This post is a continuation of the previous post about van der Vaart’s Theorem 7.2. As before, these are just my personal notes, with no guarantee of correctness nor insight. (In fact, if you’re reading this and find an error or misunderstanding, I would love to hear from you!)

## Notation

\begin{aligned} h & =\sqrt{n}\left(\theta-\theta_{0}\right)=\textrm{Rescaled parameter}\\ \Pi_{n} & =\textrm{Prior on }h\textrm{ (changes with n because the scaling changes)}\\ \Pi_{n}^{C} & =\textrm{Prior on }h\textrm{ conditioned on set }C\\ \bar{H}_{n} & =\sqrt{n}\left(\bar{\Theta}_{n}-\theta_{0}\right)\textrm{ (random variable taking values }h\textrm{)}\\ P_{\bar{H}_{n}\vert\vec{x}_{n}} & =\textrm{Posterior of rescaled parameter}\\ P_{\bar{H}_{n}\vert\vec{x}_{n}}^{C} & =\textrm{Posterior of rescaled parameter conditioned on set }C\\ P_{n,h} & =\textrm{Distribution of the data given that the rescaled parameter is }h\\ P_{n,C} & =P_{n,h}\textrm{ where }h\textrm{ is averaged over the prior }\Pi_{n}^{C}.\end{aligned}

I find vdV’s notation kind of cumbersome. I kind of hate to do it, but I find it much easier to follow vdV’s reasoning in my own notation, which is as follows.

\begin{aligned} \bar{H}_{n} & \rightarrow h\quad\textrm{ (I don't distinguish between the random variable and its values)}\\ \Pi_{n} & \rightarrow P\left(h\right)\quad\textrm{ (I don't notate the n dependence)}\\ \Pi_{n}^{C} & \rightarrow P\left(h\vert C\right)\\ P_{\bar{H}_{n}\vert\vec{x}_{n}} & \rightarrow P\left(h\vert x\right)\\ P_{\bar{H}_{n}\vert\vec{x}_{n}}^{C} & \rightarrow P\left(h\vert x,C\right)\\ P_{n,h} & \rightarrow P\left(x\vert h\right)\\ P_{n,C} & \rightarrow P\left(x\vert C\right)=\int P\left(x\vert h\right)P\left(h\vert C\right)dh.\end{aligned}

## Regions

The proof divides the domain of $$h$$ (and, equivalently, $$\theta$$) into several overlapping regions. I found it helpful to make a list of all these different divisions. The main division is

\begin{aligned} R_{1} & =h:\left\Vert h\right\Vert _{2}\le M_{n}:=C_{n}\\ R_{23} & =h:\left\Vert h\right\Vert _{2}>M_{n}:=C_{n}^{c}\end{aligned}

for $$M_{n}\rightarrow\infty$$. The proof is in two distinct parts: one part for $$R_{23}$$, which shows that the integral over $$R_{23}$$ goes to zero by the testability assumption for any $$M_{n}\rightarrow\infty$$, and one part for $$R_{1}$$, which shows that the integral over $$R_{1}$$ goes to zero for any fixed $$M$$, and so for some $$M_{n}\rightarrow\infty$$. Conceptually, the integral over $$R_{1}$$ determines the rate at which $$M_{n}\rightarrow\infty$$, and the proof for $$R_{23}$$ works for that particular $$M_{n}$$. As long as $$M_{n}\rightarrow\infty$$, then $$R_{1}$$ eventually covers the whole $$h$$ space.

Obviously, the rate at which $$M_{n}$$ goes to infinity will determine the rate at which various parts of the proof go to zero. $$M_{n}$$ going to infinity fast is good for $$R_{23}$$ and bad for $$R_{1}$$, and vice-versa. Note that the proof does not actually imply that the integral over region $$R_{23}$$ goes to zero exponentially fast for any $$M_{n}\rightarrow\infty$$! On the first term on the RHS of the third equation of page 142 is of order $$e^{-n}$$. The $$o\left(1\right)$$ term is not guaranteed to have any particular rate by the proof as-written. However, as I will discuss below, I believe you can choose $$M_{n}$$ carefully so that $$M_{n}$$ grows more slowly than $$\sqrt{n}$$ but still to ensure that this term also goes to zero exponentially fast, though I’m not sure I see how to guarantee that such an $$M_{n}$$ would work for $$R_{1}$$.

