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How to determine if random variables x,y,z are independent?

  1. Oct 31, 2016 #1
    Let ##f(x,y,z)=x^2e^{-x-xy-xz}##, if ##x,y,z>0## and ##f(x,y,z)=0## otherwise. Are the continuous random variables ##x,y,z## independent or not?

    Intuitively they are not independent. I calculated the marginal density functions:

    ##f_x(x)=\iint_{\Omega} f(x,y,z) dydz=e^{-x}##

    ##f_y(y)=\iint_{\Omega} f(x,y,z) dxdz=(y+1)^{-2}##

    ##f_z(z)=\iint_{\Omega} f(x,y,z) dxdy=(z+1)^{-2}##

    Now we observe that if ##x,y,z>0##,

    ##(x,y,z)=x^2e^{-x-xy-xz}\neq{e^{-x}}(y+1)^{-2}(z+1)^{-2}##. Thus ##x,y,z## are not independent.
    Is this correct?
    Is there easier method to check if they are independent as this way is 'a bit tedious'?
     
  2. jcsd
  3. Oct 31, 2016 #2

    mfb

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    Staff: Mentor

    If they are independent, then f(2,2,1)/f(1,2,1) = f(2,1,1)/f(1,1,1), which is easy to check. Picking the right values is the interesting part, because you don't gain anything if the equality holds.
     
  4. Oct 31, 2016 #3
    Did you arbitrarily pick the values of x,y and z? I checked and the equation holds. Is the original post o.k.?
     
  5. Oct 31, 2016 #4

    mfb

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    2016 Award

    Staff: Mentor

    I picked them arbitrarly. I get $$\frac{4 e^{-8}}{e^{-4}} = \frac{4 e^{-6}}{e^{-3}}$$
    Which is wrong.

    Your method works as well, but it is more work.
     
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