Posterior mean of multivariate normal distribution in Murphy's Probabilistic Machine Learning In the new Probabilistic Machine Learning book by Kevin P. Murphy, to formulate the posterior mean of a multivariate normal distribution, he first defined the likelihood as

However, I can't seem to understand how to go from Equation 4.188 to 4.189 and subsequently to 4.190. Also, how can we simply replace the observations with their mean?
It would be an enormous help if somebody can explain it to me or perhaps guide me how to derive this. Thank you so much!
 A: I don't get it either. Here is what I have so far:
First, in equation (4.189) I think the normalization factor has to be to the power of $N$.
Next, I concentrate on the sum in the exponent and presume $\boldsymbol\mu$ is zero. Then this is just the sum of square norms over the $\mathbf y_i$ for the norm given by $\Sigma^{-1}$, i.e.:
$$
\sum_{n=1}^N\mathbf y_n^T\mathbf\Sigma^{-1}\mathbf y_n = \sum_{n=1}^N\|\mathbf y_n\|^2.
$$
Next, the sum of squares is equal to the sum of squared distances $\boldsymbol \delta_n:= \mathbf y_n - \bar{\mathbf y}$ to the mean $\bar{\mathbf y}$ plus $N$ times the square of the mean, i.e.:
$$
\sum_{n=1}^N\|\mathbf y_n\|^2 = \sum_{n=1}^N\|\boldsymbol \delta_n\|^2 + N\|\bar{\mathbf y}\|.
$$
If I substitute this into the exponent, I get
$$
\begin{align}
\exp\left(-\frac{1}{2}\sum_{n=1}^N\mathbf y_n^T\mathbf\Sigma^{-1}\mathbf y_n\right) &= \exp\left(-\frac{1}{2}\sum_{n=1}^N\boldsymbol \delta_n^T\Sigma^{-1}\boldsymbol \delta_n\right)\; \exp\left(-\frac{1}{2} \bar{\mathbf y}^T\left(\frac{1}{N}\Sigma\right)^{-1}\bar{\mathbf y}\right)\\
&= C \exp\left(-\frac{1}{2}\sum_{n=1}^N\boldsymbol \delta_n^T\Sigma^{-1}\boldsymbol \delta_n\right)\; N(\bar{\mathbf y}|0,\frac{1}{N}\Sigma )
\end{align}
$$
which is different.
A: From the likelihood, where the normalizing constant should be $(2\pi)^{-N\cdot D/2}\begin{vmatrix}\!\Sigma\!\end{vmatrix}^{-N/2}$, we see that
$$\mathbf{Y}\sim\mathcal{N}_{D\cdot N}\left({\left(\mu,\ldots,\mu\right)}^\top,\mathrm{blockdiag}\left(\Sigma,\ldots,\Sigma\right)\right)$$
for the vector $\mathbf{Y}={(Y_1,\ldots,Y_N)}^\top$.
Thus, we have
$$\bar{Y}\sim\mathcal{N}_{D}\left(\mu,\frac{1}{N}\Sigma\right).$$
Considering the likelihood of $\bar{y}$ as function of $\mu$, we get
$$
\begin{align} 
\mathcal{L}\left(\mathbf{\mu};\bar{y}\right)&\overset{\mu}{\propto}\exp\left(-\frac{1}{2}\left(N\bar{y}^\top \Sigma^{-1}\bar{y}-2\mu^\top\Sigma^{-1}N\bar{y}+N\mu^\top\Sigma^{-1}\mu\right)\right)\\
&\overset{\mu}{\propto} \exp\left(-\frac{1}{2}\left(-2\mu^\top\Sigma^{-1}N\bar{y}+N\mu^\top\Sigma^{-1}\mu\right)\right)\\
&= \exp\left(-\frac{1}{2}\sum_{n=1}^N\left(-2\mu^\top\Sigma^{-1}y_n+\mu^\top\Sigma^{-1}\mu\right)\right)\\
&\overset{\mu}{\propto} \exp\left(-\frac{1}{2}\sum_{n=1}^N\left(y_n^\top\Sigma^{-1}y_n-2\mu^\top\Sigma^{-1}y_n+\mu^\top\Sigma^{-1}\mu\right)\right)\\
&\overset{\mu}{\propto} \mathcal{L}\left(\mathbf{\mu};\mathbf{y}\right).
\end{align}
$$
Maybe this is what the excerpt from the book is trying to convey.