In the sixth paragraph of page 142, the region $$R_{23}$$ is split into two regions:

\begin{aligned} R_{2'} & =h:M_{n}\le\left\Vert h\right\Vert _{2}\le\sqrt{n}D\\ R_{3'} & =h:\sqrt{n}D\le\left\Vert h\right\Vert _{2}.\end{aligned}

Note that this is the same split as in Lemma 10.3, but with $$D$$ instead of $$\varepsilon$$. Note that $$D$$ is chosen to uniformly bound the prior in a region of $$\theta_{0}$$; $$\varepsilon$$ is chosen for different reasons as we soon discuss.

Lemma 10.3 also divides $$R_{23}$$ into two further regions:

\begin{aligned} R_{2}= & h:M_{n}\le\left\Vert h\right\Vert _{2}\le\sqrt{n}\varepsilon\\ R_{3}= & h:\sqrt{n}\varepsilon\le\left\Vert h\right\Vert _{2}\\ R_{23} & =R_{2}\bigcup R_{3}.\end{aligned}

The constant $$\varepsilon$$ is chosen to control the first-order Taylor series expansion of $$P_{\theta}\dot{\ell}_{\theta_{0}}^{L}$$.

Of course, if $$M_{n}$$ grows faster than a $$\sqrt{n}$$ rate, then $$R_{2}$$ and $$R_{2'}$$ will be empty for sufficiently large $$n$$. The division is only necessary when $$M_{n}$$ grows more slowly than $$\sqrt{n}$$. And indeed this is the interesting case, as we can see when we consider the corresponding regions for $$\theta$$:

\begin{aligned} R_{1} & =\theta:\left\Vert \theta-\theta_{0}\right\Vert _{2}\le\frac{M_{n}}{\sqrt{n}}\\ R_{2} & =\theta:\frac{M_{n}}{\sqrt{n}}\le\left\Vert \theta-\theta_{0}\right\Vert _{2}\le\varepsilon\\ R_{3} & =\theta:\varepsilon\le\left\Vert \theta-\theta_{0}\right\Vert .\end{aligned}

Again, recall that the integral over $R_{2}\bigcup R_{3}$ goes to zero exponentially fast. If $$M_{n}=o\left(\sqrt{n}\right)$$, this means the region that defines the integral to any order higher than the integral over $$R_{23}$$ (which I believe can be made arbitrarily fast) is determined by a region in $$\theta$$ space that shrinks to a point. If we want to do something more interesting with our BCLT than simply show convergence in total variation (e.g., to use the Taylor series expansions of the relevant terms to develop an asymptotic expansion of a posterior expectation), then we need $$R_{1}$$ to shrink to zero. So slowly growing $$M_{n}$$ will be the hard and interesting case.

## Bottom of page 141 “For $$U$$, a ball of fixed radius...”

Recall that $$U$$ is the domain of $$h$$, so $$h\in U\Rightarrow\left\Vert \theta-\theta_{0}\right\Vert \in U/\sqrt{n},$$ and fixed $$U$$ implies a shrinking ball for $$\theta$$. Note that contiguity will be invoked once for $$R_{23}$$ and once for $$R_{1}$$. For region $$R_{23}$$, the region $$U$$ is a bit of a red herring, chosen simply to swap around the order of integration of an average posterior integral. For region $$R_{1}$$, the size of $$U$$ is given by $$M$$, and recall that we will require $$M$$ to actually go to infinity, so this may seem like a sleight of hand. However, the argument will be that the integral over $$R_{1}$$ goes to zero for fixed $$M$$ of any size, and so for some sequence $$M_{n}\rightarrow\infty$$. But I might observe that $$U$$ must be allowed to be arbitrarily large.

I think vdV is appealing to the dominated convergence theorem to show that $$P_{n,h_{n}}\triangleleft\triangleright P_{n,0}\Rightarrow P_{n,U}\triangleleft\triangleright P_{n,0}$$, since for any indicator function $$A$$

\begin{aligned} \lim_{n\rightarrow\infty}P_{n,U}A & =\lim_{n\rightarrow\infty}\Pi_{n}^{U}P_{x,h}A\quad\textrm{(Defintion)}\\ & =\Pi_{n}^{U}\lim_{n\rightarrow\infty}P_{x,h}A\quad\textrm{(Dominated convergence theorem)}\\ & =\Pi_{n}^{U}\lim_{n\rightarrow\infty}P_{x,0}A\quad\textrm{(Contiguity)}\\ & =\lim_{n\rightarrow\infty}P_{x,0}A\quad\textrm{(No more h dependence).}\end{aligned}

Now, in my application of the dominated convergence theorem, the measure $$\Pi_{n}^{U}$$ is actually changing with $$n$$, which means we can’t apply dominated convergence directly (see the comment on “The integrand converges to zero pointwise” below). However, for some $$n$$ sufficiently large, the density $$\pi_{n}^{U}$$ is bounded above and below by constants.

