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Distributional compatibility for change of measures

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Abstract

In this paper, we characterise compatibility of distributions and probability measures on a measurable space. For a set of indices \(\mathcal{J}\), we say that the tuples of probability measures \((Q_{i})_{i\in \mathcal{J}} \) and distributions \((F_{i})_{i\in \mathcal{J}} \) are compatible if there exists a random variable having distribution \(F_{i}\) under \(Q_{i}\) for each \(i\in \mathcal{J}\). We first establish an equivalent condition using conditional expectations for general (possibly uncountable) \(\mathcal{J}\). For a finite \(n\), it turns out that compatibility of \((Q_{1},\dots ,Q_{n})\) and \((F_{1},\dots ,F _{n})\) depends on the heterogeneity among \(Q_{1},\dots ,Q_{n}\) compared with that among \(F_{1},\dots ,F_{n}\). We show that under an assumption that the measurable space is rich enough, \((Q_{1},\dots ,Q_{n})\) and \((F_{1},\dots ,F_{n})\) are compatible if and only if \((Q_{1},\dots ,Q _{n})\) dominates \((F_{1},\dots ,F_{n})\) in a notion of heterogeneity order, defined via the multivariate convex order between the Radon–Nikodým derivatives of \((Q_{1},\dots ,Q_{n})\) and \((F_{1},\dots ,F_{n})\) with respect to some reference measures. We then proceed to generalise our results to stochastic processes, and conclude the paper with an application to portfolio selection problems under multiple constraints.

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Notes

  1. We thank Marcel Nutz for suggesting the second and third connections, and Fabio Maccheroni for helpful discussions leading to the fourth connection.

  2. We thank Freddy Delbaen for pointing out the preprint and for very useful discussions.

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Acknowledgements

The authors are grateful to the Editor, the Associate Editor, two referees, Michel Baes, Fabio Bellini, Paul Embrechts, Fabio Maccheroni, Tiantian Mao, Alfred Müller, Marcel Nutz, Jan Obłoj, Sidney Resnick, Ludger Rüschendorf, Alexander Schied and Xiaolu Tan for various helpful suggestions and discussions on an earlier version of the paper. J. Shen acknowledges financial support from the China Scholarship Council. Y. Shen and R. Wang acknowledge financial support by the Natural Sciences and Engineering Research Council (NSERC 2014-04840, RGPIN-2018-03823, RGPAS-2018-522590) of Canada. R. Wang is also grateful to FIM at ETH Zurich for supporting his visit in 2017, during which part of this paper was written.

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Appendix

Appendix

1.1 A.1 Details for Example 3.8

Note that \(\frac{\mathrm{d}Q_{1}}{\mathrm{d}Q_{2}}\) is uniform on \([0,2]\) under \(Q_{2}=\lambda \), and \(\frac{\mathrm{d}F_{1}}{ \mathrm{d}F_{2}}\) is also uniform on \([0,2]\) under \(F_{2}=\lambda \). Thus we have

$$ \left .\left (\frac{\mathrm{d}F_{1}}{\mathrm{d}\lambda },\frac{ \mathrm{d}F_{2}}{\mathrm{d}\lambda }\right )\right |_{\lambda }\stackrel{ {{\mathrm{d} }}}{=}\left . \left (\frac{\mathrm{d}Q_{1}}{\mathrm{d} \lambda },\frac{\mathrm{d}Q_{2}}{\mathrm{d}\lambda }\right )\right | _{\lambda }. $$

Therefore \((F_{1},F_{2})\preceq _{\mathrm{h}}(Q_{1},Q_{2})\).

