Asymmetric Coding Region

In section , it was assumed that $\theta^\prime$ lies uniformly distributed between $\theta-\frac{1}{2}s(\theta)$ and $\theta+\frac{1}{2}s(\theta)$. As suggested by Dowe [#!Dowe:private!#], this assumption will be skipped in this section. Instead an asymmetric coding region will be assumed. The coding region is defined to be between $\theta - \delta_{lower}(\theta) = \theta - \delta_l(\theta)$and $\theta + \delta_{upper}(\theta) = \theta + \delta_u(\theta)$. That is, the coding region is $R = [\theta-\delta_l(\theta), \theta+\delta_u(\theta)]$

Again, a two part message is involved. The first part can be approximated using the midpoint rule (subsection ) by

$$-\log_e \int_ {\theta - \delta_l} ^ {\theta+\delta_u} h(\the... ...\approx -\log_e [h(\theta)(\delta_l(\theta)+\delta_u(\theta))]$$ (24)

For the second part of the message, it can be expanded Taylor series (subsection ) up to the second order.

$$\begin{split} -\log_e f(x\vert\theta^\prime) &= -\log_e f(x\... ...\partial \theta^2 } (-\log_e f(x\vert\theta)) + ... \end{split}$$ (25)

The zeroth term is obvious. Hence the first order term of the Taylor series approximation (after taking the expectation value of $\theta^\prime$ from $\theta - \delta_l(\theta)$ to $\theta + \delta_u(\theta)$) becomes

$$\frac{1}{\delta_l+\delta_u} \int_ {\theta - \delta_l} ^ {\th... ...{2}\frac{\partial}{\partial \theta } (-\log_e f(x\vert\theta))$$ (26)

The second term is

$$\frac{1}{\delta_l+\delta_u} \int_ {\theta - \delta_l} ^ {\th... ...frac{\partial^2}{\partial \theta^2 } (-\log_e f(x\vert\theta))$$ (27)

This results in an expression for the total minimum message length -

$$\begin{split} \textrm{MesgLen} &= -\log_e [ h(\theta)(\delta... ...ta_u^2-\delta_u\delta_l+\delta_l^2}{6} B(x, \theta) \end{split}$$ (28)

where $A(x, \theta) = \frac{\partial}{\partial \theta }(-\log_e f(x\vert\theta))$and $B(x, \theta) = \frac{\partial^2}{\partial \theta^2 }(-\log_e f(x\vert\theta))$.

Performing a variable substitution $\delta_u = r \delta_l$ where r lies within the domain $[0, \infty)$.

$$\begin{split} \textrm{MesgLen} &= -\log_e [ h(\theta) \delta... ...ad + \frac{(r^2-r+1)\delta_l^2}{6} B(x, \theta) \\ \end{split}$$ (29)

When minimised with respect to $\delta_l(\theta)$, the parameter $\delta_l(\theta)$ can be derive

$$\begin{split} 0 &= 2(r^2-r+1)B(x,\theta) \delta_l(\theta)^2 ... ...)^2+48(r^2-r+1)B(x,\theta)}}{4(r^2-r+1)B(x,\theta)} \end{split}$$ (30)

As in subsection , the expected values of $A(x,\theta)$ and $B(x,\theta)$ are required for $\delta_l(\theta)$ in order for the receiver to be able to decode the message.

$$\delta_l(\theta) = \frac{3(1-r)A(\theta) \pm \sqrt{9(r^2-2r+1)A(\theta)^2+48(r^2-r+1)B(\theta)}}{4(r^2-r+1)B(\theta)}$$ (31)

Therefore the message length is -

$$\begin{split} \textrm{MesgLen} &= -\log_e [ h(\theta) \delta... ...(r^2-r+1)\delta_l(\theta, r)^2}{6} B(x, \theta) \\ \end{split}$$ (32)

At present, the derivation of an estimator for the asymmetric coding region remains uncompleted. It is not clear whether this problem can be solved in a satisfactory manner. However, a possible solution might be to find the value of r that minimises equation (). Then the message length can also be minimised with respect to $\theta$ and a parameter estimate derived. These steps may have to be performed numerically.