Almost sure convergence in density of infinite dimensional product measure of normal distributions unsing Kakutani's theorem

Question

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.

Answer 1

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 :)

Answer 2

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.)