GLRT of exponential distribution and critical region

Question

Find the Generalized likelihood ratio test (GLRT) for $H_0: \lambda = \lambda_0$ when $H_A: \lambda \ne \lambda_0$ for $X_1 ... X_n$ taken from $X \sim Exp(\lambda;x)$, with a test size of $0.06$, obtain the critical region.

Here's what I have tried:

$L(\lambda_0;x) = \prod_{i=1}^nf(x_i;\lambda_0) = \prod_{i=1}^n\lambda_0 e^{-\lambda_0 x_i}=\lambda_0^ne^{-\lambda_0 \sum_{i=1}^nx_i}$

$\mathbf{L}(\lambda_0;x) = \log L(\lambda_0;x) = n\log(\lambda_0) - \lambda_0\sum_{i=1}^nx_i$

$\frac{\partial\mathbf{L}(\lambda_0;x)}{\partial\lambda_0} = \frac{n}{\lambda_0}-\sum_{i=1}^nx_i \implies \frac{n}{\lambda_0}-\sum_{i=1}^nx_i=0 \implies \lambda_0 = \frac{1}{\bar{x}}$

$\mathbf{L}(\hat{\lambda};x) = \left(\frac{1}{\bar{x}}\right)^ne^{-\left(\frac{1}{\bar{x}} \right) \sum_{i=1}^nx_i}$

Then applying the likelihood ratio:

$$\Lambda :=\frac{L(\lambda_0;x)}{\mathbf{L}(\hat{\lambda};x)} = \frac{\lambda_0 ^ne^{-\lambda_0 \sum_{i=1}^nx_i}}{\left(\frac{1}{\bar{x}}\right)^ne^{-\left(\frac{1}{\bar{x}} \right) \sum_{i=1}^nx_i}}= \left(\bar{x}\lambda_0 e^{1-\lambda_0\bar{x}} \right)^n$$

We can find the critical region by:

$$\begin{align} C &= \{x : \lambda \le k\} \\ &= \{x : \left(\bar{x}\lambda_0 e^{1-\lambda_0\bar{x}} \right)^n \le k\} \\ \end{align} $$

However, I am unsure on how to proceed from here to obtain the critical region/

Edit:

I have been doing further reading and it seems that the critical region can be found by the following, Wilks Theorem $-2\log(\Lambda) \approx \chi^2_{1, 1-\alpha}$

And as such: $$\begin{align} -2\log(\Lambda) &= -2n\log\left(\bar{x}\lambda_0 e^{1-\lambda_0\bar{x}} \right) \\ &= -2n \left(\log(\bar{x})+\log(\lambda_0) + 1-\lambda_0\bar{x} \right) \le \chi^2_{1, 0.94} = 0.2571744 \\ &= \log(\bar{x})+\log(\lambda_0) + 1-\lambda_0\bar{x} \ge -\frac{\chi^2_{1, 0.94}}{2n} \end{align} $$

Link to the reading: likelihood ratio test

Answer 1

[Not enough reputation to comment, so I write in an answer]

Hint: consider the function $g(y)=y\lambda_0e^{1-\lambda_0 y}$ for $y>0$. Take the derivative to help draw the curve. You'll find a region of $y$ such that $g(y)\le k^{1/n}$ by drawing a horizontal line intersecting the curve. After getting a form of the region based on $y$, you can forget about $k$, and determine the boundary (intersection points of the horizontal line and the curve) using the distribution of $y$. Here $y$ is your $\bar{x}$.

By the way, $\lambda_0$ is a specific value of $\lambda$ and it's generally not recommended to write $\hat{\lambda}_0$, because there is nothing unknown to estimate. Instead, I'll write $\hat{\lambda}$ as the value that maximizes your log-likelihood function $\mathbf{L}(\lambda;x)$. The likelihood ratio has nothing to do with the parameter $\lambda$ of the exponential distribution, so I'll suggest using another letter, say $\Lambda := \frac{L(\lambda_0;x)}{L(\hat{\lambda};x)}$, where the denominator is the likelihood function evaluated at $\hat{\lambda}$, not $\mathbf{L}(\hat{\lambda};x)$.