Two Functions of Two Random Variables. Two Functions of Two RV.s X and Y are two random variables...
-
Upload
spencer-robert-montgomery -
Category
Documents
-
view
230 -
download
2
Transcript of Two Functions of Two Random Variables. Two Functions of Two RV.s X and Y are two random variables...
Two Functions of Two Random Variables
Two Functions of Two RV.s
X and Y are two random variables with joint p.d.f
and are functions
define the new random variables:
How to determine the joint p.d.f
with in hand, the marginal p.d.fs and can be easily determined.
).,( yxf XY
).,(
),(
YXhW
YXgZ
),( yxg),,( yxh
?),( wzfZW
),( wzfZW )(zfZ)(wfW
Two Functions of Two RV.s
For given z and w,
where is the region in the xy plane such that :
wzDyx XYwz
ZW
dxdyyxfDYXP
wYXhzYXgPwWzZPwzF
,),( , ,),(),(
),(,),()(,)(),(
wzD ,
zyxg ),(wyxh ),(
x
y
wzD ,
wzD ,
X and Y are independent uniformly distributed rv.s in
Define
Determine
Obviously both w and z vary in the interval Thus
two cases:
.0or 0 if ,0),( wzwzFZW
).,( wzfZW
).,0(
).,0(
. ),max( ,),min(,),( wYXzYXPwWzZPwzFZW wzD ,
X
Y
),( ww
),( zz
zwb )(
X
Y
wy
),( ww
),( zz
zwa )(
Example
).,max( ),,min( YXWYXZ
Solution
For
For
With
we obtain
Thus
, , ),(),(),(),( zwzzFzwFwzFwzF XYXYXYZW
,zw
,zw
. , ),(),( zwwwFwzF XYZW
,)( )(),(2
xyyxyFxFyxF YXXY
2
2 2
(2 ) / , 0 ,( , )
/ , 0 .ZW
w z z z wF z w
w w z
.otherwise,0
,0,/2),(
2 wzwzfZW
Example - continued
X
Y
wy
),( ww
),( zz
zwa )(
X
Y
),( ww
),( zz
zwb )(
Also,
and
and are continuous and differentiable functions,
So, it is possible to develop a formula to obtain the joint p.d.f directly.
,0 ,12
),()(
z
zdwwzfzf
z ZWZ
.0 ,2
),()(
0 2
w
wdzwzfwf
w
ZWW
),( yxg ),( yxh
),( wzfZW
Example - continued
Let’s consider:
Let us say these are the solutions to the above equations:
.),( ,),( wyxhzyxg iiii
z
(a)
w
),( wz
z
www
zz
(b)
x
y1
2
i
n
),( 11 yx
),( 22 yx
),( ii yx
),( nn yx
.),( ,),( wyxhzyxg
),,( , ),,( ),,( 2211 nn yxyxyx
Consider the problem of evaluating the probability
This can be rewritten as:
To translate this probability in terms of we need to evaluate the
equivalent region for in the xy plane.
. ),(,),(
,
wwYXhwzzYXgzP
wwWwzzZzP
.),(, wzwzfwwWwzzZzP ZW
),,( yxf XY
wz
The point A with coordinates (z,w) gets mapped onto the point with coordinates
As z changes to to point B in figure (a),
- let represent its image in the xy plane.
As w changes to to C,
- let represent its image in the xy plane.
A
),( ii yx
zz
B
ww
C
(a)
w
Az
ww
B
zz
C D
(b)
y
Aiy B
xix
CD
i
Finally D goes to
represents the equivalent parallelogram in the XY plane with area
The desired probability can be alternatively expressed as
Equating these, we obtain
To simplify this, we need to evaluate the area of the parallelograms in terms of
DCBA
.i
,D
.),(),( i
iiiXYi
i yxfYXP
.),(),(
i
iiiXYZW wz
yxfwzf
.wzi
.i
let and denote the inverse transformations, so that
As the point (z,w) goes to
- the point
- the point
- the point
Hence x and y of are given by:
Similarly, those of are given by:
,),( Ayx ii
,),( Bwzz ,),( Cwwz
.),( Dwwzz
,),(),( 1111 z
z
gxz
z
gwzgwzzg i
.),(),( 1111 z
z
hyz
z
hwzhwzzh i
1g 1h
).,( ),,( 11 wzhywzgx ii
. , 11 ww
hyw
w
gx ii
B
C
The area of the parallelogram is given by
From the figure and these equations,
so that
or
.cos ,sin
,sin ,cos
11
11
ww
gCAz
z
hBA
ww
hCAz
z
gBA
wzz
h
w
g
w
h
z
gi
1111
w
h
z
h
w
g
z
g
z
h
w
g
w
h
z
g
wzi
11
11
1111 det
( , )J z w
DCBA . cos sinsin cos
)sin(
CABACABA
CABAi
This is the Jacobian of the transformation
Using these in , and we get
where represents the Jacobian of the original transformation:
1 1
1 1
( , ) det .
