For independent and identically distributed samples $[y_1,...,y_m]$ where $y$ is uniformly distributed between $[0,\theta]$ with $0 \lt \theta \lt \infty$, finding the method of moments estimator for $\theta$ is very straightforward using the first moment, $E[Y]$, and its natural estimator.
I am interested in knowing whether it possible to use the natural estimator for the second moment, $\frac{1}{m}\sum_{i=1}^{m}y_i^2$, to arrive a the same estimate of $\theta$ and whether or not its properties are the same as the estimator using the first moment.
Since $Var(Y) = \frac{\theta^2}{12} = E[Y^2] - E^2[Y] = E[Y^2] - \frac{\theta^2}{4}$ we have that $E[Y^2] = \frac{\theta^2}{3}$.
Then using the method of moments theory and the known support for $\theta$, the invertible function operating on $E[Y^2]$ to produce $\theta$ is $\sqrt{3E[Y^2]}$. Because we do not know the exact value of $E[Y^2]$, we use its natural estimator given above to produce the estimate. $$\hat{\theta} = \sqrt{\frac{3}{m}\sum_{i=1}^{m}y_i^2}$$
I would also like to show whether or not this is biased, but I am not sure if the way I would like to go about it is correct. I have that... $$E[\hat{\theta}] = \sqrt{\frac{3}{m}} E\left[\sqrt{\sum_{i=1}^{m}y_i^2}\right] $$
$$\sqrt{\frac{3}{m}}E\left[\sqrt{\sum_{i=1}^{m}y_i^2}\right] = \sqrt{\frac{3}{m}}\sqrt{\frac{m\theta^2}{3}} = \theta$$
Particularly, I am unsure of whether the following equality holds which I see as being the only way that the estimator could be unbiased. $$E\left[\sqrt{\sum_{i=1}^{m}y_i^2}\right] = \sqrt{\frac{m\theta^2}{3}}$$