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I am reading http://www.yisongyue.com/courses/cs159/lectures/LinUCB.pdf and come across this slide

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What has been confusing me boils down to showing that multivariate Gaussian is conjugate to itself given a Gaussian likelihood. I did quite a bit of searching on both this site and on Google and found the Kevin Murphy's slides showing the univariate case. I tried to generalize to the finite-dimension case but failed in matching the coefficients.

Can somebody help me in deriving the posterior form given the prior and the likelihood? Thank you!

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Prior:

$$p_0(\theta_a) \varpropto \text{exp}(-\frac{1}{2}\theta_{a}^{T} (\lambda I_d)^{-1} \theta_a) = \text{exp}(-\frac{1}{2} \frac{1}{\lambda}\theta_{a}^{T} \theta_a)$$

MLE: \begin{align*} p(y_{t,a} | \theta_a) &= (2 \pi)^{\frac{n_{t,a}}{2}} |\sum|^{\frac{1}{2}} \text{exp}(-\frac{1}{2} (y_{t,a} - D_{t,a} \theta_{a})^{T} I_{n_{t,a}}^{-1} (y_{t,a} - D_{t,a} \theta_a)) \\ & \varpropto \text{exp}(-\frac{1}{2}(y_{t,a} - D_{t,a} \theta_{a})^{T} I_{n_{t,a}}^{-1} (y_{t,a} - D_{t,a} \theta_a)) \end{align*}

Hence, \begin{align*} p_{t,a}(\theta_a) &\varpropto \text{exp}(-\frac{1}{2} \frac{1}{\lambda} \theta_{a}^{T} \theta_a) ~ \text{exp}(-\frac{1}{2}(y_{t,a} - D_{t,a} \theta_{a})^{T} I_{n_{t,a}}^{-1} (y_{t,a} - D_{t,a} \theta_a)) \\ &\varpropto \text{exp}(-\frac{1}{2} (\frac{1}{\lambda} \theta_{a}^{T} \theta_a + y_{t,a}^{T} y_{t,a} - y_{t,a}^{T} D_{t,a} \theta_a - \theta_{a}^{T} D_{t,a}^{T} y_{t,a} + \theta_{a}^{T} D_{t,a}^{T} D_{t,a} \theta_a)) \end{align*}

By definition, we also have \begin{align*} p_{t,a}(\theta_a) &\varpropto \text{exp}(-\frac{1}{2}(\theta_a - \hat{\theta_{a}})^{T} A_{t,a} (\theta_a - \hat{\theta_a})) \\ &\varpropto \text{exp}(-\frac{1}{2}(\theta_a^T A_{t,a} \theta_a - \theta_a^T A_{t,a} \hat{\theta_a} - \hat{\theta_a}^T A_{t,a} \theta_a + \hat{\theta}^T A_{t,a} \hat{\theta})) \end{align*}

And so we have \begin{align*} \text{exp}(-\frac{1}{2} (\frac{1}{\lambda} \theta_{a}^{T} \theta_a + y_{t,a}^{T} y_{t,a} - y_{t,a}^{T} D_{t,a} \theta_a - \theta_{a}^{T} D_{t,a}^{T} y_{t,a} + \theta_{a}^{T} D_{t,a}^{T} D_{t,a} \theta_a)) &= \text{exp}(-\frac{1}{2}(\theta_a^T A_{t,a} \theta_a - \theta_a^T A_{t,a} \hat{\theta_a} - \hat{\theta_a}^T A_{t,a} \theta_a + \hat{\theta}^T A_{t,a} \hat{\theta})) \\ \frac{1}{\lambda} \theta_{a}^{T} \theta_a + y_{t,a}^{T} y_{t,a} - y_{t,a}^{T} D_{t,a} \theta_a - \theta_{a}^{T} D_{t,a}^{T} y_{t,a} + \theta_{a}^{T} D_{t,a}^{T} D_{t,a} \theta_a &= \theta_a^T A_{t,a} \theta_a - \theta_a^T A_{t,a} \hat{\theta_a} - \hat{\theta_a}^T A_{t,a} \theta_a + \hat{\theta}^T A_{t,a} \hat{\theta} \end{align*}

Equating the "second coefficient" of $\theta_a$ $$\theta_a^{T} (\frac{1}{\lambda} I_d)\theta_a + \theta_a^T D_{t,a}^T D_{t,a} \theta_a = \theta_a^T A_{t,a} \theta_a$$ Hence $A_{t,a} = D_{t,a}^T D_{t,a} + \frac{1}{\lambda} I_d$,

Similarly, equating the "first coefficient" of $\theta_a$, we have $$\hat{\theta_a}^T = A_{t,a}^{-1} b_{t,a}$$

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