**3. Inference**

**Proposition 6.** *Let T*1, ... , *Tn be a random sample of size n of the T* ∼ MSHN(*<sup>σ</sup>*, *q*) *distribution. Then for q* > 2*, the moment estimators of σ and q are given by*

$$
\hat{\sigma}\_M = \frac{\sqrt{\pi} \,\mathrm{T}}{2^{\frac{1}{q} + \frac{1}{2}} \Gamma\left(\frac{\bar{q}\_M - 1}{\bar{q}\_M}\right)} ,\tag{9}
$$

$$2\pi \overline{T}^2 \Gamma\left(\frac{\hat{q}\_M - 2}{\hat{q}\_M}\right) - 2\overline{T^2} \Gamma^2\left(\frac{\hat{q}\_M - 1}{\hat{q}\_M}\right) = 0,\tag{10}$$

*where T is the mean of the sample and T*<sup>2</sup> *is the mean of the sample for the square of the observations.*

**Proof.** From Proposition 5, and considering the first two equations in the moments method, we have

$$
\overline{T} = \frac{2^{\frac{1}{q} + \frac{1}{2}}}{\sqrt{\pi t}} \sigma \Gamma\left(\frac{q - 1}{q}\right) \quad \text{and} \quad \overline{T^2} = 2^{\frac{2}{q}} \sigma^2 \Gamma\left(\frac{q - 2}{q}\right).
$$

.

Solving the first equation above for *σ* we obtain *σM* given in (9). Substituting *σM* in the second equation above, we obtain the result given in (10).