## Top of page 142, “By writing out the conditional densities...”

I think this is a tad easier to see in simpler notation. For just this section, let $P=P\left(h\vert x\right)$. We want to show that conditioning on the set $$C$$ can’t make a big difference if the probability of $$C$$ is small. To show this, we need to split everything into $$C$$ and $$C^{c}$$ and see what we get. Doing this, the first equation on page 142 is equivalent to

\begin{aligned} P\left(B\right)-P\left(B\vert C\right) & =P\left(B\bigcap C\right)+P\left(B\bigcap C^{c}\right)-\frac{P\left(B\bigcap C\right)}{P\left(C\right)}\\ & =P\left(B\bigcap C^{c}\right)+\frac{P\left(B\bigcap C\right)}{P\left(C\right)}\left(P\left(C\right)-1\right)\\ & =P\left(B\vert C^{c}\right)P\left(C^{c}\right)-\frac{P\left(B\bigcap C\right)}{P\left(C\right)}P\left(C^{c}\right)\\ & =\left(P\left(B\vert C^{c}\right)-P\left(B\vert C\right)\right)P\left(C^{c}\right).\end{aligned}

Now, $\left|P\left(B\vert C^{c}\right)-P\left(B\vert C\right)\right|\le2$ (they’re both probabilities, and so each is bounded by $$1$$), which gives the second equation on page 142. You can in fact get rid of the factor of $$2$$ because

\begin{aligned} P\left(B\vert C^{c}\right)-P\left(B\vert C\right) & \le1\textrm{ and }\\ -\left(P\left(B\vert C^{c}\right)-P\left(B\vert C\right)\right) & \le1\Rightarrow\\ \left|P\left(B\vert C^{c}\right)-P\left(B\vert C\right)\right| & \le1.\end{aligned}

## Third equation on page 142

This is just

\begin{aligned} P_{n,U}P_{\bar{H}_{n}\vert\vec{X}_{n}}\left(C_{n}^{c}\right) & =P_{n,U}P_{\bar{H}_{n}\vert\vec{X}_{n}}\left(C_{n}^{c}\right)\left(1-\phi_{n}+\phi_{n}\right)\\ & =P_{n,U}P_{\bar{H}_{n}\vert\vec{X}_{n}}\left(C_{n}^{c}\right)\left(1-\phi_{n}\right)+P_{n,U}P_{\bar{H}_{n}\vert\vec{X}_{n}}\left(C_{n}^{c}\right)\phi_{n}\\ & \le P_{n,U}P_{\bar{H}_{n}\vert\vec{X}_{n}}\left(C_{n}^{c}\right)\left(1-\phi_{n}\right)+P_{n,U}\phi_{n}.\end{aligned}

Now, we know by assumption that $$P_{n,U}\phi_{n}=o\left(1\right)$$, but we do not know how fast. So the fact that we show that $$P_{n,U}P_{\bar{H}_{n}\vert\vec{X}_{n}}\left(C_{n}^{c}\right)\left(1-\phi_{n}\right)$$ goes to zero exponentially fast does not guarantee that the whole integral over $$R_{23}$$ goes to zero exponentially fast. Note that Lemma 10.3 only proves the exponential rate of convergence for $$P_{\theta}^{n}\left(1-\phi_{n}\right)$$, where $\left\Vert \theta-\theta_{0}\right\Vert \ge M_{n}/\sqrt{n}$.

## Third equation on page 142 for moments

Note that we will have

\begin{aligned} \int g\left(\theta,x\right)\mathbb{I}\left(\theta>\frac{M_{n}}{\sqrt{n}}\right)p\left(\theta\vert x\right)d\theta & \le\\ \sqrt{\int g\left(\theta,x\right)^{2}p\left(\theta\vert x\right)d\theta}\sqrt{\int\mathbb{I}\left(\theta>\frac{M_{n}}{\sqrt{n}}\right)p\left(\theta\vert x\right)d\theta} & ,\end{aligned}

so the moments will be controlled $$\int\mathbb{I}\left(\theta>\frac{M_{n}}{\sqrt{n}}\right)p\left(\theta\vert x\right)d\theta$$ by vdV’s $$P_{\bar{H}_{n}\vert\vec{X}_{n}}\left(C_{n}^{c}\right)$$ under the condition of square integrability.