Next, we argue that \((Q_{1},Q_{2})\) and \((F_{1},F_{2})\) are not compatible. Suppose for a contradiction that they are. By Theorem 2.2, there exists a random variable \(X\) on \((\Omega , \mathcal{A})\) with a uniform distribution on \([0,1]\) under \(Q_{2}= \lambda \) such that

$$ \frac{\mathrm{d}F_{1}}{\mathrm{d}\lambda }(X)=\mathbb{E}^{\lambda } \left [\left .\frac{\mathrm{d}Q_{1}}{\mathrm{d}\lambda }\right | X\right ]. $$

In addition,

$$ \left .\frac{\mathrm{d}F_{1}}{\mathrm{d}\lambda }(X)\right |_{\lambda }\stackrel{{{\mathrm{d} }}}{=}\left . \frac{\mathrm{d}Q_{1}}{ \mathrm{d}\lambda }\right |_{\lambda }, $$

and therefore

$$ \frac{\mathrm{d}F_{1}}{\mathrm{d}\lambda }(X)= \frac{\mathrm{d}Q_{1}}{ \mathrm{d}\lambda } \qquad \mbox{$\lambda $-almost surely.} $$

From the definition of \(F_{1}\) and \(Q_{1}\), we have \(|4X(t)-2|=2t\) for \(\lambda \)-almost all \(t\in [0,1]\). It follows that \(X(t)=(1+t)/2\) or \(X(t)=(1-t)/2\) for \(\lambda \)-almost all \(t\in [0,1]\). Write

$$ A=\bigg\{ t\in [0,1]: X(t)=\frac{1+t}{2}\bigg\} , \qquad B=\left \{t\in [0,1]: X(t)=\frac{1-t}{2}\right \} $$

and

$$ C=\left \{\frac{1-t}{2}:t\in A\right \}. $$

As \(X\) is \(\mathcal{B}([0,1])\)-measurable with distribution \(F_{2}\) under \(\lambda \), we have \(A,B \in \mathcal{B}([0,1])\) and \(\lambda (A)=\lambda (B)=1/2\). Note that \(\lambda (C)=1/4\); however, \(\lambda (C\cap X(A\cup B))=0\), contradicting the fact that \(X\) has a uniform distribution on \([0,1]\) under \(\lambda \).

1.2 A.2 Proof of Theorem 3.10

(Necessity) Assume that \((Q_{1},\dots ,Q_{n})\) and \((F_{1},\dots ,F _{n})\) are almost compatible. This means that for any \(\varepsilon >0\), there exists \((F_{1,\varepsilon },\dots ,F_{n,\varepsilon })\) with \(D _{\mathrm{KL}}(F_{i,\varepsilon }| F_{i})<\varepsilon \) for \(i=1,\dots ,n\), and such that \((Q_{1},\dots ,Q_{n})\) is compatible with \((F_{1,\varepsilon },\dots ,F_{n,\varepsilon })\). Define probability measures

$$\begin{aligned} F_{\varepsilon } &=\frac{1}{n}(F_{1,\varepsilon }+\cdots +F_{n,\varepsilon }), \\ F &=\frac{1}{n}(F_{1}+\cdots +F_{n}), \\ Q &=\frac{1}{n}(Q_{1}+\cdots +Q_{n}). \end{aligned}$$

Note that the distribution of \(X_{\varepsilon }\) under \(Q\) is \(F_{\varepsilon }\), where \(X_{\varepsilon }\) is the random variable defining the compatibility between \((Q_{1},\dots ,Q_{n})\) and \((F_{1,\varepsilon },\dots ,F_{n,\varepsilon })\). Moreover, for \(i=1,\dots ,n\), we have \(F_{i,\varepsilon }\ll F_{\varepsilon }\), \(Q_{i}\ll Q\), \(\mathrm{d}F_{i,\varepsilon }/\mathrm{d}F_{\varepsilon }\leqslant n\) and \(\mathrm{d}Q_{i}/\mathrm{d}Q\leqslant n\). For \(\varepsilon >0\), by Lemma 3.3,

$$ \left .\left (\frac{\mathrm{d}F_{1,\varepsilon }}{\mathrm{d}F_{\varepsilon }},\dots , \frac{\mathrm{d}F_{n,\varepsilon }}{\mathrm{d}F_{\varepsilon }} \right )\right |_{F_{\varepsilon }}\preceq _{\mathrm{cx}}\left . \left (\frac{\mathrm{d}Q_{1}}{\mathrm{d}Q},\dots ,\frac{\mathrm{d}Q _{n}}{\mathrm{d}{Q}}\right )\right |_{Q}. $$

