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Suppose that internet users access a particular website according to a Poisson process with rate $\lambda$ per hour, but $\lambda$ is unknown. The website maintainer believes that $\lambda$ has a continuous distribution with p.d.f.

$$f(\lambda) =\begin{cases} 2e^{-2\lambda} & \text{for}\ \lambda > 0 \\ 0 & \text{otherwise} \end{cases}$$

Let $X$ be the number of users who access the website during a one-hour period. If $X = 1$ is observed, find the conditional p.d.f. of $\lambda$ given $X = 1$.

My work: The joint p.f./p.d.f of $X$ and $\lambda$ should be the Poisson probability function with parameter being $\lambda \times f(\lambda)$, i.e.

$$\lambda_0 = e^{-\lambda}\frac{\lambda^{x}}{x!}2e^{-2\lambda}$$

Can someone help me proceed?

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The conditional density function is given by $$ f_{\lambda \mid X=x}(t, x) = \frac{p_{X=x \mid \lambda=t}(x, t) f_{\lambda}(t)}{p_X(x)} $$ where $p_{X=x \mid \lambda = t}(x, t)$ is the Poisson$(\lambda=t)$ mass function evaluated at $x$, and $p_X(x)$ is the marginal probability that $X = x$, found by taking the integral of the numerator with respect to $\lambda$ from $0$ to $\infty$. As another hint, the gamma distribution is the conjugate prior for the mean of a Poisson distribution, so you should find that $f_{\lambda \mid X=x}$ also belongs to the gamma family.

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  • $\begingroup$ Correct me if I am wrong. What I took as $\lambda_0$ is exactly the numerator. And the denominator is its integral. $\endgroup$ – Qwerty Dec 9 '15 at 4:39
  • $\begingroup$ Yes, that's correct. $\endgroup$ – dsaxton Dec 9 '15 at 13:50

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