# 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.