## Fourth equation on page 142, “Manipulating again the expression...”

I find this easier to see in my own notation. By re-arranging the order of the integrals (i.e. by Fubini’s theorem),

\begin{aligned} & P_{n,U}P_{\bar{H}_{n}\vert\vec{x}_{n}}\left(C_{n}^{c}\right)\left(1-\phi_{n}\right)\\ & =\int P\left(h\vert U\right)\left(\int P\left(x\vert h\right)\left(\left(1-\phi_{n}\left(x\right)\right)\int P\left(\tilde{h}\vert x\right)\mathbb{I}\left(\tilde{h}\in C_{n}^{c}\right)d\tilde{h}\right)dx\right)dh\\ & =\frac{1}{P\left(U\right)}\int\int\int P\left(h\right)\mathbb{I}\left(h\in U\right)P\left(x\vert h\right)\frac{P\left(x\vert\tilde{h}\right)P\left(\tilde{h}\right)}{P\left(x\right)}\mathbb{I}\left(\tilde{h}\in C_{n}^{c}\right)\left(1-\phi_{n}\left(x\right)\right)d\tilde{h}dxdh\\ & =\frac{1}{P\left(U\right)}\int\int\int P\left(\tilde{h}\right)\mathbb{I}\left(\tilde{h}\in C_{n}^{c}\right)\frac{P\left(C_{n}^{c}\right)}{P\left(C_{n}^{c}\right)}P\left(x\vert\tilde{h}\right)\left(1-\phi_{n}\left(x\right)\right)\frac{P\left(x\vert h\right)P\left(h\right)\mathbb{I}\left(h\in U\right)}{P\left(x\right)}dhdxd\tilde{h}\\ & =\frac{P\left(C_{n}^{c}\right)}{P\left(U\right)}\int P\left(\tilde{h}\vert C_{n}^{c}\right)\left(\int P\left(x\vert\tilde{h}\right)\left(1-\phi_{n}\left(x\right)\right)\left(\int\frac{P\left(x\vert h\right)P\left(h\right)\mathbb{I}\left(h\in U\right)}{P\left(x\right)}dh\right)dx\right)d\tilde{h}\\ & =\frac{\Pi_{n}\left(C_{n}^{c}\right)}{\Pi_{n}\left(U\right)}P_{n,C}P_{\bar{H}_{n}\vert\vec{x}_{n}}\left(U\right)\left(1-\phi_{n}\right).\end{aligned}

Basically, the trick of integrating the data distribution over the prior allows us to swap a test in $$U$$ and a posterior probability of $$C_{n}^{c}$$ for a posterior probability of $$U$$ and a test in $$C_{n}^{c}$$. I have to say, this is pretty slick. After dropping the $$\Pi_{n}\left(C_{n}^{c}\right)$$ and $$P_{\bar{H}_{n}\vert\vec{x}_{n}}\left(U\right)$$ terms (which are bounded above by $$1$$) you get an upper bound on $$P_{n,U}P_{\bar{H}_{n}\vert\vec{x}_{n}}\left(C_{n}^{c}\right)$$ which depends only on the prior, which is remarkable.

Note that for this trick to work, the prior must be proper. This is not a stated assumption as far as I can see, but I believe it is necessary for this argument to go through.

## “The integrand converges to zero pointwise, but this is not enough.”

I would like to hear a professional’s explanation of this comment. Note that if the domain and integrating distribution were fixed, then pointwise convergence would be enough by the dominated convergence theorem, since the integrand is bounded. So I believe the difficulty is in the fact that everything you’re integrating over is changing in $$n$$. If you asked me to guess beforehand whether there is a version of the dominated convergence theorem that was capable of dealing with that I would probably guess yes, but the fact that vdV goes this route instead strongly suggests not.

## “Here $$\Pi_{n}\left(U\right)$$...is bounded below...”

Because $$\theta_{0}+U/\sqrt{n}$$ is shrinking, and so eventually $$\Pi_{n}\left(U\right)$$ is strictly bounded below, $$\Pi_{n}\left(U\right)$$ is bounded below by a constant times the volume of a $$k-$$dimensional sphere of radius $$\sqrt{n}$$, which is proportional to $$1/\sqrt{n}^{k}$$. It is $$k$$-dimensional because $$\theta\in\mathbb{R}^{k}$$.

