2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Oh, of course! Thanks! $\endgroup$ – Elizabeth_Banks Jan 18 at 23:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.