CDF of correlated mixed random variables

  • Thread starter nikozm
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  • #1
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Hello,

i m trying to evaluate the following:

r*x - r*y ≤ g, where r,x,y are nonnegative random variables of different distribution families and g is a constant nonnegative value.

Then, Pr[r*x - r*y ≤ g] = Pr[r*x ≤ g + r*y] = ∫ Fr x(g + r*y)*fr*y(y) dy, where F(.) and f(.) denote CDF and PDF, respectively.

The above formula works for independent variables. I m not sure if it works for correlated variables (the above are correlated due to the "r" variable in both "r*x" and "r*y").

Any help would be useful.

Thanks
 

Answers and Replies

  • #2
35,647
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I would express the inequality as ##r(x-y) \leq g## and find the distribution for x-y first, afterwards you just have two variables left.
 
  • #3
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However, the distribution of (x-y) is unreachable..Is there an other way to solve the problem ?

Thanks
 
  • #4
Stephen Tashi
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However, the distribution of (x-y) is unreachable..Is there an other way to solve the problem ?

Thanks

I think you'll get a better answer if you explain completely what facts are known about the random variables.
 
  • #5
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Ok, assume that x and y are independent and identically distributed variables following a PDF as given bellow:
fz(z)=Exp[-1/z]/(z^2), where z ε {x,y}

(Thus, they follow an inverse exponential distribution..)
 
  • #6
35,647
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I don't understand what is unreachable about the difference of two of those distributions.
I guess it is not a nice expression, but it is still something you can write down.
 
  • #7
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The difference of (x-y) when both of them (x and y) follow the above distribution (i.e., see fz(z)) can not be obtained in closed-form. Moreover, both x and y are i.i.d. and real positive random numbers.

What else...?
 
  • #9
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Ok, assume that (x-y) ≥ 0, and thus r*(x-y) ≥ 0.

Thanks
 
  • #10
Stephen Tashi
Science Advisor
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Ok, assume that (x-y) ≥ 0,

If that is assumed then x and y are not independent.
 

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