# Bayes estimator under LINEX loss?

Suppose we are trying to estimate a real valued parameter under the linear exponential (LINEX) loss. Suppose we have some prior $g$ that has finite mean, variance and a moment generating function.

What is the Bayes estimator without being given data?

What about when data is normal with mean $\theta$ (unknown) and variance 1, under a normal prior $g$? Here the real valued parameter is the mean.

• What is LINEX loss?
– Sycorax
Commented Nov 3, 2014 at 3:44
• Linear exponential loss. See, for instance, www2.isye.gatech.edu/~brani/isyebayes/bank/handout4.pdf: Commented Nov 3, 2014 at 3:51
• Can't one just take the derivative of $\int [e^{c(a - \theta)} - c(a - \theta) - 1] f(\theta) \ d\theta$ with respect to $a$ and set equal to $0$?
– guy
Commented Nov 3, 2014 at 17:19
• What is f, the prior pdf on theta? I am studying for a prelim and a little in over my head on this stuff, so I may be missing something obvious. Commented Nov 3, 2014 at 17:34
• @GregorianFunk $f$ is an arbitrary distribution on $\theta$. Once you find the answer for an arbitrary distribution, you can just plug the posterior in. For example, if it were squared error loss you would show that the minimizer of $\int (\theta - a)^2 f(\theta) \ d\theta$ is $a = \int \theta f(\theta) \ d\theta$ and conclude that the optimal decision rule is to take $a = E[\theta \mid X]$ if $X$ is the data.
– guy
Commented Nov 6, 2014 at 19:26

For an arbitrary density $f(\theta)$, the expected LINEX loss is

$$E[L(\theta, a)] = \int [e^{c(a - \theta)} - c(a - \theta) - 1] f(\theta) \ d\theta.$$

Differentiating with respect to $a$ shows that the minimum occurs when

$$\int [c e^{c(a - \theta)} - c] f(\theta) \ d\theta = 0.$$

Solving this gives

$$a = \frac{-\log \int e^{-c\theta} f(\theta) \ d\theta}{c} = \frac{-\kappa(-c)}{c}$$

where $\kappa(\cdot)$ denotes the cumulant-generating function.

I'm going to answer a modified version of your follow-up question where we assume $\theta$ has a flat prior and $X \sim N(\theta, 1)$. Then, $[\theta \mid X = x] \sim N(x, 1)$. The cumulant-generating function of an $N(x, 1)$ distribution is $\kappa(t) = xt + t^2 / 2$, so this gives

$$a(X) = X - c / 2.$$