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Let $X_1,...,X_n$ be a random sample and $\lambda >0$ be a parameter, with $X_i |\lambda \sim Poisson (\lambda)$ and $\lambda \sim Gamma(\alpha, \beta) (\lambda)=\dfrac {1}{\Gamma(\alpha) \beta^\alpha} \lambda ^{\alpha-1} e^{-\lambda / \beta}$.

I want to find a Bayesian test of hypothesis for $H_0 : \lambda \le \lambda_0$ against $H_1: \lambda > \lambda_1$ .

Now I know that a simple rule is that accept $H_0 :$ If $P(\lambda \le \lambda_0|Y=y) \ge P(\lambda > \lambda_0 | Y=y) $ .

Now $Y=\sum_{i=1}^n X_i$ is a sufficient statistics , and the posterior pdf of $\lambda |Y=y$ is a constant (depending on $y$ but independent of $\lambda$ ) times $\lambda ^{y+\alpha -1} e^{-n\lambda -\frac {\lambda}{ \beta}}$ , so $\lambda|Y=y \sim Gamma (y+\alpha, \dfrac {\beta}{n\beta+1}), so $ $P(\lambda \le \lambda_0|Y=y) \ge P(\lambda > \lambda_0 | Y=y)$ is equivalent to saying

$\dfrac {1}{\Gamma (y+\alpha ) } \gamma (y+\alpha,n\lambda_0 +\frac {\lambda_0}{ \beta})\ge 1-\dfrac {1}{\Gamma (y+\alpha ) } \gamma (y+\alpha,n\lambda_0 +\frac {\lambda_0}{ \beta})$, so

$\lambda_0 \ge $ median of $Gamma (y+\alpha, \dfrac {\beta}{n\beta+1}) $ .

But I don't know what to do further. Am I even on the right track ?

Please help .

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  • $\begingroup$ Yes, the rule you chose is equivalent to deciding based of which side of $\lambda_0$ the median of the posterior is on, and it seems like you've computed the posterior correctly. Assuming this is the rule you want, I don't know that there's anything you can do further until it's actually time to put some numbers in. There's no closed form expression for that median. $\endgroup$ – spaceisdarkgreen Jan 13 at 6:42

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