Maximum of dependent exponential random variables

[chased]

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?

 

[Focus]

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 that's helpful

 

[chased]

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?

 

[winterfors]

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 $$Z_{\max } = \max \left( {Z_1<br /> ,Z_2 ,Z_3 ,Z_4 } \right)$$ where
$$\begin{array}{l}<br /> Z_1 = f_1 ({\rm {\bf X,Y}}) = \left| {X_1 + X_2 + X_3 } \right|^2 + \left|<br /> {Y_1 + Y_2 + Y_3 } \right|^2 \\<br /> Z_2 = f_2 ({\rm {\bf X,Y}}) = \left| {X_1 - X_2 + X_3 } \right|^2 + \left|<br /> {Y_1 - Y_2 + Y_3 } \right|^2 \\<br /> Z_3 = f_3 ({\rm {\bf X,Y}}) = \left| {X_1 + X_2 - X_3 } \right|^2 + \left|<br /> {Y_1 + Y_2 - Y_3 } \right|^2 \\<br /> Z_4 = f_4 ({\rm {\bf X,Y}}) = \left| {X_1 - X_2 - X_3 } \right|^2 + \left|<br /> {Y_1 - Y_2 - Y_3 } \right|^2 \\<br /> \end{array}$$You can write a symbolic solution to the PDF of any function $$Z=f(X)$$ of a stochastic variable $$X$$ as
$$p(z) = \int {p({ {x}})\delta \left( {z - f \left(<br /> {x} \right)} \right)d{ {x}}}$$
where $$\delta(z)$$ is Dirac's delta function.

Thus, the PDF of $$Z_{\max } = \max \left(<br /> {Z_1 ,Z_2 ,Z_3 ,Z_4 } \right)$$ can be written as

$$p(z_{\max } ) = \int {p({\rm {\bf z}})\delta \left( {z_{\max } - \max \left(<br /> {z_1 ,z_2 ,z_3 ,z_4 } \right)} \right)d{\rm {\bf z}}} \qquad , \qquad (1)$$
​

where
$$p({\rm {\bf z}}) = \int {\int {p({\rm {\bf x}},{\rm {\bf y}})\prod\limits_{k<br /> = 1}^4 {\delta \left( {z_k - f_k ({\rm {\bf x}},{\rm {\bf y}})} \right)}<br /> d{\rm {\bf x}}} d{\rm {\bf y}}}$$

$$p({\rm {\bf x}},{\rm {\bf y}}) = p(x_1 )p(x_2 )p(x_3 )p(y_1 )p(y_2 )p(y_3 )$$

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

If you had already calculated $$p({\rm {\bf z}})$$, the integral in (1) can be
calculated by dividing the four-dimensional space of $${\rm {\bf z}}$$ into
four parts, over each of which $$\max \left( {z_1 ,z_2 ,z_3 ,z_4 } \right)$$
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.