I have gradually come to appreciate how much insight can be found in Asymptotic Statistics by van der Vaart’s (henceforth vdV). I have come to appreciate this only gradually because I have only gradually become able to actually read the damn thing. I am still able to read it only with some effort, an effort which is usually worthwhile. But I often wish there were some kind of Cliff’s Notes for the book.

In the sections that I read carefully, I fill the margins with notes to make things easier for myself the next time I revisit the book. For a few key sections I’m finding that these notes need more space than I have (books like Asymptotic statistics should be printed with bigger margins). So I thought I’d type up the points that I found difficult, and then I thought I might post these write-ups here in case someone else also finds them useful. I can’t guarantee that my explanations are correct, since I do find this material a little difficult at times. And of course, my notes are not going to be adequate for everyone, since they are targeted to the areas where I feel personally shaky. Also some of my notes are obvious and not insightful, but I’m including them because I think I will find them useful the next time I read the proof. If anything here is useful to you I’d be glad, and if you find mistakes I would also be happy to hear about them.

For this post I’m going to address the proof of Theorem 7.2. My real goal is the proof of Theorem 10.1, which I will leave for a future (longer) post, but which relies heavily on Theorem 7.2. I am looking at the Cambridge University Press 2007 edition.

### Definitions.

I find it really useful to have definitions written out in one place rather than within the text. So here are the definitions for the proof of Theorem 7.2:

\begin{aligned} I_{\theta} & :=P_{\theta}\dot{\ell}_{\theta}\dot{\ell}_{\theta}^{T}\textrm{ (covariance of the score under }\theta\mathrm{)}\\ p_{n} & :=p_{\theta+h_{n}/\sqrt{n}}\\ p & :=p_{\theta}\\ g & :=h^{T}\dot{\ell}_{\theta}\\ W_{ni} & :=2\left(\sqrt{\frac{p_{n}}{p\left(X_{i}\right)}}-1\right). \end{aligned}

Recall that van der Vaart uses the notation that, for a random variable $$x$$ distributed according to the distribution $$P$$,

\begin{aligned} Px & =\mathbb{E}_{P}\left[x\right]=\int xdP\left(x\right).\end{aligned}

### The sequence $$h_n$$.

Theorem 7.2 holds for any $$h_n$$ congerving to $$h$$, and the log likelihood is evaluated at $$\theta + h_n / \sqrt{n}$$. This is as much as to say that the theorem holds for any sequence $$\theta_n$$ such that $$\sqrt{n} (\theta_n - \theta) \rightarrow h$$. Of course, this necessarily implies that $$\theta_n \rightarrow \theta$$, so that the Theorem only governs the behavior of the log likelihood ratio in a neighborhood of $$\theta$$. In theorem 10.1, vdV will simply take $$h_n \equiv h$$.

### “By continuity of the inner product...” (third equation on page 94).

Here, vdV is referring to the inner product between random variables which are measurable with respect to the dominating measure, $$\mu$$. Let $$x$$ and $$y$$ be two such variables, then the inner product is

\begin{aligned} \left\langle x,y\right\rangle & :=\int xyd\mu. \end{aligned}

This inner product defines a Hilbert space that’s referred to as $$L_{2}\left(\mu\right)$$ (because $$\sqrt{\left\langle x,x\right\rangle }=\sqrt{\int x^{2}d\mu}$$ is the $$L_{2}$$ norm of $$x$$ and $$\mu$$ is the measure with respect to which the inner product is taken). Any inner product is continuous in its arguments (using its own topology) since

\begin{aligned} \left\langle x+\epsilon_{n},y+\delta_{n}\right\rangle & =\left\langle x,y\right\rangle +\left\langle \epsilon_{n},y\right\rangle +\left\langle x,\delta_{n}\right\rangle +\left\langle \epsilon_{n},\delta_{n}\right\rangle \\ & \le\left\langle x,y\right\rangle +\left\Vert y\right\Vert _{2}\left\Vert \epsilon_{n}\right\Vert _{2}+\left\Vert x\right\Vert _{2}\left\Vert \delta_{n}\right\Vert _{2}+\left\Vert \epsilon_{n}\right\Vert _{2}\left\Vert \delta_{n}\right\Vert _{2}.\end{aligned}

He is using the decomposition

\begin{aligned} \frac{1}{2}g\sqrt{p} & =\sqrt{n}\left(\sqrt{p_{n}}-\sqrt{p}\right)+\epsilon_{n}\\ 2\sqrt{p} & =\sqrt{p}+\sqrt{p}=\sqrt{p}+\sqrt{p_{n}}+\delta_{n},\end{aligned}

where $$\left\Vert \epsilon_{n}\right\Vert _{2}$$ and $$\left\Vert \delta_{n}\right\Vert _{2}$$ both go to zero in $$L_{2}\left(\mu\right)$$ by Eq. 7.1.

Often in probability you need to use something like the dominated convergence theorem to exchange integration and limits like this, but here the inner product structure makes things easier.

### Equation 7.4.

\begin{aligned} \mathbb{E}\left[\sum_{i=1}^{n}W_{ni}+\frac{1}{4}Pg^{2}\right] & =o\left(1\right).\end{aligned}

\begin{aligned} \mathbb{E}\left[\frac{1}{\sqrt{n}}\sum_{i=1}^{n}g\left(X_{i}\right)\right] & =\frac{n}{\sqrt{n}}\mathbb{E}\left[g\right]=0.\end{aligned}

Combining gives Equation 7.4.

