# Alternate definitions of risk in decision theory?

In decision theory, given a family of distributions $$(P_{\theta})_{\theta \in \Theta}$$ on the data space $$\mathcal{X}$$, the risk $$R_L(\theta, \delta)$$ of an estimator $$\delta$$ for a given loss function $$L(\theta, d)$$ is given by $$R_L(\theta, \delta) = \int_{\mathcal{X}} L(\theta, \delta(x)) \, dP_{\theta}(x)$$ I have only seen cases where the loss function $$L$$ is nonnegative, making the risk function $$R_L$$ effectively an $$L^1$$-norm. Just considering the $$L^1$$-norm of $$L$$ seems a bit restrictive.

For example if we wanted to talk about the tails of $$L(\theta, \delta(X))$$ when $$X \sim P_{\theta}$$, why not use for example an Orlicz norm, $$\|\cdot\|_{\psi}$$? Then we could look for an estimator that minimizes $$R^{\psi}_L(\theta, \delta) = \|L(\theta, \delta(X))\|_{\psi}$$ In the case of $$\psi_1$$, $$\psi_2$$ or a similar norm, this would imply we are looking for an estimator such that $$L(\theta, \delta(X))$$ has (at least for a minimum assumed decay rate $$\exp(-t)$$, $$\exp(-t^2)$$ or similar) the fastest decaying tails.

I haven't been able to find anything, but I am guessing that the choice of loss function can account for this, at least to within an arbitrarily small error?

• By definition, norms cannot be negative. Moreover, this formulation works for all valid loss functions, which also (by definition) cannot be negative.
– whuber
Commented Apr 2, 2023 at 17:15
• @whuber What I am asking is if instead of the integral of $L$, which is the $L^1$-norm of $L$ since it is nonnegative, why not take for example $\psi_1$-norm of $L$. Commented Apr 2, 2023 at 17:23
• The integral is the expectation of $L.$ That's the key.
– whuber
Commented Apr 2, 2023 at 17:24
• Re the edit: I can't figure out what you might mean by "the tails of $L,$" because $L$ is not a probability distribution. If you want to weight extreme values of $L$ differently then simply change $L$!
– whuber
Commented Apr 2, 2023 at 17:31
• @whuber Ah so you can account for that by changing the loss. I was kind of expecting that. But that is good to know. I was thinking maybe it was because you couldn't as easily make a Bayesian connection. However, if you need just modify the loss, then that is not a problem. I will think about how to do this. Thanks! Commented Apr 2, 2023 at 18:09