I'm trying to follow a calculation made in a paper(section 2 from the supplementary contents of Likelihood based observability analysis and confidence intervals for predictions of dynamic models. Kreutz C, Raue A and Timmer J. BMC Systems Biology 6, 120, 2012) for determining a $prediction$ confidence interval (PCI) for a prediction made from a very simple differential equation.
Before diving into the specifics I wanted to briefly give you the background of the calculation. Also, thank you in advance for reading though and sticking with the problem.
------------------------The bit I understand ------------------
The ODE in question:
$[1]$ $$\frac{dx}{dt}=-θ$$
where $x(0)=1$, we have a single data point $y=0.9$ for $x$ at time point $t=1$ (with assumed Gaussian measurement noise $N(0,0.1^2)$.
The solution to this equation is:
$[2]$
$$ x(t)=-exp(-θt) $$
The log-likelihood is:
$[3]$
$$ LL(y|θ)=log[\frac{1}{\sqrt{2*pi*sigma^2}}exp(\frac{-0.9 - exp(-θ))^2}{2*sigma^2})] $$
$[4]$
$$=-50(0.9-exp(-θ))^2+c $$
Where $ c = \frac{1}{\sqrt{2*pi*sigma^2}}$
The maximum likelihood estimate is obtained by:
$[5]$
$$\frac{\partial{LL(y|θ)}}{\partial{θ}}=0 $$
They then analytically derive that the $θ$ estimate $θhat$ is:
$$-log(0.9)=0.1054 $$
-------------The bit I don't understand----------
I now understand how the above was obtained thanks to a previous question I asked on this site. Now I need to understand the next bit - which is predicting the confidence intervals assuming the desired prediction is given by $z=θ^2$
$[6]$
$$ \{z(θ)|LL(y|θ) <= LL(y|θhat) + icdf(chisq_1,alpha)\}$$
(that is <= is smaller than or equal to and icdf(chisq_1,alpha) is the alpha quantile of the chi squared distribution with 1 degrees of freedom (which is 3.8415) .
They now say: For the boundaries of the parameter confidence interval it holds:
$[7]$ $$ LL(y|θ) = LL(y|θhat) + icdf(chisq_1,alpha) $$
$question 1$: Why does it hold? Have they derived this from some theorem I'm currently unaware of or is it derived from these calculations? Should I just take this as fact?
if and only if
$[8]$
$$ 50(0.9-exp(-θ))^2 + c = c + 3.8415 $$
$Question2$: clearly they have substituted the log likelihood function on the left here, but how has the estimated version on the right now become c?
And this implies that:
$[9]$
$ θ=-0.1631$ or $θ = 0.4734$
$Question3:$ how have they arrived to these figures? Substituting the calculated $θ$ estimate does not work and neither does canceling $c$ and solving for $θ$
Therefore, they have calculated that the $parameter$ confidence interval is $θ $ in the set $[-0.1631,0.4734]$. They then go on to say that the $prediction$ confidence interval for the $z=θ^2$ is $z$ in the set $[0,0.2241]$.
$Question 4:$ How have they used the $parameter$ confidence interval to calculate the $prediction$ confidence internal?