Let $Y$ be a random variable with $Pois(\theta)$ distribution and the parameter $\theta$ be a realization of a random variable $\Theta$ with a priori distribution $Exp(\lambda)$.

The task is to calculate $P(\Theta \le c| Y=0)$.

I got stuck. So far I have:

$Y|\Theta \sim Poiss(\Theta)$, $\Theta \sim Exp(\lambda)$, $\pi(\theta) = \lambda e^{-\lambda \theta}$, $\lambda>0, P(Y=k|\Theta = \theta) = \frac{\theta^k}{k!} e^{-\theta}$.

$P(\Theta \le c | Y=0) = \frac{P(Y=0, \Theta \le c)}{P(Y=0)}$

$P(Y=0) = \int P(Y=0|\theta) \pi(\theta) d\theta = \int \limits_0^{\infty} e^{-\theta} \lambda e^{-\lambda \theta} d\theta = \frac{\lambda}{\lambda+1}$

$P(Y=0, \Theta \le c) = P(Y=0|\Theta \le c) P(\Theta \le c)$

$P(\Theta \le c) = 1 - e^{-\lambda c}$.

I got stuck at $P(Y=0|\Theta \le c)$. How to calculate it? Please help.


I think it'd be easier to calculate the following:$$P(\Theta \leq c| Y = 0)=\int_{0}^c f_{\Theta|Y=0}(\theta)d\theta$$

where $$f_{\Theta|Y=0}(\theta)=\frac{P(Y=0|\Theta=\theta)f_\Theta(\theta)}{P(Y=0)}=\frac{e^{-\theta}\lambda e^{-\lambda\theta}}{P(Y=0)}$$ which is easy to integrate.


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