This is very basic, but I have been stuck on this problem for a while. Suppose $Y_1, \dots, Y_n|\alpha,\beta\sim Gamma(\alpha, \beta)$ is iid with $\alpha$ known. I want conjugate prior for $\beta$ and the posterior.
My work:
$p(\beta|y_1, \dots, y_n)=p(y_1, \dots, y_n|\beta)p(\beta)=(\prod_{i=1}^{n} y_i)^{\alpha-1} \exp(-\beta\sum_{i} y_i) p(\beta)$
$\propto \exp(-\beta\sum_{i} y_i) p(\beta)$.
Therefore, the conjugate prior for $\beta$ would be gamma$(\alpha_0, \beta_0)$.
In this case, we can derive the posterior as:
$p(\beta|y_1,\dots, y_n)\propto \beta^{\alpha_0 -1} \exp(-\beta(\sum_{i} y_i+\beta_0))$.
So, the posterior is gamma$(\alpha_0 , \sum_{i} y_i+\beta_0)$.
However, Wikipedia says the posterior should be gamma$(\alpha_0 +n\alpha, \sum_{i} y_i+\beta_0)$. I don't understand where does $n\alpha$ come from. I also don't understand why my derivation is not correct. Thanks in advance!
I think you may have missed an important term from the likelihood
The density of a Gamma distribution with parameters $\alpha$ and $\beta$ is $\frac{1}{\Gamma(\alpha)}\beta^{\alpha }y^{\alpha-1}\exp(-\beta y)$ when $y \gt 0$
so the likelihood for the observations here is proportional (with terms involving $\beta$) to $$\beta^{\alpha n}\exp\left(-\beta \sum_{i=1}^{n} y_i\right)$$
while your prior for $\beta$ is proportional (with terms involving $\beta$) to $$\beta^{\alpha_0-1}\exp(-\beta_0 \beta)$$
so the posterior for $\beta$ is proportional to $$\beta^{\alpha_0+\alpha n-1}\exp\left(-\beta\left(\beta_0+\sum_{i=1}^{n} y_i\right)\right)$$
which is proportional to the density of a Gamma distribution with parameters $\alpha_0+\alpha n$ and $\beta_0+\sum_{i=1}^{n} y_i$