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    9. Two Functions of Two RandomVariables

    In the spirit of the previous lecture, let us look at animmediate generalization: Suppose X and Y are two random

    variables with joint p.d.f Given two functionsand define the new random variables

    How does one determine their joint p.d.f Obviouslywith in hand, the marginal p.d.fs andcan be easily determined.

    (9-1)

    ).,( y x f XY

    ).,(

    ),(

    Y X hW

    Y X g Z

    ),( y x g

    ),,( y xh

    (9-2)

    ?),( w z f ZW ),( w z f ZW )( z f Z )(w f W

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    Example 9.1: Suppose X and Y are independent uniformlydistributed random variables in the intervalDefine Determine

    Solution: Obviously both w and z vary in the intervalThus

    We must consider two cases: and since theygive rise to different regions for (see Figs. 9.2 (a)-(b)).

    ).,max( ),,min( Y X W Y X Z

    .0or0 if ,0),( w z w z F ZW

    ).,( w z f ZW

    ).,0(

    ).,0(

    (9-4)

    .),max( ,),min(,),( wY X z Y X P wW z Z P w z F ZW (9-5)

    z w , z w

    w z D ,

    X

    Y

    w y

    ),( ww

    ),( z z

    z wa

    )(

    X

    Y

    ),( ww

    ),( z z

    z wb )(Fig. 9.2 PILLAI

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    For from Fig. 9.2 (a), the region is represented by the doubly shaded area. Thus

    and for from Fig. 9.2 (b), we obtain

    With

    we obtain

    Thus

    , ,),(),(),(),( z w z z F z w F w z F w z F XY XY XY ZW

    w z D ,

    (9-6)

    , z w

    , z w. ,),(),( z www F w z F XY ZW (9-7)

    ,)()(),( 2 xy y x y F x F y x F Y X XY (9-8)

    2

    2 2

    (2 ) / , 0 ,( , )

    / , 0 . ZW w z z z w

    F z ww w z

    .otherwise,0

    ,0,/2),(

    2 w z w z f ZW

    (9-9)

    (9-10)

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    From (9-10), we also obtain

    and

    If and are continuous and differentiablefunctions, then as in the case of one random variable (see (5-30)) it is possible to develop a formula to obtain the joint

    p.d.f directly. Towards this, consider the equations

    For a given point (z,w), equation (9-13) can have manysolutions. Let us say

    ,0 ,12

    ),()(

    z z

    dww z f z f z ZW Z

    (9-11)

    .),( ,),( w y xh z y x g (9-13)

    .0 ,2

    ),()(

    0 2

    w

    wdz w z f w f

    w

    ZW W (9-12)

    ),( y x g ),( y xh

    ),( w z f ZW

    ),,( , ),,( ),,( 2211 nn y x y x y x PILLAI

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    represent these multiple solutions such that (see Fig. 9.3)

    (9-14).),( ,),( w y xh z y x g iiii

    Fig. 9.3

    Consider the problem of evaluating the probability

    z

    (a)

    w

    ),( w z z

    www

    z z

    .),(,),(

    ,

    wwY X hw z z Y X g z P

    wwW w z z Z z P (9-15)

    (b)

    x

    y1

    2

    i

    n

    ),( 11 y x

    ),( 22 y x

    ),( ii y x

    ),( nn y x

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    Using (7-9) we can rewrite (9-15) as

    But to translate this probability in terms of we need

    to evaluate the equivalent region for in the xy plane.Towards this referring to Fig. 9.4, we observe that the point

    A with coordinates ( z ,w) gets mapped onto the point withcoordinates (as well as to other points as in Fig. 9.3(b)).As z changes to to point B in Fig. 9.4 (a), letrepresent its image in the xy plane. Similarly as w changesto to C , let represent its image in the xy plane.

    (9-16)

    ),,( y x f XY

    .),(, w z w z f wwW w z z Z z P ZW

    w z

    A),( ii y x

    z z B

    ww C

    (a)

    w

    A z

    ww

    B

    z z

    C D

    Fig. 9.4 (b)

    y

    Ai y B

    xi x

    C D

    i

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    As the point ( z ,w) goes to the pointthe point and the pointHence the respective x and y coordinates of are given by

    and

    Similarly those of are given by

    The area of the parallelogram in Fig. 9.4 (b) isgiven by

    ,),( A y x ii ,),( Bw z z

    ,),( C ww z .),( Dww z z

    B

    ,),(),( 1111 z z g x z

    z g w z g w z z g i

    .),(),(11

    11 z z h

    y z z h

    w z hw z z h i

    (9-20)

    (9-21)

    C

    . , 11 w

    w

    h yw

    w

    g x ii (9-22)

    DC B A

    .cossinsincos

    )sin(

    C A B AC A B A

    C A B Ai (9-23)

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    But from Fig. 9.4 (b), and (9-20) - (9-22)

    so that

    and

    The right side of (9-27) represents the Jacobian ofthe transformation in (9-19). Thus