As a result, for any convex function \(f:{\mathbb{R}}^{n}\to \mathbb{R}\),

$$ \mathbb{E}^{F_{\varepsilon }}\left [f\left (\frac{\mathrm{d}F_{1, \varepsilon }}{\mathrm{d}F_{\varepsilon }},\dots , \frac{\mathrm{d}F _{n,\varepsilon }}{\mathrm{d}F_{\varepsilon }}\right ) \right ]\leqslant \mathbb{E}^{Q}\left [f\left (\frac{\mathrm{d}Q_{1}}{\mathrm{d}Q}, \dots ,\frac{\mathrm{d}Q_{n}}{\mathrm{d}{Q}}\right )\right ]. $$

For \(i=1,\dots ,n\),

$$ \frac{\mathrm{d}F_{i,\varepsilon }}{\mathrm{d}F_{\varepsilon }}=\frac{ \mathrm{d}F_{i}}{\mathrm{d}F}\frac{\mathrm{d}F_{i,\varepsilon }/ \mathrm{d}F_{i}}{\mathrm{d}F_{\varepsilon }/\mathrm{d}F}. $$
(A.1)

By Pinsker’s inequality, for any two probability measures \(P\) and \(Q\) defined on the same probability space, the total variation distance between them is dominated by \(\sqrt{\frac{1}{2}D_{\mathrm{KL}}(P| Q)}\). Since \(D_{\mathrm{KL}}(F _{i,\varepsilon }| F_{i})\) converges to 0, \((F_{i,\varepsilon })\) thus converges to \(F_{i}\) in total variation, which is equivalent to \((\mathrm{d}F_{i,\varepsilon }/\mathrm{d}F_{i})\) converging in \(L^{1}(F_{i})\) to 1. Hence for any sequence \(\varepsilon _{m}\downarrow 0\), there exists a subsequence, which we still denote as \(\varepsilon _{m}\downarrow 0\) by a slight abuse of notation, such that \(( \mathrm{d}F_{i,\varepsilon _{m}}/\mathrm{d}F_{i})\) converges to 1 \(F_{i}\)-almost surely. It is easy to check that \((\mathrm{d}F_{ \varepsilon _{m}}/\mathrm{d}F)\) converges to 1 as well. Equation (A.1) then implies that

$$ \frac{\mathrm{d}F_{i,\varepsilon _{m}}}{\mathrm{d}F_{\varepsilon _{m}}} \longrightarrow \frac{\mathrm{d}F_{i}}{\mathrm{d}F} \qquad F_{i}\text{-almost surely}. $$
(A.2)

On any set \(B\in \mathcal{B}(\mathbb{R})\) such that \(F_{i}(B)=0\) but \(F(B)>0\), suppose \((\mathrm{d}F_{i,\varepsilon }/\mathrm{d}F_{\varepsilon })\) does not converge to \(\mathrm{d}F_{i}/\mathrm{d}F=0\) in probability under \(F|_{B}\), the measure \(F\) restricted to \(B\). Then there exist \(\delta >0\) and a subsequence of \((\varepsilon _{m})\) (again denoted as \((\varepsilon _{m})\)) such that \(P^{F|_{B}}[\mathrm{d}F_{i,\varepsilon _{m}}/\mathrm{d}F_{\varepsilon _{m}}>\delta ]\geqslant c\) for some constant \(c>0\). Since \((F_{\varepsilon _{m}})\) converges to \(F\) in total variation, we have \(P^{F_{\varepsilon _{m}}|_{B}}[\mathrm{d}F_{i, \varepsilon _{m}}/\mathrm{d}F_{\varepsilon _{m}}>\delta ]\geqslant c/2\) for \(m\) large enough. Hence \(F_{i,\varepsilon _{m}}(B)\geqslant \delta P^{F_{\varepsilon _{m}}|_{B}}[\mathrm{d}F_{i,\varepsilon _{m}}/ \mathrm{d}F_{\varepsilon _{m}}>\delta ]\geqslant \frac{c\delta }{2}\), which contradicts the fact that \((F_{i,\varepsilon _{m}})\) converges to \(F_{i}\) in total variation. We conclude that \((\mathrm{d}F_{i,\varepsilon }/\mathrm{d}F_{\varepsilon })\) converges to \(\mathrm{d}F_{i}/ \mathrm{d}F=0\) in probability under \(F\) on the set \(\{\mathrm{d}F_{i}/ \mathrm{d}F=0\}\). Combining this result with (A.2) and taking a further subsequence allows us to replace the \(F_{i}\)-almost sure convergence in (A.2) by \(F\)-almost sure convergence.

