density with parameters (µu , µv , σu2 , σv2 , ρ).
To finish, note that
u − µu v − µv
FU|V (u|v) = P {U ≤ u | V = v} = P X ≤
σu σv
u − µu v − µv
= FX|Y
.
σu σv
Differentiate [du] to find that
1
fU|V (u|v) =
f
σu X|Y
u − µu v − µv
,
σu σv
which is the asserted formula, but (X, Y ) have now been respectively replaced by (U, V ),
and (µx , µy , σx , σy ) by (µu , µv , σu , σv ).
22. (b) This is Lemma 1 above, after standardization.
22. (c) Check that f (x, y) is a product of a function of x and a function y when ρ = 0.
[This independence case, for normals, is if and only if. It states that when (X, Y ) are
jointly normal in the sense of this problem, then they are independent if and only if they
are socalled uncorrelated; i.e., ρ = 0. For nonnormal random variables, this statement is
false, although sadly “used” too often in applied settings.]
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