    .cos ,sin

    ,sin ,cos

    11

    11

    ww

    g C A z z h B A

    wwh

    C A z z g

    B A

    (9-25)

    (9-24)

    w z z

    h

    w

    g

    w

    h

    z

    g i

    1111 (9-26)

    wh

    z h

    w g

    z g

    z

    h

    w

    g

    w

    h

    z

    g

    w z

    i

    11

    11

    1111 det (9-27)

    ( , ) J z w

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    Next we shall illustrate the usefulness of the formula in(9-29) through various examples:

    Example 9.2: Suppose X and Y are zero mean independentGaussian r.vs with common varianceDefine whereObtain

    Solution: Here

    Since

    if is a solution pair so is From (9-33)

    ),/(tan , 122 X Y W Y X Z .2

    ).,( w z f ZW

    .2

    1),(

    222 2/)(2

    y x

    XY e y x f (9-32)

    ,2/||),/(tan),(;),(122

    w x y y xhw y x y x g z (9-33)),( 11 y x ).,( 11 y x

    .tanor,tan w x yw x y

    (9-34)

    .2/|| w

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    Substituting this into z , we get

    and

    Thus there are two solution sets

    We can use (9-35) - (9-37) to obtain From (9-28)

    so that

    .cos or,sec tan1 222 w z xw xw x y x z (9-35)

    .sintan w z w x y (9-36)

    .sin ,cos ,sin ,cos 2211 w z yw z xw z yw z x (9-37)

    ).,( w z J

    ,cossin

    sincos),( z

    w z ww z w

    w y

    z y

    w x

    z x

    w z J (9-38)

    .|),(| z w z J (9-39) PILLAI

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    We can also compute using (9-31). From (9-33),

    Notice that agreeing with (9-30).Substituting (9-37) and (9-39) or (9-40) into (9-29), we get

    Thus

    which represents a Rayleigh r.v with parameter and

    ),( y x J

    .11

    ),(22

    2222

    2222

    z y x

    y x x y x y

    y x

    y

    y x

    x

    y x J (9-40)

    |,),(|/1|),(| ii y x J w z J

    .

    2|| ,0 ,

    ),(),(),(22 2/

    2

    2211

    w z e z

    y x f y x f z w z f z

    XY XY ZW

    (9-41)

    ,0 ,),()(22 2/

    2

    2/

    2/

    z e z dww z f z f z ZW Z

    (9-42)

    ,2

    ,2|| ,1

    ),()(

    0

    wdz w z f w f ZW W (9-43)PILLAI

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    which represents a uniform r.v in the intervalMoreover by direct computation

    implying that Z and W are independent. We summarize theseresults in the following statement: If X and Y are zero meanindependent Gaussian random variables with common

    variance, then has a Rayleigh distribution andhas a uniform distribution. Moreover these two derived r.vsare statistically independent. Alternatively, with X and Y asindependent zero mean r.vs as in (9-32), X + jY represents acomplex Gaussian r.v. But

    where Z and W are as in (9-33), except that for (9-45) to holdgood on the entire complex plane we must haveand hence it follows that the magnitude and phase of

    ).2/,2/(

    22

    Y X )/(tan 1

    X Y

    , jW Ze jY X (9-45)

    )()(),( w f z f w z f W Z ZW (9-44)

    , W PILLAI

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    a complex Gaussian r.v are independent with Rayleighand uniform distributions ~ respectively. Thestatistical independence of these derived r.vs is an interesting

    observation.Example 9.3: Let X and Y be independent exponentialrandom variables with common parameter .Define U = X + Y , V = X - Y . Find the joint and marginal

    p.d.f of U and V .Solution: It is given that

    Now since u = x + y, v = x - y, always and there isonly one solution given by

    Moreover the Jacobian of the transformation is given by

    .0 ,0 ,1

    ),( /)(2 y xe y x f y x XY

    (9-46)

    .2

    ,2

    vu y

    vu x

    (9-47)

    ,|| uv

    ),( U

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    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.

    As we show below, the general transformation formula in(9-29) making use of two functions can be made useful even

    when only one function is specified.

    2 11

    1 1),( y x J

    ,||0 ,21),( /2 uvevu f uUV (9-48)

    ,0 ,2

    1),()( /2

    /2

    ueudvedvvu f u f uu

    u

    uu

    u UV U

    (9-49)

    / | |/2 | | | |

    1 1( ) ( , ) , .

    2 2u v

    V UV v v f v f u v du e du e v

    (9-50)

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    Auxiliary Variables :

    Suppose

    where X and Y are two random variables. To determine by making use of the above formulation in (9-29), we can

    define an auxiliary variable

    and the p.d.f of Z can be obtained from by proper

    integration.

    Example 9.4: Suppose Z = X + Y and let W = Y so that thetransformation is one-to-one and the solution is given

    by

    ),,( Y X g Z

    Y W X W or

    (9-51)

    (9-52)

    ),( w z f ZW

    . , 11 w z xw y

    )( z f Z

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    The Jacobian of the transformation is given by

    and hence

    or

    which agrees with (8.7). Note that (9-53) reduces to theconvolution of and if X and Y are independentrandom variables. Next, we consider a less trivial example.

