# Calculating conditional probability $P(\Theta \le c | Y=0)$

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.