# Fisher information in a hierarchical model

Given the following hierarchical model, $$X \sim {\mathcal N}(\mu,1),$$ and, $$\mu \sim {\rm Laplace}(0, c)$$ where $\mathcal{N}(\cdot,\cdot)$ is a normal distribution. Is there a way to get an exact expression for the Fisher information of the marginal distribution of $X$ given $c$. That is, what is the Fisher information of: $$p(x | c) = \int p(x|\mu) p(\mu|c) d\mu$$ I can get an expression for the marginal distribution of $X$ given $c$, but differentiating w.r.t. $c$ and then taking expectations seems very difficult. Am I missing something obvious? Any help would be appreciated.

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I had a try at it myself, but it is beyond my abilities. Absolute value functions ruin everything! You are basically stuck with numerical methods. –  probabilityislogic Mar 28 '11 at 12:40
@probability You can obtain an expression for the integrand simply by splitting the integral into regions $\mu \ge 0$ and $\mu \lt 0$; no absolute values are needed. But the result is a messy rational function of $x$, $exp(-x^2)$, and error functions, and so is unlikely to be integrable in closed form. –  whuber Mar 28 '11 at 17:23
@whuber - that is what I meant by "hopeless". Not that the integral is impossible, but the fisher information is impossible. Because you have to take the expected value over $X$ of a ratio of two of these types of integral –  probabilityislogic Mar 28 '11 at 22:02
A lower bound for the Fisher information in this case is $1/(1+2c^2)$. Is it possible to get a tighter uppper bound on the Fisher information than the general $1 + 1/c^2$? –  emakalic Apr 3 '11 at 23:56