g g
z wJ z w
h h
z w
,),(|),(|
1),(|),(|),(
iiiXY
iiiiiXYZW yxf
yxJyxfwzJwzf
|),(|
1 |),(|
ii yxJwzJ
( , )i iJ x y
.det),(
, ii yyxx
ii
y
h
x
h
y
g
x
g
yxJ
).,( ),,( 11 wzhywzgx ii
.),(),(
i
iiiXYZW wz
yxfwzf
(*)
Example 9.2: Suppose X and Y are zero mean independent Gaussian r.vs with common variance
Define where
Obtain
Here
Since
if is a solution pair so is Thus
),/(tan , 122 XYWYXZ
.2
).,( wzfZW
.2
1),(
222 2/)(2
yx
XY eyxf
,2/||),/(tan),(;),( 122 wxyyxhwyxyxgz
),( 11 yx ).,( 11 yx
.tanor ,tan wxywx
y
.2/|| w
Example
Solution
Substituting this into z, we get
and
Thus there are two solution sets
so that
.cos or ,sec tan1 222 wzxwxwxyxz
.sintan wzwxy
.sin ,cos ,sin ,cos 2211 wzywzxwzywzx
:is ),( wzJ,
cossin
sincos),( z
wzw
wzw
w
y
z
y
w
x
z
x
wzJ
.|),(| zwzJ
Example - continued
Also is
Notice that here also
Using (*),
Thus
which represents a Rayleigh r.v with parameter
),( yxJ
.11
),(22
2222
2222
zyx
yx
x
yx
y
yx
y
yx
x
yxJ
|,),(|/1|),(| ii yxJwzJ
.2
|| ,0 ,
),(),(),(
22 2/2
2211
wzez
yxfyxfzwzf
z
XYXYZW
,0 ,),()(22 2/
2
2/
2/
zez
dwwzfzf zZWZ
,2
Example - continued
Also,
which represents a uniform r.v in
Moreover,
So Z and W are independent.
To summarize, If X and Y are zero mean independent Gaussian random variables with common variance, then
- has a Rayleigh distribution
- has a uniform distribution.
- These two derived r.vs are statistically independent.
).2/,2/(
22 YX )/(tan 1 XY
)()(),( wfzfwzf WZZW
,2
|| ,1
),()(
0
wdzwzfwf ZWW
Example - continued
Alternatively, with X and Y as independent zero mean Gaussian r.vs with
common variance, X + jY represents a complex Gaussian r.v. But
where Z and W are as in
except that for the abovementioned, to hold good on the entire complex plane we must have
The magnitude and phase of a complex Gaussian r.v are independent with Rayleigh and uniform distributions respectively.
,jWZejYX
, W
),( ~ U
,2/||),/(tan),(;),( 122 wxyyxhwyxyxgz
Example - continued
Let X and Y be independent exponential random variables with common parameter .
Define U = X + Y, V = X - Y.
Find the joint and marginal p.d.f of U and V.
It is given that
Now since u = x + y, v = x - y, always and there is only one solution given by
.0 ,0 ,1
),( /)(2
yxeyxf yxXY
.2
,2
vuy
vux
,|| uv
Example
Solution
Moreover the Jacobian of the transformation is given by
and hence
represents the joint p.d.f of U and V. This gives
and
Notice that in this case the r.vs U and V are not independent.
2 11
1 1 ),(
yxJ
, || 0 ,2
1),( /
2 uvevuf u
UV
,0 ,2
1),()( /
2
/2
ueu
dvedvvufuf uu
u
uu
u UVU
/ | |/2 | | | |
1 1( ) ( , ) , .
2 2u v
V UVv vf v f u v du e du e v
Example - continued
As we will show, the general transformation formula in (*) making use of two functions
can be made useful even when only one function is specified.
Auxiliary Variables
Suppose
X and Y : two random variables.
To determine by making use of the above formulation in (*), we
can define an auxiliary variable
and the p.d.f of Z can be obtained from by proper integration.
),,( YXgZ
YWXW or
),( wzfZW
)(zfZ
Z = X + Y
Let W = Y so that the transformation is one-to-one and the solution is given by
The Jacobian of the transformation is given by
and hence
or
This reduces to the convolution of and if X and Y are independent random variables.
1 1 0
1 1 ),( yxJ
),(),(),( 11 wwzfyxfyxf XYXYZW
,),(),()( dwwwzfdwwzfzf XYZWZ
)(zf X )(zfY
. , 11 wzxwy
Example
Let and be independent.
