# How do I show this equation involving the Gaussian pdf? [duplicate]

Im reading this paper momentarily, and there is one equation (9) in section 3.1. that I just can't wrap my head around yet:

\begin{align} \mathcal{N}(\textbf{y}_d;\textbf{0},\pmb{\Phi \Phi}^T + \tau^{-1}\textbf{I}_N) = \int \mathcal{N}(\textbf{y}_d;\pmb{\Phi}\textbf{w}_d, \tau^{-1}\textbf{I}_N) \mathcal{N}(\textbf{w}_d; \textbf{0}, \textbf{I}_k) d\textbf{w}_d, \end{align}

where $$\pmb{\Phi \Phi^T}$$ is our covariance function of a Gaussian Process, $$\pmb{\Phi}$$ is our feature matrix, $$(\textbf{X,Y})$$ our data with $$\textbf{W}_2 = [\textbf{w}_d]_{d=1}^D$$ and $$\textbf{w}_d \sim \mathcal{N}(\textbf{0}, \textbf{I}_K)$$, with a constant $$\tau$$.

If I would want to show this equation mathematically, I would use the pdf of the joint normal distribution:

\begin{align} \mathcal{N}(\textbf{y}_d;\textbf{0},\pmb{\Phi \Phi}^T + \tau^{-1}\textbf{I}_N) = \frac{1}{\sqrt{(2\pi)^D \text{det}(\pmb{\Phi \Phi}^T + \tau^{-1}\textbf{I}_N)}} \exp \left(-\frac{1}{2}(\textbf{y}_d - \textbf{0})^T (\pmb{\Phi \Phi}^T + \tau^{-1}\textbf{I}_N)^{-1}(\textbf{y}_d - \textbf{0})\right) \end{align} and then actually I don't know how to proceed to end up on the left side.. I also tried starting from right to left, but similarly I get stuck..

Anyone has an idea how to go about showing this equation?

Cheers

EDIT:

I tried showing the equation from right to left:

\begin{align} &\int \mathcal{N}(\textbf{y}_d;\pmb{\Phi}\textbf{w}_d, \tau^{-1}\textbf{I}_N) \mathcal{N}(\textbf{w}_d; \textbf{0}, \textbf{I}_k) d\textbf{w}_d \\ &= \int \left( \frac{1}{\sqrt{(2\pi)^D \text{det}(\tau^{-1}\textbf{I}_N) }} \exp \left( -\frac{1}{2}(\textbf{y}_d - \pmb{\Phi}\textbf{w}_d)^T (\tau^{-1}\textbf{I}_N)^{-1} (\textbf{y}_d - \pmb{\Phi}\textbf{w}_d) \right) \\ \cdot \frac{1}{\sqrt{(2\pi)^D \text{det}(\textbf{I}_K) }} \exp \left( -\frac{1}{2}(\textbf{w}_d - \textbf{0})^T (\textbf{I}_K)^{-1} (\textbf{w}_d - \textbf{0}) \right) \right) d\textbf{w}_d \\ &= \int \frac{1}{\sqrt{(2\pi)^{2D} \cdot N \cdot \tau{-1} }} \exp \left( -\frac{1}{2}(\textbf{y}_d - \pmb{\Phi}\textbf{w}_d)^T (\tau \textbf{I}_N) (\textbf{y}_d - \pmb{\Phi}\textbf{w}_d) -\frac{1}{2}\textbf{w}_d^T \textbf{I}_K \textbf{w}_d \right) d\textbf{w}_d \\ &= \int \frac{\sqrt{\tau}}{(2\pi)^D \cdot \sqrt{N}} \exp \left( -\frac{1}{2}\tau\textbf{y}_d^2 + \frac{1}{2}\tau2\textbf{y}_d\pmb{\Phi}\textbf{w}_d - \frac{1}{2}\tau\pmb{\Phi}^2\textbf{w}_d^2 - \frac{1}{2}\textbf{w}_d^2 \right) d\textbf{w}_d \\ &= \frac{\sqrt{\tau}}{(2\pi)^D \cdot \sqrt{N}} \exp \left( -\frac{1}{2}\tau\textbf{y}_d^2 + \frac{1}{2}\tau2\textbf{y}_d\pmb{\Phi}\textbf{w}_d - \frac{1}{2}\tau\pmb{\Phi}^2\textbf{w}_d^2 - \frac{1}{2}\textbf{w}_d^2 \right) \\ &\cdot \frac{1}{\frac{1}{2}\tau2\textbf{y}_d\pmb{\Phi}} \cdot \frac{-1}{\tau\pmb{\Phi}^2\textbf{w}_d} \cdot \frac{-1}{\textbf{w}_d} \end{align}

All correct so far? And now?

• The product of the integrands is obtained by adding the arguments of the exponential functions. Simply complete the square and perform the integration.
– whuber
Commented Jun 12, 2020 at 13:12
• @whuber thanks for your comment! I edited my question with how far I got into the calculations, and I just saw you linked another answer on how to complete the square, thanks friend! Commented Jun 12, 2020 at 16:34
• Dear friend @whuber, I continued working on the equation, taking your helpful thoughts into account. I opened a new question for the actual integration, perhaps you could take a look at it, I'd be grateful, cheers Commented Jun 15, 2020 at 10:29