Solving for a parameter in an inequality with $\log(\text{parameter})$ and the parameter

Question

I need to find $\lambda$ such that $$6\log(1-e^{-\lambda}) + 5\log(\lambda) - 67.6\lambda \geq -35.1208\text{.}$$ Assume that all you have available is a scientific calculator. On a timed qualifying exam, how would you go about solving for $\lambda$?

Assuming that I interpreted this particular solution correctly, I believe the solution to this question is $\lambda \in (0.171, 0.475)$. If I had more context, I would provide it, but this is all I'm working from a qualifying exam solution.

If the statistical context (although I doubt it would be helpful) is necessary, here it is.

Consider $X_1, \dots, X_{20} \overset{\text{iid}}{\sim}\text{Exp}(\lambda)$; i.e., they follow the PDF $$f_{X}(x) = \lambda e^{-\lambda x}\mathbf{1}_{(0, \infty)}(x)\text{.}$$ However, what is actually observed is $U_1, \dots, U_{20}$ where $$U_i = \begin{cases} 1, & X_i \leq 1 \\ X_i & X_i \in (1, 5) \\ 5, & X_i \geq 5\text{.} \end{cases}$$ A sample of 20 observations is 1, 5, 2.58, 1.98, 4.10, 2.13, 1, 1.35, 5, 1, 2.99, 1.11, 1, 5, 1, 2.99, 3.37, 1, 5, and 5. The loglikelihood calculated at $\hat{\lambda} = 0.297$ is $-33.2$, with a first derivative approximately $0$ when evaluated at $\hat\lambda$, and a second derivative approximately $-169.2$ when evaluated at $\hat\lambda$.

Find a an approximate 95% large-sample confidence interval for $\lambda$ based on the inversion of likelihood ratio tests of point null hypotheses about $\lambda$.

My solution: we have, after doing some work, $\{\lambda: \ell(\lambda) \geq -35.1208\}$ where $\ell$ is the loglikelihood, is a 95% CI for $\lambda$, i.e., $$\ell(\lambda) = 6\log(1-e^{-\lambda})+5\log(\lambda)-22.6\lambda-45\lambda$$ (assuming that my counting and summing is all correct).

Answer 1

The best I think that we can do is to note $\lambda > 0$. Furthermore, the derivative is given by $$ \frac{6}{e^\lambda -1} + \frac{5}{\lambda} - 67.6. $$ This tells us that for $\lambda > \lambda_c$ where $\lambda_c$ is the critical point, the function is always decreasing. It is clear $\lambda_c \ll 1$ and so we approximate by the taylor series to obtain $\lambda_c \approx 11/67.6 = 0.16$. It is clear then that $\lambda > 1$ is not valid. We can do better though, consider $\lambda = 1/2$, then using the taylor series again we see $$ 11 \log(\lambda) - 67.6 + 35. 1208 < 0. $$ Finally, we can also use the small $\lambda$ approximation to estimate the first zero by $11 \log \lambda - 67.6\lambda + 35.12 =0$. Exponentiating and using taylor series gives $\lambda = 0.059$. Hence we can crudely estimate $\lambda \in (0.059,0.5)$.

However the actual result (which I have differs from yours) is $\lambda \in (0.06,0.32)$

Answer 2

First, let me rewrite the first equation as $$f=6 \log \left(1-e^{-\lambda }\right)+5 \log (\lambda )-\frac{338 }{5}\lambda +\frac{43901}{1250}$$ and say that we look for the zero's of it.

Assuming that $\lambda$ is small, make a Taylor expansion around $\lambda=0$ to get $$f=11 \log (\lambda )-\frac{353 }{5}\lambda+\frac{43901}{1250}+O\left(\lambda ^2\right)$$ Ignoring the higher order terms, let $\lambda=e^x$ and consider solving $$g=11x-\frac{353 }{5}e^x+\frac{43901}{1250}=0$$ the solution of which being given in terms of Lambert function $$x_1=-W\left(-\frac{353}{55 e^{3991/1250}}\right)-\frac{3991}{1250}$$ $$x_2=-W_{-1}\left(-\frac{353}{55 e^{3991/1250}}\right)-\frac{3991}{1250}$$ The argument is quite small $(-\frac{353}{55 e^{3991/1250}}\approx-0.26351)$ and the series expansions given in the Wikipedia page would converge quite fast (for sure, you must be aware of them) and then$$x_1\approx -2.804\qquad x_2\approx -1.134$$ which correspond to $$\lambda_1\approx 0.0606 \qquad \lambda_2\approx 0.3200$$ while the "exact" solutions would be $$\lambda_1\approx 0.0605523 \qquad \lambda_2\approx 0.320571$$

On a (limited) timed qualifying exam,, knowing the expansions of Lambert function, should I be able to do it ? I cannot answer. However, all of that is doable with a scientific pocket calculator.

I hope and wish this be of some help.