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

  • Thread starter lep11
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  • #1
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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'?
 

Answers and Replies

  • #2
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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.
 
  • #3
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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.
Did you arbitrarily pick the values of x,y and z? I checked and the equation holds. Is the original post o.k.?
 
  • #4
35,635
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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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