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Given the quartic loss: $$ l(\theta, \hat{\theta}) = (\theta - \hat{\theta})^4 $$ what is the estimator that minimizes the Bayes risk?: $$ R(\hat{\theta}|y) = \int l(\theta, \hat{\theta}) p(\theta|y)d\theta $$

For the quadratic loss, plugging the loss into the above equation, taking the derivative w.r.t. $ \hat{\theta} $, and equating to 0, we quickly see the estimator which maximizes the loss is the posterior mean. However for the quartic loss, it's not so obvious after the above steps and arriving to $ E_{p(\theta|y)}[(\theta-\hat{\theta})^3] = 0 $. I feel like the answer should be the posterior mean as well, but I don't know how to arrive to that conclusion.

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Unless the posterior is symmetric around $\mathbb{E}[\theta|y]$ there is no reason for the estimator to be equal to $\mathbb{E}[\theta|y]$. It is one of the solutions to a third degree polynomial equation.

Take the following counter-example of a Gamma $\mathfrak G(\alpha,\beta)$ posterior. The posterior risk function is then $$R(\pi,\hat\theta)=\hat\theta^4-4\hat\theta^3\mathbb E^\pi[\theta|y]+6\hat\theta^2\mathbb E^\pi[\theta^2|y]-4\hat\theta\mathbb E^\pi[\theta^3|y]+\mathbb E^\pi[\theta^4|y]$$ and $$\mathbb E^\pi[\theta^k|y]=\beta^{-k}(k+\alpha)\cdots\alpha$$ hence $$R(\pi,\hat\theta)=\hat\theta^4-4\hat\theta^3\alpha\beta^{-1}+6\hat\theta^2\alpha(\alpha+1)\beta^{-2}-4\hat\theta\alpha(\alpha+1)(\alpha+2)(\alpha+3)\beta^{-3}+\mathbb E^\pi[\theta^4|y]$$ Without loss of generality, we can take $\beta=1$. Here is an illustration of the risk function when $\alpha=3.1415...$:

enter image description here

The solution is clearly different from $\alpha/\beta$, the posterior mean.

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