# Is this a typo on P.75, Theorem 5.52 of the book "Asymptotic Statistics" by Van der Vaart?

Let $$\Theta$$ be a compact metric space, $$\theta \in \Theta.$$ Let $$m_{\theta}:\mathbb{R}^d\to \mathbb{R}: x\mapsto m_{\theta}(x)$$ be a family of measurable function indexed by $$\theta \in \Theta.$$ Let $$P(f):=E[f(X)],$$ where $$X:\Omega \to \mathbb{R}^d, f:\mathbb{R}^d\to \mathbb{R}.$$ So below, $$P(m_{\theta}-m_{\theta_0})$$ should be $$E[m_{\theta}(X)-m_{\theta_0}(X)].$$ Also below:

$$\theta_0:= arg max_{\theta \in \Theta} P(m_\theta(X))= arg max_{\theta \in \Theta} E[m_{\theta}(X)].$$

I'm having to look into a few results in M-estimation from the book "Asymptotic Statistics (2000)" by Van der Vaart, and I'm looking at this theorem on P.75: (I think) here $$\theta_0$$ is a value of the parameter $$\theta$$ that maximizes the maximum likelihood estimate. Below,

It seems to me that sup in the first line is a typo, as the quantities on the LHS there are nonpositive and bounded by something strictly negative, however, at $$\theta=\theta_0,$$ the quantities on the LHS achieves $$0,$$ that's not $$\le C\delta^{\alpha}.$$ So did he intend to write inf here instead of sup?

First point: Since $$d(\theta, \theta_0)<\delta$$ it should be possible to select $$\theta = \theta_0$$ which gives $$\sup=0>-C\cdot0^\alpha$$.
Without being specific for this problem, if $$x^*=\arg\max a(x)$$ then $$a(x) - a(x^*) \leq 0$$ Taking inf typically means that an upper bound is easily achievable. The posted requirement (with the modification $$d(\theta, \theta_0)>\delta$$ ) seems reasonable: $$\sup_{|x-x^*|>\delta} a(x) - a(x^*) \leq C\delta^\alpha;$$ when $$x$$ is not identical to $$x^*$$, there will be a distance between the function values.
• Thanks! But then I wonder if $\sup_{|x-x^*|>\delta} a(x) - a(x^*) \leq C\delta^\alpha;$ will help us prove the theorem that book seems to prove. But the line I pointed out in my OP does have a typo, correct? Dec 2, 2023 at 23:48
• I do not know the context and can not help with those questions. But, if $C>0$ and $m$ is allowed to be linear then it looks like the first assumption never can be true. Dec 2, 2023 at 23:56