    Example 9.5: Let and be independent.Define

    1 1 0

    1 1),( y x J

    ),(),(),( 11 ww z f y x f y x f XY XY ZW

    ,),(),()( dwww z f dww z f z f XY ZW Z (9-53)

    )( z f X )( z f Y

    )1,0( U X )1,0( U Y

    ).2cos(ln2 2/1 Y X Z (9-54)PILLAI

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    Find the density function of Z .Solution: We can make use of the auxiliary variable W = Y in this case. This gives the only solution to be

    and using (9-28)

    Substituting (9-55) - (9-57) into (9-29), we obtain

    ,,

    1

    2/)2sec(1

    2

    w ye x w z (9-55)

    (9-56)

    .)2(sec

    10

    )2(sec

    ),(

    2/)2sec(2

    12/)2sec(2

    11

    11

    2

    2

    w z

    w z

    ew z

    w x

    ew z

    w y

    z y

    w x

    z x

    w z J

    (9-57)

    2sec(2 ) / 22( , ) sec (2 ) ,

    , 0 1,

    z w ZW f z w z w e

    z w

    (9-58)

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    and

    Let so that Noticethat as w varies from 0 to 1, u varies from toUsing this in (9-59), we get

    which represents a zero mean Gaussian r.v with unit

    variance. Thus Equation (9-54) can be used asa practical procedure to generate Gaussian random variablesfrom two independent uniformly distributed randomsequences.

    221 1 tan(2 ) / 2/ 2 2 0 0

    ( ) ( , ) sec (2 ) . z w z Z ZW f z f z w dw e z w e dw

    tan(2 )u z w 22 sec (2 ) .du z w dw .

    , ,21

    2

    21)( 2/

    1

    2/2/ 222 z eduee z f z u z Z

    (9-59)

    (9-60)

    ).1,0( N Z

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    Example 9.6 : Let X and Y be independent identically distributedGeometric 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 random variables.Solution: (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 We have P ( Z = m, W = n) = P {min ( X , Y ) = m, X Y = n}. But

    Thus

    },,21,0,{ }.,21,0,{

    .,2,1,0 ,)()( k pq K Y P k X P k

    negative.is enonnegativis ),min(

    Y X W Y X X Y X W Y X Y Y X Z

    ),,),(min(

    ),,),(min(

    )}(,,),{min(),(

    Y X nY X mY X P

    Y X nY X mY X P

    Y X Y X nY X mY X P nW m Z P

    (9-61)

    PILLAI(9-62)

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    ,2,1,0 ,2,1,0 ,

    0,0,)()(

    0,0,)()(

    ),,(

    ),,(),(

    ||22 nmq p

    nm pq pqnmY P m X P

    nm pq pqmY P nm X P

    Y X nmY m X P

    Y X nm X mY P nW m Z P

    nm

    nmm

    mnm

    (9-63)

    PILLAI

    represents the joint probability mass function of the random variables Z and W . Also

    Thus Z represents a Geometric random variable sinceand),1(1 2 q pq

    .,2,1,0 ,)1(

    )1()1(

    )221(

    ),()(

    2

    222

    222

    ||22

    12

    mqq p

    q pqq p

    qqq p

    qq pnW m Z P m Z P

    m

    mm

    m

    n n

    nm

    qq

    (9-64)

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    2 2 | |

    0 0

    2 | | 2 4 2 | | 12

    | |1

    1

    ( ) ( , )

    (1 )

    , 0, 1, 2, .

    m n

    m m

    n n

    nq

    pq

    P W n P Z m W n p q q

    p q q q p q

    q n

    PILLAI

    (9-65)

    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.(b) Let Z = min ( X , Y ) , R = max ( X , Y ) min ( X , Y ).

    In this case both Z and R take nonnegative integer valuesProceeding as in (9-62)-(9-63) we get

    ),()(),( nW P m Z P nW m Z P (9-66)

    (9-67).,21,0,

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    .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

    nmq pnmq p

    nm pq pq

    nm pq pq pq pq

    Y X nmY m X P Y X mY nm X P

    Y X nmY m X P Y X nm X mY P

    Y X nY X Y X mY X P

    Y X nY X Y X mY X P n Rm Z P

    m

    nm

    mnm

    nmmmnm

    (9-68)

    Eq. (9-68) represents the joint probability mass function of Z and R in (9-67). From (9-68),

    ,2,1,0 ,)1(

    1)21(},{)(

    2

    0

    22

    1

    22 2

    mqq p

    q pqq pn Rm Z P m Z P

    m

    n

    m

    n

    nm pq

    (9-69)PILLAI

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    (9-71)

    and

    From (9-68)-(9-70), we get

    which proves the independence of the random variables Z and R defined in (9-67) as well.

    )()(),( n R P m Z P n Rm Z P

    .,2,1 ,

    0 ,},{)(

    12

    1

    0 nq

    nn Rm Z P n R P

    nm q

    p

    q p

    (9-70)