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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

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1 Answer 1

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[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 $\mathbf{L}(\hat{\lambda};x)$.

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  • $\begingroup$ Thanks for the hint, I'll look into it some further. I have edited my post from a reading, not sure if you're familiar with it? I might have got it totally wrong though. Also, thanks for the tips on notation also. I'll make an update later today on your hint and will upload it to my post for progress. $\endgroup$ Mar 8 at 16:24
  • $\begingroup$ I have taking the first derivative and I get $e^{1-\lambda_0 y}(2-\lambda_0 y)$ by setting this equal to $0$ I find that I get $\bar{x} = y = \frac{2}{\lambda_0}$ how do I proceed from here? $\endgroup$ Mar 8 at 16:36
  • $\begingroup$ @dollarbill Are you familiar with drawing the function curve with the help of derivatives? You got the first derivative, which contains some important information about the function. The first part, $e^{1-\lambda_0y}$, is always positive, so the sign of $g'$ is solely determined by the second part, $2-\lambda_0 y$, where $\lambda_0>0$. Then you can see that $g'(y)>0$ for all $0<y<2/\lambda_0$, and $g'(y)\le 0$ otherwise. What do these tell you about the shape of $g$? $\endgroup$
    – Min
    Mar 9 at 7:01
  • $\begingroup$ @dollarbill About your update. That is the asymptotic distribution of $-2\log\Lambda$, which approximately holds when you have a finite sample. In this case, actually, you can derive an exact distribution, which is true for any finite sample size. $\endgroup$
    – Min
    Mar 9 at 7:04
  • $\begingroup$ @dollarbill In addition, you may confuse a distribution itself with critical values under a distribution. $-2\log\Lambda\overset{approx.}{\sim} \chi^2_1$, where $\sim$ means the random variable on the left side follows a distribution whose name is indicated on the right, and the critical region is the set of all viable values of $\bar{x}$ such that $-2\log\Lambda\ge \chi^2_{1,\alpha}$. $\endgroup$
    – Min
    Mar 9 at 7:15

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