Let me recall the basics of the EM algorithm first. When looking for the maximum likelihood estimate of a likelihood of the form$$\int f(x,z|\beta)\text{d}z,$$ the algorithm proceeds by iteratively maximising (M) expected (E) complete log-likelihoods, which results in maximising (in $\beta$)at iteration $t$ the function $$Q(\beta|\beta_i)=\int \log f(x,z|\beta) f(z|x,\beta_t)\text{d}z$$ The algorithm must therefore starts by identifying the latent variable $z$ and its conditional distribution.

In your case it seems that the latent variable is $\varpi$ made of the $w_i$'s while the parameter of interest is $\beta$. If you process both $\beta$ and $\varpi$ as latent variables there is no parameter left to optimise. However, this also means that the prior on $\beta$ is not used.

If we look more precisely at the case of $w_i$, its conditional distribution is given by$$f(w_i|x_i,y_i,\beta)\propto\sqrt{w_i}\exp\left\{-w_i(y_i-\beta^Tx_i)^2/2\sigma^2\right\}\times w_i^{a-1}\exp\{-bw_i\}$$ which qualiifies as a $$\mathcal{G}\left(a+1/2,b+(y_i-\beta^Tx_i)^2/2\sigma^2\right)$$distribution.

The completed log-likelihood being$$\sum_i \frac{1}{2}\left\{\log(w_i)- w_i(y_i-\beta^Tx_i)^2/\sigma^2\right\}$$ the part that depends on $\beta$ simplifies as$$\sum_iw_i(y_i-\beta^Tx_i)^2/2\sigma^2$$and the function $Q(\beta|\beta_t)$ is proportional to \begin{align*}\mathbb{E}\left[\sum_iw_i(y_i-\beta^Tx_i)^2\Big|X,Y,\beta_t\right]&=\sum_i\mathbb{E}[w_i|X,Y,\beta_t](y_i-\beta^Tx_i)^2\\&=\sum_i\frac{a+1/2}{b+(y_i-\beta_t^Tx_i)^2/2\sigma^2}(y_i-\beta^Tx_i)^2\end{align*} Maximising this function in $\beta$ amounts to a weighted linear regression, with weights $$\frac{a+1/2}{b+(y_i-\beta_t^Tx_i)^2/2\sigma^2}$$

Let me recall the basics of the EM algorithm first. When looking for the maximum likelihood estimate of a likelihood of the form$$\int f(x,z|\beta)\text{d}z,$$ the algorithm proceeds by iteratively maximising (M) expected (E) complete log-likelihoods, which results in maximising (in $\beta$)at iteration $t$ the function $$Q(\beta|\beta_i)=\int \log f(x,z|\beta) f(z|x,\beta_t)\text{d}z$$ The algorithm must therefore starts by identifying the latent variable $z$ and its conditional distribution.

In your case it seems that the latent variable is $\varpi$ made of the $w_i$'s while the parameter of interest is $\beta$. If you process both $\beta$ and $\varpi$ as latent variables there is no parameter left to optimise. However, this also means that the prior on $\beta$ is not used.

If we look more precisely at the case of $w_i$, its conditional distribution is given by$$f(w_i|x_i,y_i,\beta)\propto\sqrt{w_i}\exp\left\{-w_i(y_i-\beta^Tx_i)^2/2\sigma^2\right\}\times w_i^{a-1}\exp\{-bw_i\}$$ which qualiifies as a $$\mathcal{G}\left(a+1/2,b+(y_i-\beta^Tx_i)^2/2\sigma^2\right)$$distribution.

The completed log-likelihood being$$\sum_i \frac{1}{2}\left\{\log(w_i)- w_i(y_i-\beta^Tx_i)^2/\sigma^2\right\}$$ the part that depends on $\beta$ simplifies as$$-\sum_iw_i(y_i-\beta^Tx_i)^2/2\sigma^2$$and the function $-Q(\beta|\beta_t)$ is proportional to \begin{align*}\mathbb{E}\left[\sum_iw_i(y_i-\beta^Tx_i)^2\Big|X,Y,\beta_t\right]&=\sum_i\mathbb{E}[w_i|X,Y,\beta_t](y_i-\beta^Tx_i)^2\\&=\sum_i\frac{a+1/2}{b+(y_i-\beta_t^Tx_i)^2/2\sigma^2}(y_i-\beta^Tx_i)^2\end{align*} Maximising this function in $\beta$ amounts to a weighted linear regression, with weights $$\frac{a+1/2}{b+(y_i-\beta_t^Tx_i)^2/2\sigma^2}$$

Let me recall the basics of the EM algorithm first. When looking for the maximum likelihood estimate of a likelihood of the form$$\int f(x,z|\beta)\text{d}z,$$ the algorithm proceeds by iteratively maximising (M) expected (E) complete log-likelihoods, which results in maximising (in $\beta$)at iteration $t$ the function $$Q(\beta|\beta_i)=\int \log f(x,z|\beta) f(z|x,\beta_t)\text{d}z$$ The algorithm must therefore starts by identifying the latent variable $z$ and its conditional distribution.

