3rd order moment of normal random variable

Question

Let $h=[h_1\,\, h_2]'$ be a normal random vector such that $$h\sim \mathcal{N}(h;\, 0,\, \text{diag(1/4,1/4)})$$ since $h$ is distributed symmetrically with respect its mean, is it correct to say that the 3rd order momements are zero? $$\mathbb{E}[h_1^3]=\mathbb{E}[h_1^2 h_2]=\mathbb{E}[h_1 h_2^2]=\mathbb{E}[h_2^3]=0 ?$$

Answer 1

I expand my comment into a full-fledged answer. We consider the first moment of 3rd order, \begin{align} E[h_1^3] = \int_{-\infty}^{\infty} h_1^3 c_0 \exp\left(-\frac{h_1^2}{8}\right) d h_1 \\ = -\int_0^{\infty} h_1^3 c_0 \exp\left(-\frac{h_1^2}{8}\right) d h_1 + \int_{0}^{\infty} h_1^3 c_0 \exp\left(-\frac{h_1^2}{8}\right) d h_1 \\ = 0, \end{align} and using similar arguments we arrive at the conclusion that $$ E[h_2^3] = 0. $$ Now for $ E[h_1^2 h_2] $ we must use the full joint distribution to compute the moment. \begin{align} E[h_1^2 h_2] = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} h_1^2 h_2 \exp\left(-\frac{1}{2}\begin{pmatrix} h_1 & h_2 \end{pmatrix} \begin{pmatrix} 1/4 & 0 \\ 0 & 1/4 \end{pmatrix} \begin{pmatrix} h_1 \\ h_2 \end{pmatrix}\right) dh_1 dh_2 \\ = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} h_1^2 h_2 \exp\left(-\frac{h_1^2}{8} - \frac{h_2^2}{8}\right) dh_1 dh_2 \\ = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} h_1^2 h_2 \exp\left(-\frac{h_1^2}{8}\right)\exp \left( - \frac{h_2^2}{8}\right) dh_1 dh_2 \\ = \int_{-\infty}^{\infty} h_2 \exp \left( - \frac{h_2^2}{8}\right) \underbrace{\int_{-\infty}^{\infty} h_1^2 \exp\left(-\frac{h_1^2}{8}\right) dh_1}_{=c} dh_2 \\ = c \int_{-\infty}^{\infty} h_2 \exp \left( - \frac{h_2^2}{8}\right) d h_2 \\ = 0 \end{align} We do a similar operation for $E[h_2^2 h_1]$. This operation is identical to the above with $h_1, h_2$ interchanged, but yields the conclusion that $$ E[h_2 h_1] = 0. $$