## “The total variation distance...can be expressed in the form...”

For two densities $$p\left(\theta\right)$$ and $$q\left(\theta\right)$$,

\begin{aligned} \left\Vert P-Q\right\Vert _{TV} & =\int\left|q\left(\theta\right)-p\left(\theta\right)\right|d\theta\\ & =\int\left|1-\frac{p\left(\theta\right)}{q\left(\theta\right)}\right|q\left(\theta\right)d\theta\\ & =\int1\left(p\left(\theta\right)>q\left(\theta\right)\right)\left(\frac{p\left(\theta\right)}{q\left(\theta\right)}-1\right)q\left(\theta\right)d\theta+\\ & \quad\int1\left(p\left(\theta\right)\le q\left(\theta\right)\right)\left(1-\frac{p\left(\theta\right)}{q\left(\theta\right)}\right)q\left(\theta\right)d\theta\\ & =\int\left(1-1\left(p\left(\theta\right)\le q\left(\theta\right)\right)\right)\left(\frac{p\left(\theta\right)}{q\left(\theta\right)}-1\right)q\left(\theta\right)d\theta+\\ & \quad\int1\left(p\left(\theta\right)\le q\left(\theta\right)\right)\left(1-\frac{p\left(\theta\right)}{q\left(\theta\right)}\right)q\left(\theta\right)d\theta\\ & =2\int1\left(p\left(\theta\right)\le q\left(\theta\right)\right)\left(1-\frac{p\left(\theta\right)}{q\left(\theta\right)}\right)q\left(\theta\right)d\theta+\\ & \quad\int\left(\frac{p\left(\theta\right)}{q\left(\theta\right)}-1\right)q\left(\theta\right)d\theta\\ & =2\int\left(1-\frac{p\left(\theta\right)}{q\left(\theta\right)}\right)^{+}q\left(\theta\right)d\theta.\end{aligned}

One particularly nice thing about this expression is that the integrand, $$\left(1-\frac{p\left(\theta\right)}{q\left(\theta\right)}\right)^{+}$$, is bounded between $$0$$ and $$1$$.

## “It follows that.…” Last equation on page 142.

For this section, let $$\psi\left(\cdot\right):=dN^{C}\left(\Delta_{n,\theta_{0}},I_{\theta_{0}}^{-1}\right)\left(\cdot\right)$$ for compactness. Note that the map $$x\mapsto\left(1-x\right)^{+}$$ is convex. Then the last equation can be written as

\begin{aligned} \int\left(1-\frac{\psi\left(h\right)\int1_{C}\left(g\right)P\left(x\vert g\right)P\left(g\right)dg}{1_{C}\left(h\right)P\left(x\vert h\right)P\left(h\right)}\right)^{+}P\left(h\vert x,C\right)dh & =\\ \int\left(1-\int\left(\frac{\psi\left(h\right)}{1_{C}\left(h\right)P\left(x\vert h\right)P\left(h\right)}\frac{1_{C}\left(g\right)P\left(x\vert g\right)P\left(g\right)}{\psi\left(g\right)}\right)\psi\left(g\right)dg\right)^{+}P\left(h\vert x,C\right)dh & \le\\ \textrm{ (Jensen applied to }\int\left(\cdot\right)\psi\left(g\right)dg\textrm{)}\\ \int\int\left(1-\frac{\psi\left(h\right)}{\psi\left(g\right)}\frac{1_{C}\left(g\right)P\left(x\vert g\right)P\left(g\right)}{1_{C}\left(h\right)P\left(x\vert h\right)P\left(h\right)}\right)^{+}\psi\left(g\right)P\left(h\vert x,C\right)dhdg & =\\ \textrm{(The indicators are one a.s.} \psi\textrm{ and }P\left(h\vert x,C\right) \textrm{)} \\ \int\int\left(1-\frac{\psi\left(h\right)}{\psi\left(g\right)}\frac{P\left(x\vert g\right)P\left(g\right)}{P\left(x\vert h\right)P\left(h\right)}\right)^{+}\psi\left(g\right)P\left(h\vert x,C\right)dhdg & \le\\ \textrm{(} \psi\left(g\right) \textrm{is bounded above on }C \textrm{)} \\ \left(\sup_{g\in C}\psi\left(g\right)\right)\int\int\left(1-\frac{\psi\left(h\right)}{\psi\left(g\right)}\frac{P\left(x\vert g\right)P\left(g\right)}{P\left(x\vert h\right)P\left(h\right)}\right)^{+}P\left(h\vert x,C\right)1_{C}\left(g\right)dhdg & .\end{aligned}