For any convex function \(f:{\mathbb{R}}^{n}\to \mathbb{R}\),

$$ \mathbb{E}^{F_{\varepsilon _{m}}}\left [f\left (\frac{\mathrm{d}F_{1, \varepsilon _{m}}}{\mathrm{d}F_{\varepsilon _{m}}},\dots , \frac{ \mathrm{d}F_{n,\varepsilon _{m}}}{\mathrm{d}F_{\varepsilon _{m}}}\right ) \right ]=\int f\left (\frac{\mathrm{d}F_{1,\varepsilon _{m}}}{ \mathrm{d}F_{\varepsilon _{m}}},\dots , \frac{\mathrm{d}F_{n, \varepsilon _{m}}}{\mathrm{d}F_{\varepsilon _{m}}}\right )\mathrm{d}F _{\varepsilon _{m}}. $$

Since \(\frac{\mathrm{d}F_{i,\varepsilon _{m}}}{\mathrm{d}F_{\varepsilon _{m}}}\in [0,n]\) and \(f\) is convex, hence continuous, \(|f(\frac{ \mathrm{d}F_{1,\varepsilon _{m}}}{\mathrm{d}F_{\varepsilon _{m}}}, \dots , \frac{\mathrm{d}F_{n,\varepsilon _{m}}}{\mathrm{d}F_{ \varepsilon _{m}}})|\) is bounded. Let \(b\) be an upper bound of it. Because \((F_{\varepsilon _{m}})\) converges in total variation to \(F\), we have

$$\begin{aligned} &\left |\int f\left (\frac{\mathrm{d}F_{1,\varepsilon _{m}}}{ \mathrm{d}F_{\varepsilon _{m}}},\dots , \frac{\mathrm{d}F_{n, \varepsilon _{m}}}{\mathrm{d}F_{\varepsilon _{m}}}\right )\mathrm{d}F _{\varepsilon _{m}}-\int f\left (\frac{\mathrm{d}F_{1,\varepsilon _{m}}}{ \mathrm{d}F_{\varepsilon _{m}}},\dots , \frac{\mathrm{d}F_{n, \varepsilon _{m}}}{\mathrm{d}F_{\varepsilon _{m}}}\right )\mathrm{d}F\right | \\ &\leqslant 2b\delta (F_{\varepsilon _{m}},F)\longrightarrow 0, \end{aligned}$$
(A.3)

where \(\delta (\cdot , \cdot )\) is the total variation distance. Moreover, dominated convergence gives

$$ \int f\left (\frac{\mathrm{d}F_{1,\varepsilon _{m}}}{\mathrm{d}F_{ \varepsilon _{m}}},\dots , \frac{\mathrm{d}F_{n,\varepsilon _{m}}}{ \mathrm{d}F_{\varepsilon _{m}}}\right )\mathrm{d}F\longrightarrow \int f\left (\frac{\mathrm{d}F_{1}}{\mathrm{d}F},\dots , \frac{ \mathrm{d}F_{n}}{\mathrm{d}F}\right )\mathrm{d}F. $$
(A.4)

