Assume that each $Y_i$ has a Poisson distribution with mean $\lambda_i$ where: $$log(\lambda_i)=\alpha+\beta x_i$$ Moreover, we assume $Y_1,...,Y_n$ are independent (given the parameters). For the Bayessian approach, we are going to use the prior for $(\alpha,\beta)$, where $\alpha$ and $\beta$ are independent, and $\alpha \sim N(3,25)$ and $\beta \sim N(0,1)$.
Now I have to write up an unnormalized expression for posterior density. Bayes rule give us: $$p(\alpha,\beta|Y_1,...,Y_n)\propto p(\alpha)p(\beta) p(Y_1,...,Y_n|\alpha,\beta)\\ $$ And then I get $$p(Y_1,...,Y_n|\alpha,\beta)\propto \exp(-\sum_i \lambda_i)\prod_{i}{\lambda_{i}}^{Y_i}\\ = \exp(- n\alpha-\beta\sum_i x_i)\prod_{i}{(\alpha+\beta x_i)}^{Y_i} $$ and $$p(\alpha)=\frac{1}{\sqrt{25}\sqrt{2\pi}}e^{-\frac{1}{2}(\frac{\alpha-3}{\sqrt{25}})^2}=\frac{1}{5 \sqrt{\pi}}e^{-\frac{(\alpha-3)^2}{50}}$$ and $$p(\beta)=\frac{1}{\sqrt{1}\sqrt{2\pi}}e^{-\frac{1}{2}(\frac{\beta-0}{\sqrt{1}})^2}=\frac{1}{ \sqrt{\pi}}e^{-\frac{\beta^2}{2}}$$
Therefore we get that: $$p(\alpha,\beta|Y_1,...,Y_n)\propto \frac{1}{5 \sqrt{\pi}}e^{-\frac{(\alpha-3)^2}{50}}\frac{1}{ \sqrt{\pi}}e^{-\frac{\beta^2}{2}}\exp(- n\alpha-\beta\sum_i x_i)\prod_{i}{(\alpha+\beta x_i)}^{Y_i}$$ Now I have to find out what distribution $p(Y_1,...,Y_n|\alpha,\beta)$ is. Can anyone help me?