Likelihood test ratio for random sample with p.d.f. $f(x\mid \theta) = e^{-(x-\theta)}$

Question

I don't know how to find the max region of the denominator since doing the usual gives me a constant as the derivate of the likelihood function of theta. The problem is as follows

Let $X_1 , \ldots , X_n$ be a random sample with $$f(x\mid \theta) = e^{-(x-\theta)}$$ where $x \in [\theta,\infty)$

find $\lambda(x)$ where $H_0: \theta = \theta_0$ and $H_1: \theta > \theta_0$

so I know that the numerator since it's just a point, is equal to the maximum likelihood function on $\theta_0$

but if I try to maximize the denominator since the Likelihood function is $$ \prod^n_{i=1} e^{-(x_i-\theta)} = e^{n\theta-n \bar{x} } $$ but I can't maximize the likelihood from its derivative in the region $[\theta_0 , \infty]$

Answer 1

The main error you did is that the likelihood you stated is WRONG. You did not take into consideration that the suppport is depending on the parameter.

Let's see how to proceed in this case:

$$f(x|\theta)=e^{\theta}e^{-x}$$

$x\geq\theta$

$\theta \in \mathbb{R}$

This implies that $min(\mathbf{X})=X_{(1)}\geq\theta$

---

The likelihood is the following

$$L(\theta)=e^{n \theta}e^{-\Sigma_i X_i}\cdot\mathbb{1}_{[\theta;\infty)}(x_{(1)})$$

---

Now if we take two points $\theta'<\theta''$ we get the following likelihood ratio

$$\frac{L(\theta';\mathbf{x})}{L(\theta'';\mathbf{x})}=\frac{e^{n\theta'}\mathbb{1}_{[\theta';\infty)}(x_{(1)})}{e^{n\theta''}\mathbb{1}_{[\theta'';\infty)}(x_{(1)})}=\begin{cases} +\infty, & \text{if $x_{(1)}<\theta''$ } \\ e^{n(\theta'-\theta'')}, & \text{if $x_{(1)}\geq\theta''$ } \end{cases}$$

Thus in this case we have a "Monotone Likelihood Ratio" that is a "Non increasing LR" with respect to the Statistic $T=X_{(1)}$ and so we can apply the following theorem

Answer 2

For any real $\theta$, the likelihood function given the sample $\boldsymbol x=(x_1,\ldots,x_n)$ is

\begin{align} L(\theta\mid \boldsymbol x)&= \begin{cases}e^{-n(\bar x -\theta)}&, \text{ if }x_1,\ldots,x_n\ge \theta \\ 0 &, \text{ otherwise }\end{cases} \\&=\begin{cases}e^{-n(\bar x -\theta)}&, \text{ if }x_{(1)}\ge \theta \\ 0 &, \text{ otherwise }\end{cases}\,, \end{align}

where $x_{(1)}=\min\{x_1,\ldots,x_n\}$.

This is a non-decreasing function of $\theta$, so its maximum is reached for the maximum possible value of $\theta$, namely $\theta=x_{(1)}$.

Now suppose $\theta$ is restricted to $[\theta_0,\infty)$. If $x_{(1)}\ge \theta_0$, then $L(\theta\mid \boldsymbol x)$ is still maximized at $\theta=x_{(1)}$. But if $x_{(1)}<\theta_0$, then because of the nature of the likelihood, $L(\theta\mid \boldsymbol x)$ is maximized at the boundary point $\theta=\theta_0$. A sketch of the likelihood function makes this point clear.

So the restricted MLE of $\theta$ in this case is $$\hat\theta=\begin{cases}x_{(1)}&,\text{ if }x_{(1)}\ge \theta_0 \\ \theta_0 &,\text { if }x_{(1)}< \theta_0\end{cases}$$

The likelihood ratio test criterion is therefore $$\Lambda(\boldsymbol x)=\frac{\sup_{\theta=\theta_0}L(\theta\mid \boldsymbol x)}{\sup_{\theta\ge \theta_0}L(\theta\mid \boldsymbol x)}=\frac{L(\theta_0\mid \boldsymbol x)}{L(\hat\theta\mid \boldsymbol x)}$$

Simplify this and reject $H_0$ for small values of $\Lambda$ to obtain the likelihood ratio test. This test would coincide with the uniformly most powerful test derived here for the same problem.