Equations (A.3) and (A.4) together show that

$$\begin{aligned} \mathbb{E}^{F}\left [f\left (\frac{\mathrm{d}F_{1}}{\mathrm{d}F}, \dots , \frac{\mathrm{d}F_{n}}{\mathrm{d}F}\right ) \right ] &= \lim _{m\to \infty }\mathbb{E}^{F_{\varepsilon _{m}}}\left [f\left (\frac{ \mathrm{d}F_{1,\varepsilon _{m}}}{\mathrm{d}F_{\varepsilon _{m}}}, \dots , \frac{\mathrm{d}F_{n,\varepsilon _{m}}}{\mathrm{d}F_{ \varepsilon _{m}}}\right ) \right ] \\ &\leqslant \mathbb{E}^{Q}\left [f\left (\frac{\mathrm{d}Q_{1}}{ \mathrm{d}Q},\dots ,\frac{\mathrm{d}Q_{n}}{\mathrm{d}{Q}}\right )\right ]. \end{aligned}$$

(Sufficiency) Assume that \((F_{1},\dots ,F_{n})\preceq _{\mathrm{h}}(Q _{1},\dots ,Q_{n})\). By Lemma 3.5, this means that

$$ \left .\left (\frac{\mathrm{d}F_{1}}{\mathrm{d}F},\dots , \frac{ \mathrm{d}F_{n}}{\mathrm{d}F} \right )\right |_{F} \preceq _{ \mathrm{cx}}\left . \left (\frac{\mathrm{d}Q_{1}}{\mathrm{d}Q},\dots ,\frac{\mathrm{d}Q_{n}}{\mathrm{d}Q}\right )\right |_{Q} $$

holds for \(F=\frac{1}{n}\sum _{i=1}^{n} F_{i}\) and \(Q=\frac{1}{n}\sum _{i=1}^{n} Q_{i}\). By Lemma 3.2, there exist a probability space \((\Omega ', {\mathcal{A}}', Q')\) and random vectors \(\mathbf{Y}'=(Y _{1}',\dots ,Y_{n}'), \mathbf{Z}'=(Z_{1}',\dots ,Z_{n}')\) defined on \((\Omega ', {\mathcal{A}}', Q')\) such that

$$\begin{aligned} (Y_{1}',\dots ,Y_{n}') &\stackrel{{\mathrm{d} }}{=}\left (\frac{ \mathrm{d}Q_{1}}{\mathrm{d}Q},\dots ,\frac{\mathrm{d}Q_{n}}{ \mathrm{d}Q}\right )=:\mathbf{Y}=(Y_{1},\dots ,Y_{n}), \\ (Z_{1}',\dots ,Z_{n}') &\stackrel{{\mathrm{d} }}{=}\left (\frac{ \mathrm{d}F_{1}}{\mathrm{d}F},\dots ,\frac{\mathrm{d}F_{n}}{ \mathrm{d}F}\right )=:\mathbf{Z}=(Z_{1},\dots ,Z_{n}) \end{aligned}$$

and

$$ \mathbb{E}^{Q'}[Y_{i}'|Z_{i}']=Z_{i}',\quad i=1,\dots ,n. $$

Given \(m=0,1,\dots \), define the random vector \(\mathbf{Y}_{m}=(Y_{m,1}, \dots ,Y_{m,n})\) by

$$ Y_{m,i}=\left \{ \textstyle\begin{array}{l@{\quad }l} 0 &\quad \text{ if }Y_{i}=0, \\ \exp (2^{-m}\lfloor 2^{m}\log Y_{i}\rfloor )& \quad \text{ otherwise}, \end{array}\displaystyle \right . $$

for \(i=1,\dots ,n\). Similarly, we define \(\mathbf{Y}'_{m}\), \(\mathbf{Z}_{m}\) and \(\mathbf{Z}'_{m}\) for \(\mathbf{Y}'\), \(\mathbf{Z}\) and \(\mathbf{Z}'\), respectively. Note that

$$\begin{aligned} \mathbb{E}^{Q'}[Y_{m,i}'|Z_{m,i}'] &\in \big[\exp (-2^{-m})\mathbb{E} ^{Q'}[Y_{i}'|Z_{m,i}'], \mathbb{E}^{Q'}[Y_{i}'|Z_{m,i}']\big] \\ &\subseteq [\exp (-2^{-m})Z_{m,i}', \exp (2^{-m})Z_{m,i}'] \end{aligned}$$

for \(i=1,\dots ,n\).