## “By the dominated-convergence theorem, the double integral...converges to zero...” Top of page 143

The final equation on page 142 (also given in the last comment) is a function of $$x$$. Denote that integral $$I\left(x\right)$$, and note that the integrand is bounded between $$0$$ and $$1$$. If $$P_{n,C}I\left(x\right)=\int I\left(x\right)P\left(x\vert C\right)dx\rightarrow0$$, then the integral goes to zero in probability under $$P_{n,C}$$ since convergence in mean implies convergence in probability. And, by contiguity, convergence in $$P_{n,C}$$ in probability implies convergence in $$P_{n,0}$$ in probability.

Now,

\begin{aligned} \int I\left(x\right)P\left(x\vert C\right)dx & =\int\int\int\left(1-\frac{\psi\left(h\right)}{\psi\left(g\right)}\frac{P\left(x\vert g\right)P\left(g\right)}{P\left(x\vert h\right)P\left(h\right)}\right)^{+}P\left(h\vert x,C\right)P\left(x\vert C\right)1_{C}\left(g\right)dxdhdg\\ & =\int\int\int\left(1-\frac{\psi\left(h\right)}{\psi\left(g\right)}\frac{P\left(x\vert g\right)P\left(g\right)}{P\left(x\vert h\right)P\left(h\right)}\right)^{+}\frac{P\left(x\vert h\right)P\left(h\vert C\right)}{P\left(x\vert C\right)}P\left(x\vert C\right)1_{C}\left(g\right)dxdhdg\\ & =\int\int\int\left(1-\frac{\psi\left(h\right)}{\psi\left(g\right)}\frac{P\left(x\vert g\right)P\left(g\right)}{P\left(x\vert h\right)P\left(h\right)}\right)^{+}P\left(x\vert h\right)P\left(h\vert C\right)1_{C}\left(g\right)dxdhdg.\end{aligned}

Now, I’m not sure I see how this relates to the dominated convergence theorem, but suppose that the integrand goes to zero in probability according to the integrating measure. Letting $$P^{*}$$ denote the integrating measure, and denoting the integrand with

\begin{aligned} I\left(x,g,h\right):= & \left(1-\frac{\psi\left(h\right)}{\psi\left(g\right)}\frac{P\left(x\vert g\right)P\left(g\right)}{P\left(x\vert h\right)P\left(h\right)}\right)^{+}\end{aligned}

this would mean that, for any $$\epsilon$$,

\begin{aligned} P^{*}\left(I\left(x,g,h\right)>\epsilon\right) & =\\ \int\mathbb{I}\left(I\left(x,g,h\right)>\epsilon\right)dP^{*} & =o\left(n\right),\end{aligned}

Now,

\begin{aligned} I\left(x,g,h\right) & \le\epsilon+\mathbb{I}\left(I\left(x,g,h\right)>\epsilon\right)\Rightarrow\\ \int I\left(x,g,h\right)dP^{*} & \le\epsilon+o\left(n\right).\end{aligned}

Since this holds for any $$\epsilon$$, it means that $$\int I\left(x,g,h\right)dP^{*}\rightarrow0.$$

## “...the sequence of measures on the right is continguous with respect to the measures .…”

This follows from the dominated convergence theorem, since for any set indicator $$1_{A}$$

\begin{aligned} \lim_{n\rightarrow\infty}\int\int\int1_{A}P_{n,h}\left(dx\right)\lambda_{C}\left(dh\right)\lambda_{C}\left(dg\right) & =\\ \int\int\left(\lim_{n\rightarrow\infty}\int1_{A}P_{n,h}\left(dx\right)\right)\lambda_{C}\left(dh\right)\lambda_{C}\left(dg\right) & =\quad\textrm{(dominated convergence)}\\ \int\int\left(\lim_{n\rightarrow\infty}\int1_{A}P_{n,0}\left(dx\right)\right)\lambda_{C}\left(dh\right)\lambda_{C}\left(dg\right) & \quad\quad\textrm{(contiguity)}.\end{aligned}

