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So I have the likelihood being:

$\prod^{n}_{i=1}(\frac{\lambda^{x}e^{-\lambda}}{x!})$

which is proportional to

$\lambda^{\sum_{i=1}^{n}x_{i}}e^{-n\lambda}$

The prior is standard exponential $e^{-\lambda}$

So the posterior is

$\lambda^{\sum_{i=1}^{n}x_{i}}e^{-\lambda(n+1)}$

So then would it be a gamma with parameters $\sum_{i=1}^{n}+1$ and $n+1$

Gamecocks99
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1 Answers1

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Yes (I assume you mean $1+\sum_{i=1}^nx_i$). The exponential distribution is a special case of the Gamma distribution, so have a look here, plug in your values and you'll see that you have the right thing.

hejseb
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