Each of \(Q_{1},\dots , Q_{n}\) is atomless, and so is therefore \(Q\). As a result, we can divide \(\Omega \) into disjoint sets \(A^{m}_{k,j} \in {\mathcal{A}}\), where \(k=(k_{1},\dots ,k_{n})\in ({\mathbb{Z}} \cup \{-\infty \})^{n}\) and \(j=(j_{1},\dots ,j_{n})\in ({\mathbb{Z}} \cup \{-\infty \})^{n}\), such that \(Y_{m,i}(\omega )=\exp (k_{i}2^{-m})\) for \(\omega \in A^{m}_{k,j}\) and \(i=1,\dots ,n\), and

$$ Q[A^{m}_{k,j}]={Q'}[Y'_{m,i}=\exp (k_{i}2^{-m}), Z'_{m,i}=\exp (j_{i}2^{-m}), i=1,\dots ,n]. $$

Here we use the convention that \(\exp (-\infty )=0\) for ease of notation. Define the random vector \(\mathbf{Z}_{m}''\) on \((\Omega , {\mathcal{A}}, Q)\) by \(Z_{m,i}''(\omega )=\exp (j_{i}2^{-m})\) for \(\omega \in A^{m}_{k,j}\). Then we have \((\mathbf{Y}_{m}, \mathbf{Z}''_{m})|_{Q}\stackrel{{\mathrm{d} }}{=}(\mathbf{Y}_{m}', \mathbf{Z}_{m}')|_{Q'}\). Let \(\mathrm{Id}\) be the identity random variable on \((R, {\mathcal{B}}(\mathbb{R}))\). For \(\ell =0,1,\dots \) and \(h\in \mathbb{Z}\), denote by \(\varphi ^{m}_{\ell ,h}(z)\) the conditional probability under \(F\) of the event \(\{\mathrm{Id}\in [h2^{- \ell }, (h+1)2^{-\ell })\}\) given \(Z_{m}=z\), i.e.,

$$ \varphi ^{m}_{\ell ,h}(z)=F\Big(\mathrm{Id}\in \big[h2^{-\ell }, (h+1)2^{- \ell }\big)\big|Z_{m}=z\Big). $$

Then for any \(\ell =0,1,\dots \), \(A^{m}_{k,j}\) can be further divided into disjoint subsets \(A^{m}_{k,j,\ell ,h}\) such that \(Q[A^{m}_{k,j, \ell ,h}]=Q[A^{m}_{k,j}]\varphi ^{m}_{\ell ,h}(\exp (j2^{-m}))\). Moreover, the partitions can be made such that \(\{A^{m}_{k,j,\ell ',h} : h\in \mathbb{Z}\}\) is a refinement of \(\{A^{m}_{k,j,\ell ,h}:h \in \mathbb{Z}\}\) for any \(\ell '>\ell \) and any given \(m,k,j\). Define \(X_{m,\ell }(\omega )=h2^{-\ell }\) for \(\omega \in A^{m}_{k,j,\ell ,h}\), and \(X_{m}=\lim _{\ell \to \infty }X_{m,\ell }\). The limit exists since it is easy to check that \(X_{m,\ell }\) is increasing with respect to \(\ell \). Note that \(X_{m,\ell }\) is conditionally independent of \(\mathbf{Y}_{m}\) given \(\mathbf{Z}_{m}''\); hence \(X_{m}\) is also conditionally independent of \(\mathbf{Y}_{m}\) given \(\mathbf{Z}''_{m}\).