Define
Find the density function of Z.
Making use of the auxiliary variable W = Y,
,
,
1
2/)2sec(1
2
wy
ex wz
.)2(sec
10
)2(sec
),(
2/)2sec(2
12/)2sec(2
11
11
2
2
wz
wz
ewz
w
xewz
w
y
z
y
w
x
z
x
wzJ
)1,0( UX )1,0( UY
).2cos(ln2 2/1 YXZ
Example
Solution
Using these in (*), we obtain
and
Let so that
Notice that as w varies from 0 to 1, u varies from to
2sec(2 ) / 22( , ) sec (2 ) ,
, 0 1,
z wZWf z w z w e
z w
Example - continued
22 1 1 tan(2 ) / 2/ 2 2
0 0( ) ( , ) sec (2 ) .z wz
Z ZWf z f z w dw e z w e dw
tan(2 )u z w 22 sec (2 ) .du z w dw
.
Using this in the previous formula, we get
As you can see,
A practical procedure to generate Gaussian random variables is from two independent uniformly distributed random sequences, based on
, ,2
1
2
2
1)( 2/
1
2/2/ 222
zedu
eezf zuzZ
).1,0( ~ NZ
Example - continued
).2cos(ln2 2/1 YXZ
Let X and Y be independent identically distributed Geometric random variables with
(a) Show that min (X , Y ) and X – Y are independent random variables.
(b) Show that min (X , Y ) and max (X , Y ) – min (X , Y ) are also independent
(a) Let
Z = min (X , Y ) , and W = X – Y.
Note that Z takes only nonnegative values while W takes both positive, zero and negative values
Example
Solution
.,2 ,1 ,0 ,)()( kpqKYPkXP k
}, ,2 1, 0,{ }. ,2 1, 0,{
We have
P(Z = m, W = n) = P{min (X , Y ) = m, X – Y = n}.
But
Thus
negative. is
enonnegativ is ),min(
YXWYXX
YXWYXYYXZ
),,),(min(
),,),(min(
)} (,,),{min(),(
YXnYXmYXP
YXnYXmYXP
YXYXnYXmYXPnWmZP
Example - continued
represents the joint probability mass function of the random variables Z
and W. Also
,2 ,1 ,0 ,2 ,1 ,0 ,
0,0,)()(
0,0,)()(
),,(
),,( ),(
||22
nmqp
nmpqpqnmYPmXP
nmpqpqmYPnmXP
YXnmYmXP
YXnmXmYPnWmZP
nm
nmm
mnm
.,2 ,1 ,0 ,)1(
)1()1(
)221(
),()(
2
222
222
||22
12
mqqp
qpqqp
qqqp
qqpnWmZPmZP
m
mm
m
n n
nm
Example - continued
Thus Z represents a Geometric random variable since and
Note that
establishing the independence of the random variables Z and W.
The independence of X – Y and min (X , Y ) when X and Y are
independent Geometric random variables is an interesting observation.
),1(1 2 qpq
Example - continued
2 2 | |
0 0
2 | | 2 4 2 | | 12
| |
1
1
( ) ( , )
(1 )
, 0, 1, 2, .
m n
m m
n n
n
qpq
P W n P Z m W n p q q
p q q q p q
q n
),()(),( nWPmZPnWmZP
(b) Let
Z = min (X , Y ) , R = max (X , Y ) – min (X , Y ).
In this case both Z and R take nonnegative integer values
we get
. ,2 1, 0,
Example - continued
.0 ,,2 ,1 ,0 ,
,2 ,1 ,,2 ,1 ,0 ,2
0 ,,2 ,1 ,0 ,
,2 ,1 ,,2 ,1 ,0 ,
} ,,() ,,{
} ,,() ,,{
} ,),min(),max( ,),{min(
} ,),min(),max( ,),{min(},{
22
22
nmqp
nmqp
nmpqpq
nmpqpqpqpq
YXnmYmXPYXmYnmXP
YXnmYmXPYXnmXmYP
YXnYXYXmYXP
YXnYXYXmYXPnRmZP
m
nm
mnm
nmmmnm
This equation is the joint probability mass function of Z and R. Also we can obtain:
and
From (9-68)-(9-70), we get
This proves the independence of the random variables Z and R as well.
,2 ,1 ,0 ,)1(
1)21(},{)(
2
0
22
1
22 2
mqqp
qpqqpnRmZPmZP
m
n
m
n
nmpq
)()(),( nRPmZPnRmZP
.,2 ,1 ,
0 , },{)(
12
1
0 nq
nnRmZPnRP
nm
qp
qp
Example - continued