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Maximum of dependent exponential random variables

  1. Jul 21, 2008 #1
    Pdf (or mgf) of maximum of dependent exponential random variables ?

    max of Z1, Z2, Z3, Z4

    where

    Z1 = |X1+X2+X3|^2 + |Y1+Y2+Y3|^2
    Z2 = |X1-X2+X3|^2 + |Y1-Y2+Y3|^2
    Z3 = |X1+X2-X3|^2 + |Y1+Y2-Y3|^2
    Z4 = |X1-X2-X3|^2 + |Y1-Y2-Y3|^2

    Xi, Yi are independent zero mean normal with variance 1/2.

    So, Z1,Z2,Z3,Z4 are indentically distributed exponential random variables,
    But they are correlated.

    Can anybody help me?
     
  2. jcsd
  3. Jul 21, 2008 #2
    Sorry are the Xis and Yis independent of each other or are they all independent of everything else?

    If they are all independent of each other then you have a Chi square as the sums of normals are a normal, and square of a normal is chi-squared and sums of chi-squares are chi-squared.

    Hope thats helpful
     
  4. Jul 21, 2008 #3
    Thank you. Xis and Yis are all independent of each other.
    Yes. I have 4 Chi-square RVs. They are dependent.
    Can I find the pdf of the maximum of those?
     
  5. Jul 22, 2008 #4
    Here is a symbolic description of the solution to the problem, hope it makes
    any sense:


    You want to calculate the PDF of a variable [tex]Z_{\max } = \max \left( {Z_1
    ,Z_2 ,Z_3 ,Z_4 } \right)[/tex] where
    [tex]
    \begin{array}{l}
    Z_1 = f_1 ({\rm {\bf X,Y}}) = \left| {X_1 + X_2 + X_3 } \right|^2 + \left|
    {Y_1 + Y_2 + Y_3 } \right|^2 \\
    Z_2 = f_2 ({\rm {\bf X,Y}}) = \left| {X_1 - X_2 + X_3 } \right|^2 + \left|
    {Y_1 - Y_2 + Y_3 } \right|^2 \\
    Z_3 = f_3 ({\rm {\bf X,Y}}) = \left| {X_1 + X_2 - X_3 } \right|^2 + \left|
    {Y_1 + Y_2 - Y_3 } \right|^2 \\
    Z_4 = f_4 ({\rm {\bf X,Y}}) = \left| {X_1 - X_2 - X_3 } \right|^2 + \left|
    {Y_1 - Y_2 - Y_3 } \right|^2 \\
    \end{array}
    [/tex]


    You can write a symbolic solution to the PDF of any function [tex]Z=f(X)[/tex] of a stochastic variable [tex]X[/tex] as
    [tex]
    p(z) = \int {p({ {x}})\delta \left( {z - f \left(
    {x} \right)} \right)d{ {x}}}
    [/tex]
    where [tex]\delta(z)[/tex] is Dirac's delta function.

    Thus, the PDF of [tex]Z_{\max } = \max \left(
    {Z_1 ,Z_2 ,Z_3 ,Z_4 } \right)[/tex] can be written as


    [tex]
    p(z_{\max } ) = \int {p({\rm {\bf z}})\delta \left( {z_{\max } - \max \left(
    {z_1 ,z_2 ,z_3 ,z_4 } \right)} \right)d{\rm {\bf z}}} \qquad , \qquad (1)
    [/tex]
    where
    [tex]
    p({\rm {\bf z}}) = \int {\int {p({\rm {\bf x}},{\rm {\bf y}})\prod\limits_{k
    = 1}^4 {\delta \left( {z_k - f_k ({\rm {\bf x}},{\rm {\bf y}})} \right)}
    d{\rm {\bf x}}} d{\rm {\bf y}}}
    [/tex]

    [tex]
    p({\rm {\bf x}},{\rm {\bf y}}) = p(x_1 )p(x_2 )p(x_3 )p(y_1 )p(y_2 )p(y_3 )
    [/tex]

    and integration with respect to a vector
    [tex]\int { \cdot d{\rm {\bf x}}} [/tex]
    stands for integration over all components:
    [tex]\int\limits_{ - \infty }^\infty
    {\int\limits_{ - \infty }^\infty {\int\limits_{ - \infty }^\infty { \cdot
    dx_1 dx_2 dx_3 } } } [/tex]

    If you had already calculated [tex]p({\rm {\bf z}})[/tex], the integral in (1) can be
    calculated by dividing the four-dimensional space of [tex]{\rm {\bf z}}[/tex] into
    four parts, over each of which [tex]\max \left( {z_1 ,z_2 ,z_3 ,z_4 } \right)[/tex]
    is linear. The integral them becomes a sum of four parts, albeit with
    slightly complicated bounds.

    The resulting expression is likely to be quite messy, I'd recommend using an
    analytic math program (e.g. Matematica or Maple) to compute it.
     
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