By construction, for any \(A\in {\mathcal{B}}({\mathbb{R}}^{n})\), \(\ell =0,1,\dots \) and \(h\in \mathbb{Z}\),

$$\begin{aligned} &Q\big[\mathbf{Z}''_{m}\in A, X_{m,\ell '}\in \big[h2^{-\ell },(h+1)2^{- \ell }\big)\big] \\ &=Q[\mathbf{Z}''_{m}\in A, X_{m,\ell }=h2^{-\ell }] \\ &=\sum _{{\scriptstyle k\atop\scriptstyle j:\exp (j2^{-m})\in A}}Q[A^{m}_{k,j,\ell ,h}] \\ &=\sum _{{\scriptstyle k \atop\scriptstyle j:\exp (j2^{-m})\in A}}Q[A^{m}_{k,j}]\varphi ^{m}_{\ell ,h}\big(\exp (j2^{-m})\big) \\ &=\sum _{j:\exp (j2^{-m})\in A}Q[\mathbf{Z}''_{m}=\exp (j2^{-m})] \varphi ^{m}_{\ell ,h}\big(\exp (j2^{-m})\big) \\ &=\sum _{j:\exp (j2^{-m})\in A}F\big(\mathbf{Z}_{m}=\exp (j2^{-m}) \big)F\Big(\big[h2^{-\ell },(h+1)2^{-\ell }\big)\big|\mathbf{Z}_{m}= \exp (j2^{-m})\Big) \\ &=F\Big(\mathbf{Z}_{m}^{-1}(A)\cap \big[h2^{-\ell },(h+1)2^{-\ell } \big)\Big) \end{aligned}$$
(A.5)

for all \(\ell '\geqslant \ell \). Thus \(\mathbf{Z}_{m}\) restricted to the interval \([h2^{-\ell },(h+1)2^{-\ell })\) has the same distribution as \(\mathbf{Z}_{m}''\) restricted to the set \(X^{-1}_{m,\ell '}([h2^{- \ell },(h+1)2^{-\ell }))\). Now note that the set \(X^{-1}_{m,\ell '}([h2^{- \ell },(h+1)2^{-\ell }))\) is the same for any \(\ell '\geqslant \ell \); hence \(\mathbf{Z}_{m}\) restricted to the interval \([h2^{-\ell },(h+1)2^{- \ell })\) also has the same distribution as \(\mathbf{Z}_{m}''\) restricted to \(X^{-1}_{m}([h2^{-\ell },(h+1)2^{-\ell }))\) for all \(m=0,1, \dots \). Because the collection of sets \(\{[h2^{-\ell },(h+1)2^{- \ell }) : h\in {\mathbb{Z}}, \ell =0,1,\dots \}\) forms a basis for \({\mathcal{B}}(\mathbb{R})\), \(\mathbf{Z}_{m}\) restricted to any Borel set \(B\) has the same distribution as \(\mathbf{Z}_{m}''\) restricted to \(X_{m}^{-1}(B)\). Therefore we conclude that \(\mathbf{Z}''_{m}= \mathbf{Z}_{m}\circ X_{m}\)\(Q\)-almost surely. Moreover, by taking \(A={\mathbb{R}}^{n}\) in (A.5), it follows that

$$ Q\Big[X_{m,\ell '}\in \big[h2^{-\ell },(h+1)2^{-\ell }\big)\Big]=F \Big(\big[h2^{-\ell },(h+1)2^{-\ell }\big)\Big) $$

for all \(\ell '\geqslant \ell \). A similar reasoning as above then shows that \(F=Q\circ X_{m}^{-1}\).

