quick question about the form of a posterior distribution.

Suppose that $\theta \sim Gamma(a, b)$ and that, given $\theta$, $Y$ has CDF $$F(Y\mid\theta) = 1 - e^{-\theta(e^y - 1)},\quad \theta>0.$$

If I didn't make any mistakes differentiating then, given $\theta$, $Y$ has PDF

$$\theta e^{y - \theta(e^y - 1)}.$$

So the posterior distribution of $\theta\mid Y$ would be

\begin{align*} p(\theta \mid Y) &\propto p(Y|\theta) p(\theta) \\ &=\theta e^{y -\theta(e^y - 1)} \cdot \frac{1}{\Gamma(a)}b^a\theta^{a- 1} e^{-b\theta} \\ &\propto \theta^a e^{ - \theta(b+ e^y - 1)} \end{align*}

which seems like a $Gamma(a+1, \, b + e^y -1)$ distribution.

Does this seem correct?

I think it seems strange to me since the likelihood

$$\theta e^{y - \theta(e^{y} - 1)}$$

is not one I've dealt with before.

  • $\begingroup$ I'd have written $F_Y(y\mid\theta)$, distinguishing between capital $Y$, the random variable, and lower-case $y$, the argument to the function. Then one would understand that an expression like $F_Y(3\mid\theta)$ means $\Pr(Y\le 3\mid\theta)$. $\endgroup$ – Michael Hardy Feb 2 '14 at 19:36
  • $\begingroup$ Do you agree with my conclusions about the Gamma posterior? Thanks! $\endgroup$ – tony0485 Feb 2 '14 at 20:06

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