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CDF&PDF examples

Date post: 07-Apr-2018
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    Example 11.2.1 (Regions BOllilded by Constants) TIle joint pdf/rv(x,y:;, is given by, . . fK (R - , - y ) 1

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    ,, y1I

    2 ......... -T TTl

    - - - ---V ,

    0 , , 4 5v

    FIGURE 9.2.2

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    REgion IT: 1 < x :'0 3, y > 2 [marginal distribution Fx(x)]1 (-XL 13x )Fyy(x,2) = Fx(x) = 9 1 : +2 - 6

    REgion 1lI: x> 3,1 < y::: 2, [marginal distribution Fy(y)]1Fyy(3,y) = Fr(y) =9(-1 + 12y -11)

    By diffErentiating thE distribution function, WE can write thE density function for thEvarious regions as follows:

    o x:'Ol or y:'Ol19(8 -x -y ) 1

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    IIJ - - ; ; - - ~ _I y

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    Example 11.2.6 (Gaussian Distribution)fx(x) = ~ e - ( X - l . l f r 2 c r 2,J27T IT

    cDx(w) = ~ J o o e-(x-I.l)2rw +j on; d:x,J27T IT -0 0(11.2.24)

    (11.2.25)Completing the squares in the exponent of Eq. (11.2.25), we obtain the resultingexpression as follows:

    cDx(w) = ejUlf1-[(ifUl2)(2] { ~ J o o exp [_ (x - flu - j W ~ l ] d : x },.j27T IT -0 0 2u2The value of the integral within braces is 1 and hence the CF of a Gaussian is

    Mean:

    Variance:

    Hence

    c D ~ ( w ) = ejUl I.l-[(0-2Ul2)(2]Uflu - r?w)cDx(O) = jflu o r ml = flu

    rr,H() jUlI.l-[(0-2Ul2)!2] ( . 2)2 JOOI.l-[(0-2 002)(2]'-JJx W = e ]flu - IT W - e ITc D ~ ( O ) = _flu2 - r? o r m2 = flu2 + r?

    -;> c>yar{X] = mz - mi = cr

    (11.2.26)

    (11.2.27)

    (11.2.28)

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    Example 12.3.4 (See Example 12.2.8) The probability density functions tx{x) and g(x)and are given by

    for all x1g(x) = r - 1x(x) = [ if -1.5 < x::: 3]o otherwiseEven though g(x) is defined from - 00 to 00 , the range of t:rtx) plays a critical role in

    detenniningfy(y). Figure 12.2.12 is redrawn in Fig. 12.3.3 to indicate the ranges.1. g(x) satisfies the restrictions.2. Solving for x in y = 1/(x2 - 1), we obtain two roots: Xl = -.jl+Tl;y),

    X2 = .;I-t(1I.Y).3. I d ~ ) 1 = I ~ ( r 1)1 = 1-2(r 1il = 21f + 4. fy(y) have to be calculated in the four ranges of y as shown in Fig. 12.3.3.The density functionfy(y) is calculated as follows:Region 1: y ::: - 1, two points of intersection:

    1 (22)2 1fy(y) = IF 11 9'+'9 = 9 If 112y2 1+- y2 1+-Y Y

    R . nil . f ' .eglOn : - < Y ::: 8' no pomts 0 mtersectlOn:Jy(y) = 01 4Region llI: 8' < Y ::: S' one point of intersection:1 (2) 1 1Jy(y) = If 11 9 =9 If 11l y2 1+- y2 1+-y y

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