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Let $C=\sum_{s=1}^tx_sx_s'+aD^{-1}$, which is symmetric positive definite, vector $x,w\in\mathbb{R}^n$ and a scalar $y\in\mathbb{R}$. The integral I am trying to solve is as follows: $$\sqrt{\frac{a\sqrt{\sigma^2}}{2\pi}}\frac{\int_{\mathbb{R}^n}e^{-\frac{a\sqrt{\sigma^2}}{2}(w' x_{t+1}-y)^2}e^{-\frac{a\sqrt{\sigma^2}}{2}\sum_{s}(w'x_s-y_s)^2-\frac{a^2\sqrt{{\sigma^2}}}{2} w' D^{-1}w}\mathrm{d}w}{\int_{\mathbb{R}^n}e^{-\frac{a\sqrt{\sigma^2}}{2}\sum_{s}(w'x_s-y_s)^2-\frac{a^2\sqrt{{\sigma^2}}}{2} w' D^{-1}w}\mathrm{d}w}$$

By using the arguments mentioned here I can solve the integral in the denominator by letting $b=2\sum_{s=1}^tx_sy_s$ and $\eta=\frac{a\sqrt{\sigma^2}}{2}$:

$\int_{\mathbb{R}^n}e^{-\eta w'Cw+\eta b'w-\eta\sum_{s}y_s^2}\mathrm{d}w=e^{-\eta\sum_{s}y_s^2}\prod_{i=1}^n\int_{\mathbb{R}}e^{-\eta\mu^i(w^{i})^2+\eta w^i(Pb)^i}\mathrm{d}w^i=e^{-\eta \sum_{s}y_s^2}\prod_{i=1}^n\left(\frac{\sqrt{\pi}e^{(\eta b^i)^2/4\eta\mu^i}}{\sqrt{\eta\mu^i}}\right)=e^{-\eta \sum_{s}y_s^2}\frac{\pi^{n/2}e^{\frac{\eta^2 b'b}{4\eta (\mu^1+...+\mu^n) }}}{\sqrt{\det \eta C}}$

To solve the numerator I write:

$$\int_{\mathbb{R}^n} e^{-\eta\left(\sum_sy_s-y^2\right)+\eta\left(\sum_s y_s x_s- yx_{t+1}\right)'w-\eta w'\left(aD^{-1}+\sum_s x_tx_t'+x_{t+1}x_{t+1}'\right)w}\mathrm{d}w $$$$=\int_{\mathbb{R}^n} e^{-\eta\left(\sum_sy_s-y^2\right)+2\eta\left(\sum_s y_s x_s- yx_{t+1}\right)'w-\eta w'\left(aD^{-1}+\sum_s^{t+1} x_tx_t'\right)w}\mathrm{d}w$$

By using similar arguments as used to solve the denominator I write:

$$e^{-\eta \sum_{s}y_s^2-\eta y^2}\frac{\pi^{n/2}e^{\frac{\eta^2 b++'b++}{4\eta (\mu^1+...+\mu^n+x^1+...+x^n)}}}{\sqrt{\det 4\eta C++}}$$

So, I am left with the following:

$\sqrt{\frac{\eta}{\pi}}\left(e^{-\eta \sum_{s}y_s^2-\eta y^2}\frac{\pi^{n/2}e^{\frac{\eta^2 b++'b++}{4\eta (\mu^1+...+\mu^n+x^1+...+x^n)}}}{\sqrt{\det C++}}\div e^{-\eta \sum_{s}y_s^2}\frac{\pi^{n/2}e^{\frac{\eta^2 b'b}{4\eta (\mu^1+...+\mu^n)}}}{\sqrt{\det 4\eta C}}\right)=\sqrt{\frac{\eta}{\pi}}\frac{e^{\frac{\eta y^2+(\eta^2 b++'b++)}{{4\eta (\mu^1+...+\mu^n+x^1+...+x^n)}}-\frac{\eta^2 b'b}{{4\eta (\mu^1+...+\mu^n)}}}\sqrt{\det 4\eta C}}{\sqrt{\det 4\eta C++}}$

where $C++=aD^{-1}+\sum_s^{t+1} x_tx_t'$ and $b++=2\sum_s y_s x_s- 2yx_{t+1}$. So I write:

$$\sqrt{\frac{\eta}{\pi}}\frac{e^{\frac{\eta y^2+(\eta^2 b++'b++)}{{4\eta (\mu^1+...+\mu^n+x^1+...+x^n)}}-\frac{\eta^2 b'b}{{4\eta (\mu^1+...+\mu^n)}}}\sqrt{\det 4\eta C}}{\sqrt{\det 4\eta C++}}=\sqrt{\frac{\eta}{\pi}}\left(\frac{e^{\frac{\eta y^2 +\eta^2(b-2yx_{t+1})'(b-2yx_{t+1})}{4\eta (\mu^1+...+\mu^n+x^1+...+x^n)}-\frac{\eta^2b'b}{4\eta (\mu^1+...+\mu^n)}}\sqrt{\det C}}{\sqrt{\det (C+x_{t+1}x_{t+1}')}}\right)$$

Maybe I can proceed further by using the information given here and (assuming $x_{t+1}x_{t+1}'$ is positive definite(Ideally I would like to avoid this)) write: $$\sqrt{\frac{\eta}{\pi}}\left(\frac{e^{\frac{\eta y^2 +\eta^2(b-2yx_{t+1})'(b-2yx_{t+1})}{4\eta (\mu^1+...+\mu^n+x^1+...+x^n)}-\frac{\eta^2b'b}{4\eta (\mu^1+...+\mu^n)}}\sqrt{\det C}}{\sqrt{\det (C+x_{t+1}x_{t+1}')}}\right)=\sqrt{\frac{\eta}{\pi}}\left(\frac{e^{\frac{\eta y^2 +\eta^2(b-2yx_{t+1})'(b-2yx_{t+1})}{4\eta (\mu^1+...+\mu^n+x^1+...+x^n)}-\frac{\eta^2b'b}{4\eta (\mu^1+...+\mu^n+x^1+...+x^n)}}\sqrt{\det C}}{\sqrt{(4\eta)^n\det C\det(I+L'C^{-1}L)}}\right)=\sqrt{\frac{\eta}{\pi}}\left(\frac{e^{\frac{\eta y^2 +\eta^2(b-2yx_{t+1})'(b-2yx_{t+1})}{4\eta (\mu^1+...+\mu^n+x^1+...+x^n)}-\frac{\eta^2b'b}{4\eta (\mu^1+...+\mu^n)}}}{\sqrt{(4\eta)^n\det(I+L'C^{-1}L)}}\right)=\sqrt{\frac{\eta}{\pi}}\left(\frac{e^{\frac{\eta y^2 +\eta^2(b-2yx_{t+1})'(b-2yx_{t+1})-\eta^2b'b4\eta (x^1+...+x^n)}{4\eta (\mu^1+...+\mu^n+x^1+...+x^n)}}}{\sqrt{(4\eta)^n\det(I+L'C^{-1}L)}}\right)$$

Can someone please tell me: what I have done so far is correct and how can I proceed to simplify whats left?

Remark: I am hoping to get a Gaussian density. Once I solve, then I can analytically get the mean and variance. It may be worth noting that $C+x_{t+1}x_{t+1}$ is also symmetric and positive definite.

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