In your case it seems that the latent variable is $\varpi$ made of the $w_i$'s while the parameter of interest is $\beta$. If you process both $\beta$ and $\varpi$ as latent variables there is no parameter left to optimise. However, this also means that the prior on $\beta$ is not used.

If we look more precisely at the case of $w_i$, its conditional distribution is given by$$f(w_i|x_i,y_i,\beta)\propto\sqrt{w_i}\exp\left\{-w_i(y_i-\beta^Tx_i)^2/2\sigma^2\right\}\times w_i^{a-1}\exp\{-bw_i\}$$ which qualiifies as a $$\mathcal{G}\left(a+1/2,b+(y_i-\beta^Tx_i)^2/2\sigma^2\right)$$distribution.

The completed log-likelihood being$$\sum_i \left\{\log(w_i)- w_i(y_i-\beta^Tx_i)^2/2\sigma^2\right\}$$ the part that depends on $\beta$ simplifies as$$\sum_iw_i(y_i-\beta^Tx_i)^2/2\sigma^2$$and the function $Q(\beta|\beta_t)$ is proportional to \begin{align*}\mathbb{E}\left[\sum_iw_i(y_i-\beta^Tx_i)^2\Big|X,Y,\beta_t\right]&=\sum_i\mathbb{E}[w_i|X,Y,\beta_t](y_i-\beta^Tx_i)^2\\&=\sum_i\frac{a+1/2}{b+(y_i-\beta_t^Tx_i)^2/2\sigma^2}(y_i-\beta^Tx_i)^2\end{align*} Maximising this function in $\beta$ amounts to a weighted linear regression, with weights $$\frac{a+1/2}{b+(y_i-\beta_t^Tx_i)^2/2\sigma^2}$$

Let me recall the basics of the EM algorithm first. When looking for the maximum likelihood estimate of a likelihood of the form$$\int f(x,z|\beta)\text{d}z,$$ the algorithm proceeds by iteratively maximising (M) expected (E) complete log-likelihoods, which results in maximising (in $\beta$)at iteration $t$ the function $$Q(\beta|\beta_i)=\int \log f(x,z|\beta) f(z|x,\beta_t)\text{d}z$$ The algorithm must therefore starts by identifying the latent variable $z$ and its conditional distribution.

In your case it seems that the latent variable is $\varpi$ made of the $w_i$'s while the parameter of interest is $\beta$. If you process both $\beta$ and $\varpi$ as latent variables there is no parameter left to optimise. However, this also means that the prior on $\beta$ is not used.

If we look more precisely at the case of $w_i$, its conditional distribution is given by$$f(w_i|x_i,y_i,\beta)\propto\sqrt{w_i}\exp\left\{-w_i(y_i-\beta^Tx_i)^2/2\sigma^2\right\}\times w_i^{a-1}\exp\{-bw_i\}$$ which qualiifies as a $$\mathcal{G}\left(a+1/2,b+(y_i-\beta^Tx_i)^2/2\sigma^2\right)$$distribution.

The completed log-likelihood being$$\sum_i \frac{1}{2}\left\{\log(w_i)- w_i(y_i-\beta^Tx_i)^2/\sigma^2\right\}$$ the part that depends on $\beta$ simplifies as$$\sum_iw_i(y_i-\beta^Tx_i)^2/2\sigma^2$$and the function $Q(\beta|\beta_t)$ is proportional to \begin{align*}\mathbb{E}\left[\sum_iw_i(y_i-\beta^Tx_i)^2\Big|X,Y,\beta_t\right]&=\sum_i\mathbb{E}[w_i|X,Y,\beta_t](y_i-\beta^Tx_i)^2\\&=\sum_i\frac{a+1/2}{b+(y_i-\beta_t^Tx_i)^2/2\sigma^2}(y_i-\beta^Tx_i)^2\end{align*} Maximising this function in $\beta$ amounts to a weighted linear regression, with weights $$\frac{a+1/2}{b+(y_i-\beta_t^Tx_i)^2/2\sigma^2}$$