Dominant term in convergence in MSE The speed of convergence in MSE is typically dominated by the variance. 
Consider the ML estimator of the variance: Its variance decays as $\mathcal{O}(1/n)$ and the bias term decays as $\mathcal{O}(1/n^2)$.  
Would it be true to assert that the variance will always be the dominant term in the convergence rate?
 A: I am not sure such a general statement can be made.
Consider the standard case of an i.i.d. sample of $n$ Uniforms $U(0,\theta)$. The MLE of $\theta$ is $\hat \theta = X_{(n)}$, the maximum order statistic, and consequently, the density of the MLE is 
$$f(\hat \theta) = \frac {n}{\theta^n}\hat \theta^{n-1}$$
So
$$E(\hat \theta^2) = \int_{0}^{\theta}\hat \theta^2\frac {n}{\theta^n}\hat \theta^{n-1}d\hat \theta = \frac {n}{(n+2)\theta^n}\hat \theta^{n+2}\Big|_0^{\theta} = \frac {n}{n+2}\theta^2$$
and
$$E(\hat \theta) = \int_{0}^{\theta}\hat \theta\frac {n}{\theta^n}\hat \theta^{n-1}d\hat \theta = \frac {n}{(n+1)\theta^n}\hat \theta^{n+1}\Big|_0^{\theta} = \frac {n}{n+1}\theta$$
Therefore
$$\operatorname{Var}(\hat \theta) = E(\hat \theta^2) -[E(\hat \theta)]^2 = \frac {n}{n+2}\theta^2 - \left(\frac {n}{n+1}\right)^2\theta^2 = \frac n{(n+2)(n+1)^2}\theta^2$$
while
$$[B(\hat \theta)]^2 = \left (E(\hat \theta) - \theta\right)^2 = \left (\frac {n}{n+1} - 1\right)^2\theta^2 = \frac 1{(n+1)^2}\theta^2$$
So both are $\mathcal{O}(1/n^2)$ here.
A: A late reply, but nevertheless. In nonparametric statistic, the rate of the bias of a kernel point-estimator of a density is $O(h^4)$ and of the variance, it is $O(1/(nh))$, where $n$ is the sample size and $h$ the bandwidth. This implies that the which term dominates depends on our choice. If we choose $h$ to go very quickly to zero, then the variance is the dominating part (undersmooth case). If we oversmooth, then the bias dominates.
