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Let's say the posterior distribution of $\theta$ is Gamma with $$\alpha = 40, \qquad \beta = \frac{1}{0.5 + \sum_{i = 1}^{10} X_i}$$

What is the Bayes point estimate using the mode of the posterior distribution?

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Thanks J.D. Appreciate the edit. – icobes Mar 29 '12 at 5:49
up vote 2 down vote accepted

Wikipedia lists the mode of a gamma distribution of the mode to be $\frac{\alpha-1}{\beta}$, so your point estimate would be $39(0.5+\sum_{i=1}^{10}X_i)$.

You can also find the mode of a gamma distribution by computing the derivative of $x^{\alpha-1}e^{-\beta x}$ which is $x^{\alpha-2}e^{-\beta x}(\alpha-1-\beta x)$, setting it to zero and solving (then checking that the result is a maximum).

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Since the value of $x^{\alpha-1} e^{-\beta x}$ is $0$ at the two endpoints $0$ and $\infty$, and the function is positive between those endpoints, and is continuous, it follows that there must be a global maximum somewhere between $0$ and $\infty$. Since there's ONLY ONE point where the derivative is $0$ and it's differentiable everywhere, that's a maximum. – Michael Hardy Mar 29 '12 at 15:35

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