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

1. Oct 31, 2016

lep11

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. Oct 31, 2016

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.

3. Oct 31, 2016

lep11

Did you arbitrarily pick the values of x,y and z? I checked and the equation holds. Is the original post o.k.?

4. Oct 31, 2016

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.