Krzys’ Ostaszewski: http://www.krzysio.net
Author of the BTDT Manual (the “Been There Done That!” manual) for Course P/1
http://smartURL.it/krzysioP or http://smartURL.it/krzysioPe
Instructor for Course P/1 seminar: http://www.math.ilstu.edu/actuary/prepcourses.html
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Dr. Ostaszewski’s online exercise posted November 26, 2005
You are given that X and Y have a joint distribution with density f X,Y ( x, y ) = e !( x+y) for
x > 0 and y > 0, and f X,Y ( x, y ) = 0 otherwise. A new random variable Z is defined as
Z = e !( X +Y ). Find the density of Z.
A.
1 !2 z
ze
2
B. e !e
!z
C. ze ! z
D. ln
1
z
E. ln z
Solution.
As the region where the joint density is positive is a rectangle with sided parallel to the
axes, and
f X,Y ( x, y ) = e !( x+y) = e ! x " e ! y = f X ( x ) " fY ( y ),
X and Y are independent and both have exponential distribution with mean 1. This means
that the distribution of X + Y is gamma with parameters ! = 2 and ! = 1. Alternatively,
the density of W = X + Y can be found as a convolution of X and Y:
w
w
w
0
0
0
f X +Y ( w ) = # f X ( x ) ! fY ( w " x ) dx = # e " x ! e "(w"x ) dx = # e " z dx = we " w .
This is, of course, the gamma distribution with parameters ! = 2 and ! = 1. Now
dw
1
= ! . Therefore,
Z = e !W , so that W = ! ln Z, and
dz
z
dw
1
1
fZ ( z ) = fW ( w ( z )) !
= ( " ln z ) ! e "(" ln z ) ! " = " ln z = ln ,
dz
z
z
with 0 < z < 1. That’s answer D. We also could use the CDF technique to solve this. In
order to do this, consider the region where e !( x+y) " z in the xy-plane, shown in the figure
below:
y
! ln z
x + y = ! ln z
! ln z
x
We have to consider separately these two areas above the cut-out triangle, the region
directly above the line x + y = ! ln z for 0 < x < ! ln z and the region directly above the xaxis for x ! " ln z, and calculate
(
)
FZ ( z ) = Pr ( Z ! z ) = Pr e "( X +Y ) ! z = Pr ( " ( X + Y ) ! ln z ) = Pr ( X + Y # " ln z ) =
+* +*
$
' " ln z $ +*
'
$
'
= Pr & Y # ( " ln z ) " X ) = + & + e "( x+y) dy) dx + + & + e "( x+y) dy) dx =
!"#
&%
)(
(
(
0 % " ln z"x
" ln z % 0
positive
=
" ln z
+
0
=
+*
$ +*
'
$ +* " y '
"x
"y
e , & + e dy) dx + + e , & + e$
dy) dx =
% " ln z"x
(
% 0 PDF of EXP(1) (
" ln z
"x
" ln z
+
0
e
"x
("e
" y y-+*
y=" ln z"x
) dx + + e
+*
" ln z
"x
,1, dx =
" ln z
+
e " x , e ln z , e x dx + z = "z ln z + z.
0
Therefore,
d
1
( "z ln z + z ) = "z # " ln z +1 = " ln z.
dz
z
Hence, the CDF technique also works, but it is definitely a more complicated way than
knowing what the distribution of X + Y is, and then using a one-dimensional
transformation. Answer D, again. Finally, we can also use a multivariate transformation:
v = x, z = e !( x+y). Then the inverse transformation is x = v, y = ! ln z ! v, and its Jacobian
is
" !x !x %
" 1
0 %
$
'
!( x, y )
1
!v
!z
$
'
' = det
= det $
1 '=( .
$
$ !y !y '
(1 (
!( v, z )
z
$#
z '&
$
'
# !v !z &
This gives
fZ ( z ) = FZ! ( z ) =
"( x, y )
1
# v+ # ln z#v )) 1
=e ( (
! = e ln z ! = 1.
"( v, z )
z
z
Recall that we are given that the joint density of X and Y is positive only if x > 0 and
y > 0, and this is equivalent to v > 0 and ! ln z ! v > 0, or just 0 < v < ! ln z. Of course,
the condition z > 0 is also required for the natural logarithm to be well-defined. Therefore
fV ,Z ( v, z ) = f X,Y ( x ( v, z ), y ( v, z )) !
fZ ( z ) =
!
all v
fV ,Z ( v, z ) dv =
" ln z
! 1dv = " ln z.
0
Answer D.
© Copyright 2005 by Krzysztof Ostaszewski.
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