Are these moments estimators asymptotic unbiased? In this paper, authors consider method of moments of fitting Gumbel distribution:
We know that maximal likelihood estimators are asymptotic unbiased. But are these moments estimators asymptotic unbiased? Or consistent estimators?
Can we find the reference to say these two estimators are asymptotic normal?
 A: Yes, they are.
Equation 4 in the paper says
$$E[X]=\mu+\gamma\sigma$$
and
$$\mathrm{var}[X]= \pi^2\sigma^2/6$$
and rearranging these gives the formulas for $\mu$ and $\sigma$ as functions of the true moments
The sample moments are consistent for the true moments, so these smooth functions of the sample moments are consistent for the same functions of the true moments
A: Recall that the Gumbell distribution has finite moments. As a result, the law of large numbers implies that
$$\frac1n\sum_{i=1}^n X_i \overset p\to \mathbb E[X],\quad \frac1n\sum_{i=1}^n X_i^2 \overset p\to \mathbb E[X^2]$$
Moreover, the CLT implies that
$$\begin{pmatrix}\frac1{\sqrt n}\sum_{i=1}^n X_i - \mathbb E[X] \\ \frac1{\sqrt n}\sum_{i=1}^n X_i^2 - \mathbb E[X^2]\end{pmatrix} \overset d\to \mathcal N(0,\Sigma)$$
where
$$\Sigma = \begin{pmatrix}\mathrm{Var}(X) & \mathrm{Cov}(X,X^2)\\ \mathrm{Cov}(X,X^2) & \mathrm{Var}(X^2)\end{pmatrix}$$
Given this information, we are ready to calculate the asymptotic distributions of $\hat\mu$ and $\hat \sigma$. First, let us calculate the probability limit of $s$. We have that
$$s = \sqrt{\frac1n\sum_{i=1}^n X_i^2 - \left(\frac1n\sum_{i=1}^n X_i\right)^2} \equiv f\left(\frac1n\sum_{i=1}^n X_i^2, \frac1n\sum_{i=1}^n X_i\right)$$
where by definition, the function $f$ is $f(x,y) = \sqrt{x - y^2}$. Note that $f$ is a continuous function of its arguments so by the continuous mapping theorem, we have that
$$s = \sqrt{\frac1n\sum_{i=1}^n X_i^2 - \left(\frac1n\sum_{i=1}^n X_i\right)^2} \overset p\to \sqrt{\mathbb E[X^2] - \mathbb E[X]^2} = \sqrt{\mathrm{Var}(X)} = \pi \sigma / \sqrt 6$$
But this implies that
$$\hat\sigma = \frac{s\sqrt 6}{\pi}\overset p\to \pi \sigma / \sqrt 6\times \sqrt 6 / \pi = \sigma$$
By similar reasoning, we know that
$$\hat\mu \overset p\to \mathbb E[X] - \gamma\sigma = \mu + \gamma\sigma - \gamma\sigma = \mu$$
This shows that $\hat \mu,\hat\sigma$ are consistent. Next, we can apply delta method to show asymptotic normality of $\hat\mu,\hat\sigma$. Specifically, we note that we can represent
$$\begin{pmatrix}\hat\mu\\\hat\sigma\end{pmatrix} = G\left(\frac1n\sum_{i=1}^n X_i, \frac1n\sum_{i=1}^n X_i^2\right)$$
where $G$ is a nice and differentiable function of its arguments. The delta method thus shows that
$$\sqrt n\left[\begin{pmatrix}\hat\mu\\\hat\sigma\end{pmatrix} - \begin{pmatrix}\mu\\\sigma\end{pmatrix}\right]\overset d\to \mathcal N(0,\nabla G(\mathbb E[X],\mathbb E[X^2])^T\Sigma\nabla G(\mathbb E[X],\mathbb E[X^2]))$$
This last result shows asymptotic normality and asymptotic unbiasedness of $(\hat\mu,\hat\sigma)$. I will leave it to you to specify what $G$ is and to derive what the resulting asymptotic variance of $(\hat\mu,\hat\sigma)$.
