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In measure theory Kakutani's Theorem is used to determine if two infinite product measures are equivalent, meaning they are both absolutely continuous with respect to each other. In one of my text books on martingale theory there is an exercise where we should use Kakutani's Theorem to investigate this for the product measure $\mathcal{P}=\otimes_{n=1}^\infty \mu_n$ and $\mathcal{Q}= \otimes_{n=1}^\infty \nu_n$ where $\mu_n$ is the Gaussian measure with expectation $0$ and variance $1$ and $\nu_n$ has variance $1$ and the expectation is given by a sequence $(\alpha_n)_{n\in\mathbb{N}}$. Thus we have
$$\frac{\mathrm{d} \nu_n}{\mathrm{d} \mu_n}(x)=\frac{\exp \left(-\left(x-\alpha_n\right)^2 / 2\right)}{\exp \left(-x^2 / 2\right)}=\exp \left(\alpha_n x-\alpha_n^2 / 2\right), \quad x \in \mathbb{R}. $$ And according to Kakutani's criterion, the product measures have a common density iff. $\prod_{n=1}^\infty a_n >0$, where in our case $a_n$ is equal to

$$\begin{align} a_n & =\int_{\mathbb{R}} \sqrt{\frac{\mathrm{d} \nu_n}{\mathrm{~d} \mu_n}} \mathrm{~d} \mu_n=\int_{\mathbb{R}} \exp \left(\frac{\alpha_n x}{2}-\frac{\alpha_n^2}{4}\right) \frac{1}{\sqrt{2 \pi}} \exp \left(-x^2 / 2\right) \mathrm{d} x \\ & =\exp \left(-\alpha_n^2 / 8\right) \int_{\mathbb{R}} \frac{1}{\sqrt{2 \pi}} \exp \left(-\frac{1}{2}\left(x-\frac{\alpha_n}{2}\right)^2\right) \mathrm{d} x=\exp \left(-\alpha_n^2 / 8\right). \end{align}$$ We can use a result from calculus, that says $\prod_{n=1}^\infty a_n >0$ iff. $\sum_{n=1}^\infty \log a_n <\infty$. Using this we know that Kakutani's criterion holds iff $(\alpha_n)_{n\in\mathbb{N}}\in\ell^2$. In this case the density between the product measures is $$ \frac{\mathrm{d}\mathcal{Q}}{\mathrm{d}\mathcal{P}} (X)= \prod_{n=1}^\infty \frac{\mathrm{d}\nu_n}{\mathrm{d}\mu_n}(x_n) = \exp\left(\sum_{n=1}^\infty \alpha_n x_n - \frac{1}{2} \sum_{n=1}^\infty \alpha_n^2\right), $$ where $X = (x_n)_{n\in\mathbb{N}}\in \mathbb{R}^{\mathbb{N}}$. Now in this density we know that the right series in the exponential converges, since $(\alpha_n)_{n\in\mathbb{N}}\in\ell^2$. Where I am stuck, is reasoning in what sense the left series converges, or if it even converges at all.

Thank you for your time.

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2 Answers 2

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In keeping with the spirit of your book: let $M_k(x) = \sum_{n=1}^k a_n x_n$. Since under $P$, the $x_n$ are iid mean-zero Gaussian, $(M_k)$ is a $P$-martingale, and its variance is $\sum_{n=1}^k a_n^2$. Since $\sum a_n^2$ converges, the martingale $M_k$ is bounded in $L^2$ and thus also in $L^1$. And an $L^1$-bounded $P$-martingale, as you surely know by this point in your book, converges $P$-almost surely :)

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  • $\begingroup$ Doesn't the $L^2$ boundedness also imply $L^2$ convergence? So $(M_k)$ converges both almost surely and in $L^2(\mathcal{P})$. But I believe we are more interested in almost sure convergence, since the density need only exist almost surely and $L^2$ convergence would not help this. $\endgroup$
    – Sussyphus
    Oct 4, 2023 at 17:36
  • $\begingroup$ @Sussyphus: We already had convergence in $L^2$ of $\sum a_n x_n$, simply from the fact that the variances are summable. $\endgroup$ Oct 5, 2023 at 3:50
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Because $(\alpha_n)$ is square integrable, the variance of $\sum_n\alpha_n x_n$ is finite, so the series converges in $L^2(\mathcal P)$, hence in probability and also in distribution. By a theorem of Lévy for infinite series of independent random variables (convergence in distribution implies a.s. convergence), the series also converges $\mathcal P$-a.s. (Alternatively, you could also apply Kolmogorov's Two-Series Theorem to reach the same conclusion). This ensures that your density formula is well defined $\mathcal P$-a.s. (hence also $\mathcal Q$-a.s.)

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