Currently reading from Christopher Bishop's Pattern Recognition and Machine Learning book about parameter distribution under a bayesian linear regression.

On page 153, the author deduces that the posterior distribution over weights for a model of the form $y(\bf x, \bf w) + \varepsilon$ after $N$ observations with $y({\bf x}, {\bf w}) = {\bf w}^T{\bf \phi(x)}$, $p(\varepsilon) = \mathcal{N}(0, \beta)$ and with a conjugate prior of the form

$$ p({\bf w}) = \mathcal{N}({\bf w} | {\bf m}_0, {\bf S}_0) $$


$$ \begin{align} {\bf m}_N &= {\bf S}_N({\bf S}_0^{-1}{\bf m}_0 + \beta{\bf \Phi}^T{\bf t}) \\ {\bf S}_N^{-1} &= {\bf S}_0^{-1} + \beta{\bf \Phi}^T{\bf \Phi} \end{align} $$

Then, he continues by saying that

If we consider an infinitely broad prior ${\bf S}_0 = \alpha^{-1}{\bf I}$ with $\alpha \to 0$, the mean ${\bf m}_N$ of the posterior distribution reduces to the maximum likelihood value ${\bf w}_{\text{ML}}$

Where ${\bf w}_{\text{ML}} =({\bf \Phi}^T{\bf \Phi})^{-1}{\bf \Phi}{\bf t}$.

In doing so, I arrive at $$ {\bf m}_N = {\bf m}_0 + \alpha^{-1}\beta{\bf \Phi}^T{\bf t} + \alpha\beta^{-1}({\bf \Phi}^T{\bf \Phi})^{-1}{\bf m}_0 + ({\bf \Phi}^T{\bf \Phi})^{-1}{\bf \Phi}{\bf t} $$

For which the only way to get rid of ${\bf m}_0$ if it is a zero vector. Thus, we should consider a zero mean isotropic gaussian as a prior. In doing so I arrive at

$$ {\bf m}_N = \alpha^{-1}\beta{\bf \Phi}^T{\bf t} + ({\bf \Phi}^T{\bf \Phi})^{-1}{\bf \Phi}{\bf t} $$

Finally, if I take the limit $\alpha \to 0$ in ${\bf m}_0$, then $\lim_{\alpha \to 0^+} 1/\alpha = \infty $ and $\lim_{\alpha \to 0^-} 1/\alpha = -\infty $, which does not converge and cannot reduced ${\bf m}_N$ to ${\bf w}_{\text{ML}}$. Since we want an infinitely broad prior, ${\bf S}_0$ has to be ${\bf S}_0 = \alpha^{-1}{\bf I}$ with $\alpha \to 0$.

What's the argument I need to conclude that ${\bf m}_N$ does indeed converge to ${\bf w}_{\text{ML}}$?

  • $\begingroup$ I'm reading this part and completely dumbfounded as to how the author got these calculations for $$m_N$$ and $$S_N^{-1}$$. Care to explain how he got there? $\endgroup$
    – doctopus
    Commented Oct 22, 2019 at 22:44

1 Answer 1


From the formulas for $\mathbf{m_N}$ and $\mathbf{S_N}^{-1}$, I arrive at: $$\mathbf{S_N^{-1} m_N=\alpha I\ m_0}+\beta\mathbf{\Phi^Tt}\rightarrow (\alpha\bf I + \beta{\bf \Phi}^T{\bf \Phi})\mathbf{m_N=\alpha\bf I\ m_0}+\beta\mathbf{\Phi^Tt}$$ Taking $\alpha\rightarrow 0$ in both sides of the equation leaves us with the following: $$\beta\ {\bf \Phi}^T{\bf \Phi}\ \mathbf{m_N}=\beta\ \mathbf{\Phi^Tt}\rightarrow \bf m_N=({\bf \Phi^T\Phi})^{-1}\bf \Phi^T\bf t$$


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