For any \(A\in {\mathcal{B}}({\mathbb{R}})\) and any \(i=1,\dots ,n\),

$$ Q_{i}[X_{m}\in A]=\int _{X_{m}^{-1}(A)}Y_{i}\mathrm{d}Q. $$
(A.6)

It is easy to see that

$$ \int _{X_{m}^{-1}(A)}Y_{m,i}\mathrm{d}Q\leqslant \int _{X_{m}^{-1}(A)}Y _{i}\mathrm{d}Q\leqslant \exp (2^{-m})\int _{X_{m}^{-1}(A)}Y_{m,i} \mathrm{d}Q. $$

Moreover, noting that \(j=(j_{1}, \dots , j_{n})\) is a vector, we have

$$\begin{aligned} &\int _{X_{m}^{-1}(A)}Y_{m,i}\mathrm{d}Q \\ &=\sum _{j}Q[X_{m}\in A|\mathbf{Z}''_{m}=e^{j2^{-m}}]\sum _{k} e^{k _{i}2^{-m}}Q[\mathbf{Y}_{m}=e^{k2^{-m}}, \mathbf{Z}''_{m}=e^{j2^{-m}}] \\ &=\sum _{j}Q[X_{m}\in A|\mathbf{Z}''_{m}=e^{j2^{-m}}]Q[\mathbf{Z}''_{m}=e ^{j2^{-m}}]\mathbb{E}^{Q}[Y_{m,i}|\mathbf{Z}''_{m}=e^{j2^{-m}}] \\ &=\sum _{j}Q[X_{m}\in A|\mathbf{Z}''_{m}=e^{j2^{-m}}]Q[\mathbf{Z}''_{m}=e ^{j2^{-m}}]\mathbb{E}^{Q'}[Y'_{m,i}|\mathbf{Z}'_{m}=e^{j2^{-m}}] \\ &\geqslant \sum _{j}Q[X_{m}\in A|\mathbf{Z}''_{m}=e^{j2^{-m}}]Q[ \mathbf{Z}''_{m}=e^{j2^{-m}}]\exp (j_{i}2^{-m}-2^{-m}) \\ &=\sum _{j}F(A |\mathbf{Z}_{m}=e^{j2^{-m}})F(\mathbf{Z}_{m}=e^{j2^{-m}}) \exp (j_{i}2^{-m}-2^{-m}) \\ &\geqslant \exp (-2^{-m})\sum _{j}\exp (j_{i}2^{-m})F(A\cap \{ \mathbf{Z}_{m}=e^{j2^{-m}}\}) \\ &=\exp (-2^{-m})\int _{A} Z_{m,i} \mathrm{d}F \\ &\geqslant \exp (-2^{-m+1})\int _{A} Z_{i} \mathrm{d}F \\ &= \exp (-2^{-m+1})F_{i}(A), \end{aligned}$$

where the first equality holds since \(X_{m}\) is independent of \(\mathbf{Y}_{m}\) given \(\mathbf{Z}''_{m}\), and the fourth equality holds because \(Q\circ X_{m}^{-1}=F\) and \(\mathbf{Z}_{m}\circ X_{m}= \mathbf{Z}_{m}''\). Symmetrically,

$$ \int _{X_{m}^{-1}(A)}Y_{m,i}\mathrm{d}Q\leqslant \exp (2^{-m}) F_{i}(A). $$
(A.7)

Combining (A.6) and (A.7), we have

$$ Q_{i}[X_{m}\in A]\in [\exp (-2^{-m+1})F_{i}(A),\exp (2^{-m+1})F_{i}(A)]. $$

Since this holds for any \(A\in {\mathcal{B}}(\mathbb{R})\), we conclude that \(Q_{i}\circ X_{m}^{-1}\) is absolutely continuous with respect to \(F_{i}\) and \(({\mathrm{d}Q_{i}\circ X_{m}^{-1}})/{\mathrm{d}F_{i}} \in [\exp (-2^{-m+1}), \exp (2^{-m+1})]\). It is then easy to see that \(D_{\text{KL}}(Q_{i}\circ X_{m}^{-1}| F_{i})\) converges to 0 as \(m\to \infty \). □

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Shen, J., Shen, Y., Wang, B. et al. Distributional compatibility for change of measures. Finance Stoch 23, 761–794 (2019). https://doi.org/10.1007/s00780-019-00393-4

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