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I recently ran into the following exercise when practicing for an exam I have about Statistical Inference. The question looks very large and complex and I'm wondering if this is actually true or if it can be solved relatively easily.

The question:

At a critical stage in the development of a new airplane, a decision must be taken to continue or to abandon the project. The financial viability of the project can be measured by the parameter $\theta\in(0,1)$, the project being profitable if $\theta>1/2$. Data $x$ provide information about $\theta$. If $\theta<1/2$, the cost of continuing the project is $1/2-\theta$, whereas if $\theta>1/2$ it is zero. If $\theta>1/2$ the cost of abandoning the project is $\theta-1/2$, whereas if $\theta<1/2$ it is zero.

We denote the decision to continue the project by $d_1$ and the decision to abandon the project by $d_2$.

Show that the optimal Bayesian decision is to continue the project if the posterior mean of $\theta$ is greater than $1/2$

My attempt: The general start to problems where we have to find Bayesian decision rules is to compute the loss functions. I found the following loss functions:

$L(\theta, d_1) = \begin{cases}0 &\text{ if }\theta > 1/2\\1/2 - \theta &\text{ if }\theta < 1/2\end{cases},\,\,L(\theta, d_2) = \begin{cases}\theta - 1/2 &\text{ if }\theta > 1/2\\0 &\text{ if } \theta < 1/2\end{cases}$

Which will allow us to compute the risk functions and this in turn could help us find the decision rule minimizes the Bayes risk, and this will be the Bayes decision rule. I'm not able to get much further than that though, since we have this tricky criterion based on the posterior mean.

How can I solve this exercise? Any tips/help are appreciated.

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    $\begingroup$ I gave a full answer below but if you just want a hint, think about when the expected posterior loss under $d_1$, $E(L(\theta, d_1)\mid x) = \int L(\theta, d_1) p(\theta \mid x) d\theta$, is less than $E(L(\theta, d_2)\mid x)$. $\endgroup$ – Alex Jan 8 at 23:10
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The Bayes risk for continuing the project, $E(L(\theta, d_1) \mid x)$, is given by $$E(L(\theta, d_1) \mid x) = \int_0^1L(\theta, d_1) p(\theta \mid x) dx = \int_0^{1/2}\left(\frac{1}{2} - \theta\right)p(\theta \mid x) d\theta$$ $$= \frac{1}{2}\int_0^{1/2} p(\theta \mid x) d\theta - \int_0^{1/2}\theta p(\theta \mid x) d\theta.$$

And the Bayes risk for cancelling the project, $E(L(\theta, d_2) \mid x)$, is given by $$E(L(\theta, d_2) \mid x) = \int_0^1L(\theta, d_2) p(\theta \mid x) dx = \int_{1/2}^{1}\left(\theta - \frac{1}{2}\right)p(\theta \mid x) d\theta$$ $$= \frac{1}{2}\int_{1/2}^{1} \theta p(\theta \mid x) d\theta - \frac{1}{2}\int_0^{1/2} p(\theta \mid x) d\theta.$$ The Bayes decision rule is to continue the project when the Bayes risk of continuing is less than the risk of cancelling, i.e. when $$E(L(\theta, d_1) \mid x) < E(L(\theta, d_2) \mid x).$$ This is equivalent to $$\frac{1}{2}\int_0^{1/2} p(\theta \mid x) d\theta - \int_0^{1/2}\theta p(\theta \mid x) d\theta < \int_{1/2}^{1} \theta p(\theta \mid x) d\theta - \frac{1}{2}\int_{1/2}^{1} p(\theta \mid x) d\theta,$$ which is equivalent to $$\frac{1}{2}\int_0^{1/2} p(\theta \mid x) d\theta + \frac{1}{2}\int_{1/2}^{1} p(\theta \mid x) d\theta < \int_0^{1/2}\theta p(\theta \mid x) d\theta + \int_{1/2}^{1}\theta p(\theta \mid x) d\theta,$$ which is equivalent to $$\frac{1}{2}\int_0^{1} p(\theta \mid x) d\theta < \int_0^{1}\theta p(\theta \mid x) d\theta.$$ We know that $\int_0^{1} p(\theta \mid x) d\theta = 1$, and that $\int_0^{1}\theta p(\theta \mid x) d\theta = E(\theta \mid x)$, so the Bayes decision rule is that we should continue with the project whenever $$\frac{1}{2} < E(\theta \mid x).$$

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    $\begingroup$ I came so close with your hint but got stuck in the end anyway, so thanks alot for the full answer! Much appreciated! $\endgroup$ – S. Crim Jan 9 at 14:04
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    $\begingroup$ Glad I could help! $\endgroup$ – Alex Jan 10 at 16:53

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