## “The integrand converges to zero in probability under the latter measure...”

We are finally at the point where we only need to consider the probability limit under $$P_{n,0}$$ of the integrand. Recall that

\begin{aligned} \Delta_{n,\theta_{0}} & =I_{\theta_{0}}\left(\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\dot{\ell}_{\theta_{0}}\left(X_{i}\right)\right)\Leftrightarrow\\ I_{\theta_{0}}^{-1}\Delta_{n,\theta_{0}} & =\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\dot{\ell}_{\theta_{0}}\left(X_{i}\right)\end{aligned}

Now, letting $$C_{\mathcal{N}}$$ be the normalizing constant for the normal distribution,

\begin{aligned} \frac{p_{n,g}\left(\vec{X}_{n}\right)}{dN^{C}\left(\Delta_{n,\theta_{0}},I_{\theta_{0}}^{-1}\right)\left(g\right)} & =\\ \frac{p_{n,0}\left(\vec{X}_{n}\right)p_{n,g}\left(\vec{X}_{n}\right)}{p_{n,0}\left(\vec{X}_{n}\right)dN^{C}\left(\Delta_{n,\theta_{0}},I_{\theta_{0}}^{-1}\right)\left(g\right)} & =\\ p_{n,0}\left(\vec{X}_{n}\right)\frac{p_{n,g}\left(\vec{X}_{n}\right)}{p_{n,0}\left(\vec{X}_{n}\right)}\exp\left(\frac{1}{2}g^{T}I_{\theta_{0}}^{-1}g+\frac{1}{2}\Delta_{n,\theta_{0}}^{T}I_{\theta_{0}}^{-1}\Delta_{n,\theta_{0}}-\Delta_{n,\theta_{0}}^{T}I_{\theta_{0}}^{-1}g\right)C_{N} & =\textrm{ (Normal)}\\ C_{N}p_{n,0}\left(\vec{X}_{n}\right)\exp\left(\frac{1}{\sqrt{n}}\sum_{i=1}^{n}g^{T}\dot{\ell}_{\theta_{0}}\left(X_{i}\right)-\frac{1}{2}g^{T}I_{\theta}g+o_{P_{\theta_{0}}}\left(1\right)\right) & \quad\textrm{(Theorem 7.2)}\\ \times\exp\left(\frac{1}{2}g^{T}I_{\theta_{0}}^{-1}g+\frac{1}{2}\Delta_{n,\theta_{0}}^{T}I_{\theta_{0}}^{-1}\Delta_{n,\theta_{0}}-\Delta_{n,\theta_{0}}^{T}I_{\theta_{0}}^{-1}g\right) & =\\ C_{N}p_{n,0}\left(\vec{X}_{n}\right)\exp\left(g^{T}I_{\theta_{0}}^{-1}\Delta_{n,\theta_{0}}-\frac{1}{2}g^{T}I_{\theta}g+o_{P_{\theta_{0}}}\left(1\right)\right) & \quad\textrm{(Def of }\Delta_{n,\theta_{0}}\textrm{)}\\ \times\exp\left(\frac{1}{2}g^{T}I_{\theta_{0}}^{-1}g+\frac{1}{2}\Delta_{n,\theta_{0}}^{T}I_{\theta_{0}}^{-1}\Delta_{n,\theta_{0}}-\Delta_{n,\theta_{0}}^{T}I_{\theta_{0}}^{-1}g\right) & =\\ C_{N}p_{n,0}\left(\vec{X}_{n}\right)\exp\left(\frac{1}{2}\Delta_{n,\theta_{0}}^{T}I_{\theta_{0}}^{-1}\Delta_{n,\theta_{0}}+o_{P_{\theta_{0}}}\left(1\right)\right) & .\end{aligned}

Since this does not depend on $$g$$ except in the $$o_{P_{\theta_{0}}}\left(1\right)$$ term, everything else cancels in the ratio, and

\begin{aligned} \frac{p_{n,g}\left(\vec{X}_{n}\right)}{dN^{C}\left(\Delta_{n,\theta_{0}},I_{\theta_{0}}^{-1}\right)\left(g\right)}\frac{dN^{C}\left(\Delta_{n,\theta_{0}},I_{\theta_{0}}^{-1}\right)\left(h\right)}{p_{n,h}\left(\vec{X}_{n}\right)} & =\exp\left(o_{P_{\theta_{0}}}\left(1\right)\right).\end{aligned}