Statistics 351 (Fall 2015)
Prof. Michael Kozdron
September 23, 2015
Lecture #7: Functions of Multivariate Random Variables
Example (Chapter 1, Problem #8). Suppose that X 2 N (0, 1) and Y 2 N (0, 1) are
independent random variables. Show that X/Y 2 C(0, 1).
Solution. We start with a bivariate random vector (X, Y )0 (i.e., two random variables), but
we want the distribution of just one random variable, namely X/Y .
The trick is to let U = X/Y and to introduce an auxiliary variable V which may be
arbitrarily chosen. (Although it may be arbitrary, choose it suitably!)
Let U = X/Y and V = Y so that solving for X and Y gives
X = UV
and Y = V.
The Jacobian of this transformation is given by
@x
@u
J=
@y
@u
@x
v u
@v
=
= v.
0 1
@y
@v
The density of (U, V )0 is therefore given by
fU,V (u, v) = fX,Y (uv, v) |J| = |v|fX (uv)fY (v)
using the assumed independence of X and Y . Substituting in the corresponding densities
gives
1
1
|v| v2 (u2 +1)
2 2
2
fU,V (u, v) = |v| p e u v /2 p e v /2 =
e 2
2
2
2
provided 1 < u, v < 1. The marginal density of U is
Z 1
Z 1
Z 1
v2
|v| v2 (u2 +1)
v
2
2
fU (u) =
fU,V (u, v) dv =
e
dv = 2
e 2 (u +1) dv
2
1
1 2
Z 10
v v2 (u2 +1)
=
e 2
dv
0
since the integrand is even. Making the substitution z = v 2 (u2 + 1)/2 so that dz =
v(u2 + 1) dv gives
Z 1
1
1
z
fU (u) =
e
dz
=
(u2 + 1) 0
(u2 + 1)
for 1 < u < 1. We recognize that this is the density of a C(0, 1) random variable, and
so we conclude that U = X/Y 2 C(0, 1).
Example (Chapter 1, Problem #39). Suppose that X1 2 (a1 , b) and X2 2 (a2 , b) are independent random variables. Show that X1 /X2 and X1 +X2 are independent, and determine
their distributions.
71
Solution. Since X1 and X2 are independent, their joint density is
fX1 ,X2 (x1 , x2 ) = fX1 (x1 ) fX2 (x2 )
(
1
xa1 1 xa22
(a1 ) (a2 ) 1
=
0,
1
e x1 /b x2 /b ,
ba1 +a2
for x1 > 0, x2 > 0,
otherwise.
Let U = X1 /X2 and V = X1 + X2 so that solving for X1 and X2 gives
X1 =
UV
V
and X2 =
.
U +1
U +1
The Jacobian of this transformation is given by
@x1
@u
J=
@x2
@u
@x1
@v
=
@x2
@v
v
v(1 + u) 2 u(1 + u) 1
.
2
1 =
v(1 + u)
(1 + u)
(1 + u)2
The density of (U, V )0 is therefore given by
fU,V (u, v) = fX1 ,X2 (uv(1 + u) 1 , v(1 + u) 1 ) |J|
1
=
(uv(1 + u) 1 )a1 1 (v(1 + u) 1 )a2
(a1 ) (a2 )
1
1 ua1 1 v a1 +a2 1 v/b
=
e
(a1 ) (a2 ) ba1 +a2 (1 + u)a1 +a2
1
ba1 +a2
uv(1+u)
1 /b
v(1+u)
provided that 0 < u < 1, 0 < v < 1. The marginal density of U is
Z 1
Z 1
1
1
u a1 1
fU (u) =
fU,V (u, v) dv =
v a1 +a2 1 e
a1 +a2 (1 + u)a1 +a2
(a
)
(a
)
b
1
2
1
0
To evaluate
v a1 +a2 1 e
v/b
|v|
(1 + u)2
dv
dv
we make the substitution z = vb so that dz = 1b dv. This implies that
Z 1
Z 1
Z 1
a1 +a2 1
v/b
a1 +a2 1
z
a1 +a2
v
e
dv =
(bz)
e b dz = b
z a1 +a2 1 e
0
v/b
1 /b
dz
= ba1 +a2 (a1 + a2 ).
This now implies that
fU (u) =
1
1
ua1 1
(a1 + a2 )
u a1 1
a1 +a2
b
(a
+
a
)
=
, u > 0.
1
2
(a1 ) (a2 ) ba1 +a2 (1 + u)a1 +a2
(a1 ) (a2 ) (1 + u)a1 +a2
To find the marginal density of V we observe that since we can write the joint density as a
product of a function of u only multiplied by a function of v only, we conclude that U and
V are independent. That is,
fU,V (u, v) = fU (u) fV (v)
72
and so using the density fU (u) that we just found gives
1
1
ua1 1 v a1 +a2
(a1 ) (a2 ) ba1 +a2 (1+u)a1 +a2
(a1 +a2 )
ua 1 1
(a1 ) (a2 ) (1+u)a1 +a2
fU,V (u, v)
fV (v) =
=
fU (u)
v/b
1
v a1 +a2
(a1 + a2 )
ba1 +a2
v/b
for v > 0. Notice that V = X1 + X2 2 (a1 + a2 , b).
It is also possible to find the marginal density of V by integrating the joint density. That is,
we observe that
Z 1
Z 1
1
1
ua1 1
a1 +a2 1
v/b
fV (v) =
fU,V (u, v) du =
v
e
du.
(a1 ) (a2 ) ba1 +a2
(1 + u)a1 +a2
1
0
Making the substitution z = u/(1 + u) so that dz = (1 + u)
Z
1
0
ua1 1
du =
(1 + u)a1 +a2
1
0
u
1+u
a1 +a2
1 a2
du and u = z/(1
du =
(1 + u)2
=
We recognize this as a beta integral so that
Z 1
z a1 1 (1 z)a2
dz =
(a1 ) (a2 )
(a1 + a2 )
from which we conclude
fV (v) =
1
1
v a1 +a2 1 e
a
(a1 + a2 ) b 1 +a2
for v > 0 as before.
73
v/b
Z
Z
a1 +a2 2
0
1
z a1 1 (1
0
z) implies
z
1
z)a2
dz.
a2
dz
