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Suppose that $(x_1,..., x_n)$ is a sample from a Poisson($λ$) distribution with $λ ≥ 0$ unknown. If we use the prior distribution for $\lambda$ given by the $Gamma(a,b)$ distribution, then determine the posterior distribution of $\lambda$.


I believe I need $\pi(\lambda | s) = L(\lambda | \bar{x}) \times \pi(\lambda)$

likelihood function times prior

likelihood of poisson is given by

$L(\lambda | \bar{x}) = \prod_{x=1}^{n} \frac{\lambda^x e^{-\lambda}}{x!} = \frac{1}{(nx!)} \lambda^{\sum_{x=1}^{n} x} e^{-n\lambda} = \frac{1}{(nx!)}\lambda^{n \bar{x}}e^{-n\lambda}$

prior is given by

$\pi (\lambda)$ ~ $Gamma(a, b)$

$= \frac{b^a \lambda^{a-1} e^{-b\lambda}}{\Gamma (a)}$ which is the probability density function

so finally

$\pi(\lambda | s) = L(\lambda | \bar{x}) \times \pi(\lambda)$

not sure how to compute this.

Any help appreciated !

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The posterior density is given by $$ f_{\lambda\mid x}(\lambda\mid x)\propto L(\lambda)\times f_{\lambda}(\lambda) $$ where $f_{\lambda}$ is the prior for $\lambda$ and $L$ is the likelihood. In our case $$ L(\lambda)\times f_{\lambda}(\lambda)\propto\lambda^{\sum_{i=1}^n x_i}e^{-n\lambda}\times \lambda^{a-1}e^{-b\lambda}=\lambda^{a-1+\sum_{i=1}^n x_i}e^{-\lambda(n+b)} $$ where we dropped all the constants (including those that depend solely on $x$). Hence $$ f_{\lambda\mid x}(\lambda\mid x)=C\lambda^{a-1+\sum_{i=1}^n x_i}e^{-\lambda(n+b)} $$ where $C$ is a function of $x$. By inspection $\lambda\mid x$ is a Gamma distributed random variable i.e. $\lambda\mid x \sim \text{Gamma}(a+\sum_{i=1}^n x_i, n+b)$ as desired.

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