Joint Maximization in Extended Likelihood

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. $$$$L(\theta, u; y, u) = \theta u \exp(-u(\theta+y))$$$$ The support is u>0, $$\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 $$$$\begin{split} \frac{d l(\theta, u; y, u)}{d\theta} = 1/\theta - u \\ \frac{d l(\theta, u; y, u)}{du} = 1/u - \theta-y \\\end{split}$$$$ I get $$\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!

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 $$y=0$$.

• Thank you, forgot about the substitution part I guess. Thanks for the R-code! Commented Jan 25, 2023 at 18:06
• Good. But the code is Mathematica rather than R.
– JimB
Commented Jan 25, 2023 at 18:34