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I have this problem: In betting situations one is often interested in odds, referring in the thumbtack tossing $\theta / (1 - \theta)$. Alternatively one may consider the log-odds:

$\lambda=log(\theta/(1-\theta))$

Show that a uniform distribution for $\lambda$ implies the following distribution for $\theta$:

$p(\theta)=\theta^{-1}(1-\theta)^{-1}$

What problems are associated with this distribution if it is used as a prior in the thumbtack tossing problem?

Now I have the thumbtack thingy, I know what it will happen if we use that p as prior, but the uniform for $\lambda$ implies the p($\theta$) I don't get it, I have tried to get to the p($\theta$) from both sides but I always get to $log(\theta)-log(1-\theta)$ and can't get out, any help??

Pedro.Alonso
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1 Answers1

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$$\frac{d\lambda}{d\theta}=\frac{1}{\theta} + \frac{1}{1-\theta} = \frac{1}{\theta(1-\theta)}$$ $$f(\lambda)\,d\lambda\propto d\lambda \quad \text{(improper uniform; aka Lebesgue measure)}$$ $$g(\theta)\,d\theta =f(\lambda)\,d\lambda=f(\lambda(\theta))\frac{d\lambda}{d\theta}\,d\theta\propto \frac{1}{\theta(1-\theta)}d\theta$$ $$g(\theta)\propto\frac{1}{\theta(1-\theta)} \quad \text{(improper Haldane's prior; see also Jaynes)}$$

Zen
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