Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
Thanks J.D. Appreciate the edit. –  icobes Mar 29 '12 at 5:49

1 Answer 1

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).

share|improve this answer
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.