Joint Maximization in Extended Likelihood

Question

I am reading Generalized linear models with random effects : unified analysis via h-likelihood Chapter 4 on Extended Likelihood principle. I came across an example and I am not sure if I am understanding joint maximization correctly. Given extended likelihood where $\theta$ is a single fixed parameter, U is a single unobservable random quantity and a single observable quantity Y. \begin{equation} L(\theta, u; y, u) = \theta u \exp(-u(\theta+y)) \end{equation} The support is u>0, $\theta>0$.

When jointly maximized wrt to $\theta$ and $u$, the book says it yields $\hat{\theta} = \infty$ and $\hat{u} = 0$.

My derivations are \begin{equation} \begin{split} \frac{d l(\theta, u; y, u)}{d\theta} = 1/\theta - u \\ \frac{d l(\theta, u; y, u)}{du} = 1/u - \theta-y \\\end{split} \end{equation} I get $\hat{\theta}$ = 1/u and $\hat{u}$ = $\frac{1}{\theta + y}$ I don't understand how the jointly maximized values are $\infty$ and 0 respectively. Do we integrate out the value eg. $\int_0^\infty(1/u) du$ here to get that? Thank you!

Answer 1

A contour plot at some fixed value of $y$ should be helpful to be convincing. (And these plots are certainly not equivalent to a proper mathematical approach.)

Using Mathematica (and R could certainly be used), here is a contour plot of the log of $L$ for two different sets of $\theta$ and $u$ with $y=10$:

logL = Log[θ] + Log[u] - u (θ + y)
ContourPlot[logL /. y -> 10, {θ, 0, 10}, {u, 0, 1}, 
Contours -> Range[-10, -1, 1], ContourShading -> False, 
ContourLabels -> True, PlotPoints -> 50,
FrameLabel -> (Style[#, Bold, 18, Italic] &) /@ {"θ", "u"}]

ContourPlot[logL /. y -> 10, {θ, 1000, 10000}, {u, 0, 3/10000},
Contours -> 
   Join[Range[-1.9, -1.1, 0.1], Range[-1.009, -1.001, 0.001]], 
ContourShading -> False, PlotPoints -> 50,
FrameLabel -> (Style[#, Bold, 18, Italic] &) /@ {"θ", "u"}]

One can see that the maximum is tending towards $\theta\rightarrow \infty$ and $u \rightarrow 0$.

Making the substitution of what you produced ($\hat{\theta}=1/\hat{u}$) into

$$\hat{u}=\frac{1}{\hat{\theta}+y}$$

results in

$$\hat{u}=\frac{1}{1/\hat{u}+y}$$

The solutions to that are $\hat{u}=0$ or $y=0$.