### Taylor expansion of the logarithm (top of page 95).

Recall that $$W_{ni}$$ are all scalars, so we only need the univariate Taylor expansion of $$\log\left(1+z\right)$$, which we will evaluate at $$z=W_{ni}/2$$ and then sum. For $$\left|z\right|<1$$, by the intermediate value theorem, for some $$0\le\bar{z}\le z$$ which is a function of $$z$$,

\begin{aligned} \log\left(1+z\right) & =z-\frac{1}{2}\frac{1}{\left(1+\bar{z}\right)^{2}}z^{2}\\ & =z-\frac{1}{2}z^{2}+\frac{1}{2}\left(1-\frac{1}{\left(1+\bar{z}\right)^{2}}\right)z^{2}.\end{aligned}

Here, he is taking $$R\left(z\right)=\frac{1}{2}\left(1-\frac{1}{\left(1+\bar{z}\right)^{2}}\right)$$. We don’t know what $$\bar{z}$$ is, but we know that

\begin{aligned} \lim_{z\rightarrow0}R\left(z\right) & =\frac{1}{2}\left(1-\frac{1}{\left(1+\lim_{z\rightarrow0}\bar{z}\right)^{2}}\right)=0,\end{aligned}

which is the key property we will need. It is this property that vdV is referring to at the end of the proof when saying “By the property of the function $$R$$...”

### “It is possible to write $$nW_{ni}^{2}=g^{2}\left(X_{i}\right)+A_{ni}$$...”

At first this flummoxed me, since Eqs. 7.3 and 7.4 shows only convergence in probability of a term which is linear in $$W_{ni}$$, which does not in general tell you anything about second moments $$W_{ni}^{2}$$. I feel like my justification here is excessive, and I would be very interested to hear if there’s a more elegant way to see this result.

Here we have special extra information, however. For one, we already have assumed that $$Pg^{2}=\mathbb{E}\left[g^{2}\left(X_{i}\right)\right]$$ is finite. We also have that $$\mathbb{E}\left[W_{ni}^{2}\right]=O\left(N^{-1}\right),$$ since

\begin{aligned} \mathbb{E}\left[\left(\frac{1}{2}W_{ni}+1\right)^{2}\right] & =\mathbb{E}\left[\frac{p_{n}\left(X_{i}\right)}{p\left(X_{i}\right)}\right]\\ & =\int\frac{p_{n}\left(X_{i}\right)}{p\left(X_{i}\right)}p\left(X_{i}\right)d\mu\\ & =1,\end{aligned} and that also \begin{aligned} \mathbb{E}\left[\left(\frac{1}{2}W_{ni}+1\right)^{2}\right] & =\frac{1}{4}\mathbb{E}\left[W_{ni}^{2}\right]+\mathbb{E}\left[W_{ni}\right]+1\\ & =\frac{1}{4}\mathbb{E}\left[W_{ni}^{2}\right]+1+O\left(n^{-1}\right)\textrm{ (Second line of Eq. 7.3)}\\ \Rightarrow\mathbb{E}\left[W_{ni}^{2}\right] & =O\left(n^{-1}\right). \end{aligned}

We can thus write

\begin{aligned} \mathbb{E}\left[nW_{ni}^{2}-g^{2}\left(X_{i}\right)\right]^{2} & =\mathbb{E}\left[\left(\sqrt{n}W_{ni}-g\left(X_{i}\right)\right)\left(\sqrt{n}W_{ni}+g\left(X_{i}\right)\right)\right]^{2}\\ & \le\mathbb{E}\left[\left(\sqrt{n}W_{ni}-g\left(X_{i}\right)\right)^{2}\right]\mathbb{E}\left[\left(\sqrt{n}W_{ni}+g\left(X_{i}\right)\right)^{2}\right]\textrm{ (Cauchy-Schwartz)}\\ & =o\left(1\right)\mathbb{E}\left[\left(\sqrt{n}W_{ni}+g\left(X_{i}\right)\right)^{2}\right].\textrm{ (First line of Eq. 7.3)} \end{aligned}

We now need to show that $$\mathbb{E}\left[\left(\sqrt{n}W_{ni}+g\left(X_{i}\right)\right)^{2}\right]$$ does not diverge, which follows from Cauchy Schwartz and the bounds on $$\mathbb{E}\left[g^{2}\left(X_{i}\right)\right]$$ and $$\mathbb{E}\left[W_{ni}^{2}\right]$$:

\begin{aligned} \mathbb{E}\left[\left(\sqrt{n}W_{ni}+g\left(X_{i}\right)\right)^{2}\right] & =n\mathbb{E}\left[W_{ni}^{2}\right]+\mathbb{E}\left[g^{2}\left(X_{i}\right)\right]+2\sqrt{n}\mathbb{E}\left[W_{ni}g\left(X_{i}\right)\right]\\ & =O\left(1\right)+2\sqrt{n}\mathbb{E}\left[W_{ni}g\left(X_{i}\right)\right]\\ & \le O\left(1\right)+2\sqrt{n\mathbb{E}\left[W_{ni}^{2}\right]\mathbb{E}\left[g^{2}\left(X_{i}\right)\right]}\\ & =O\left(1\right). \end{aligned}

We thus have that $$\mathbb{E}\left[nW_{ni}^{2}-g^{2}\left(X_{i}\right)\right]^{2}=o\left(1\right)$$, which is what we wanted to show.