I'm studying for an exam and don't know how to start here:
a) Given the loss function $$L_{\varepsilon} (x,p) = -\frac{1}{\sqrt{\varepsilon}}\exp\Big(-\frac{(x-p)^2}{\varepsilon}\Big)$$, calculate $X$ with limited continuous density $f(x)$ and limited risk function $\tilde{R}(p) = \lim_{\varepsilon \rightarrow 0} R_{\varepsilon}(p)$, where $R_{\varepsilon}(p)$ is the risk function of $L_{\varepsilon}$.
b) Calculate the optimal premium $\tilde{R}(p)$ for the r.v. $X$ with $f(x) = \frac{18}{5} (1-x)-\frac{12}{5}(1-x)^2$ on $[0,1]$. (Remark: The opmital premium is calculated via minimizing $R(p)$, i.e. differentiate w.r.t. $p$ and then set the resulting term to $0$).
The idea should be kinda like the following:
$$ R_{\varepsilon}(p)=E[L_{\varepsilon}(x,p)] = \int \frac{1}{\sqrt{\varepsilon}} \exp(-\frac{(x-p)^2}{\varepsilon}) f(x) dx $$
$$ = \int \frac{1}{\sqrt{\varepsilon}} \exp(-u^2) \sqrt{\varepsilon} f(\sqrt{\varepsilon}u+p)du \rightarrow f(p) = \tilde{R}(p) \text{ as } \varepsilon \rightarrow 0$$
Given that, it is easy to solve the second part.