Bayesian statistics: Write up expression for posterior density
Solution 1:
By independence, $p(\alpha,\beta)\propto p(\alpha)p(\beta)$. Assuming $x_i$ is fixed, and $Y_i|\alpha,\beta\sim \text{Poiss}(\lambda_i)$ and $Y_1,...,Y_n$ are independent (conditional on $\alpha,\beta$), we have likelihood $$p(Y_1,...,Y_n|\alpha,\beta)\propto \exp(-\sum_i \lambda_i)\Pi_{i}{\lambda_{i}}^{Y_i}\\ = \exp(- n\alpha-\beta\sum_i x_i)\Pi_{i}{(\alpha+\beta x_i)}^{Y_i} .$$ Now use Bayes': $$p(\alpha,\beta|Y_1,...,Y_n)\propto p(Y_1,...,Y_n|\alpha,\beta)p(\alpha)p(\beta)\\ $$