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I derived a posterior distribution, which looks like this:

$$p(\alpha, \beta | y,t) \propto \frac{exp( - \sum_i (\alpha + \beta t_i))\prod_i (\alpha + \beta t_i)^{y_i}}{\prod_i y_i!}$$

How can I show this is a proper distribution of $\alpha$ and $\beta$?

What distribution would it be to make it integrate to 1?

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Since it's non-negative, integrate it out and show that the result is 1? – gt6989b Jan 31 '13 at 18:24
My question is how do you integrate it to 1 – user1769197 Jan 31 '13 at 18:25
Defined $I = \int p$, then $f = \frac{p}{I}$ is a density function of some distribution. The question is however whether $I$ is finite – Ilya Jan 31 '13 at 18:26
I m looking for an answer which can show me computation steps. If you have any idea, please elaborate a little bit more. I spent the whole afternoon to do this but failed – user1769197 Jan 31 '13 at 18:37
Got something from the answer below? – Did Feb 9 '13 at 10:36

Assume that every $t_i$ is positive and every $y_i$ nonnegative, and consider the function $$ q(a,b)=\exp\left(-\sum_{i=1}^n(a+bt_i)\right)\cdot\prod_{i=1}^n(a+bt_i)^{y_i}. $$ The question seems to be to show that the integral of $q$ on $(0,+\infty)\times(0,+\infty)$ is finite.

To do so, call $t$ the maximal $t_i$ and note that $a+bt_i\leqslant a+bt$ and $\sum\limits_{i=1}^n(a+bt_i)\geqslant a+bt$ for every $(a,b)$ hence $q(a,b)\leqslant r(a+bt)$ with $$ r(x)=\mathrm e^{-x}x^y,\qquad y=\sum_{i=1}^ny_i. $$ The change of variable $x=a+bt$, $z=a$, yields $$ \iint r(a+bt)\mathrm da\mathrm db=\int_0^{+\infty}r(x)\int_0^x\frac{\mathrm dz}{t}\mathrm dx=\frac1t\int_0^{+\infty}xr(x)\mathrm dx=\frac{(y+1)!}t. $$ Thus there exists some constant $c$ such that $cq$ is a PDF. Note that $c\geqslant t/(y+1)!$ and that, when $n=1$, $c=t/(y+1)!$.

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