the covariance of an odd/even function

m and z have the following PDF:

$f(m,z) = (2\pi^3)^{-1/2}*(m^2+z^2)^{-1/2}*e^{-(m^2+z^2)/2}$

I want to find whether m and z are independent which I'll show through the covariance, which is E[mz]-E[m]E[z]. I know that $E[mz] = 0$ because f(m,z) is an odd function.

The E[z] is = $\int_{-\infty}^{\infty} {1 \over \,\sqrt{\,2\pi^{3}\,}\,} e^{-m^2/2}$ [ $\int_{-\infty}^{\infty} z*(m^2+z^2)^{-1/2} e^{-m^2/2}dz$]dm

which also equals 0. Meaning that m and z have 0 covariance and independent.

Can I extrapolate from this and assume that any odd function (or even function) is always independent with 0 covariance?

Hang on, your joint density isn't odd, it's even in its two arguments. This is enough to imply that the covariance is zero. Actually, being even in one argument for every choice of the other argument is enough to deduce zero covariance: if $f(m,z)=f(-m,z)$ for every $m$ and $z$, then
$$E(M)=\int mf(m) dm = \int m\left( \int f(m,z)dz\right)dm = \int \left(\int mf(m,z)dm\right)dz = 0$$ and $$E(MZ)=\iint mz f(m,z)dm\,dz = \int z\left(\int mf(m,z)dm\right)dz = 0.$$
• Calculate $f_M(m):=\int f(m,z)dz$ and $f_Z(z):=\int f(m,z)dm$. Then the variables $M, Z$ are independent if and only if $f(m,z)=f_M(m)f_Z(z)$. It doesn't look like this is the case for your $f$, so I conclude you don't have independence. – grand_chat Oct 27 '16 at 21:55
• Correct, except you should write $m$ instead of $x$. – grand_chat Oct 28 '16 at 15:58