Show existence for monotone likelihood ratio (MLR) in X

Question

I am trying to show that if we have an observation with density $$f_\theta (x)=\frac{1}{2}\exp(-|x-\theta|), \ \ x\in\mathbb{R}$$ where $\theta \in \mathbb{R}$ is an unknown parameter, then the model has monotone likelihood ratio (MLR) in X.

I have tried getting the density on the form of an exponential family to show that the natural parameter is strictly increasing in $\theta$, since the result then follows. But I have not really suceeded. I have only come to the point: $$f_\theta (x)=\frac{1}{2}\exp(-\sqrt{(x^2+\theta^2-2x\theta)})$$

How do I get this into the right form?

Answer 1

Suppose you consider two different values of $\theta$, say $\theta_1 < \theta_2$, and then consider the ratio $\dfrac{f_{\theta_2} (x)}{f_{\theta_1} (x)}$:

• for $x \le \theta_1$ you get $\frac{\frac{1}{2}\exp(x-\theta_2)}{\frac{1}{2}\exp(x-\theta_1)} = \exp(\theta_1-\theta_2)$, which is a constant between $0$ and $1$
• for $x \ge \theta_2$ you get $\dfrac{\frac{1}{2}\exp(\theta_2-x)}{\frac{1}{2}\exp(\theta_1-x)} = \exp(\theta_2-\theta_1)$, which is a constant above $1$
• for $\theta_1 \le x \le \theta_2$ you get $\dfrac{\frac{1}{2}\exp(x-\theta_2)}{\frac{1}{2}\exp(\theta_1-x)} = \exp(2x-\theta_2-\theta_1)$, which is a strictly increasing function of $x$, with values increasing from the lower constant to the higher constant.

So the ratio overall is a weakly increasing function of $x$, but constant over wide intervals, and this may for example prevent you from arguing there is a unique optimal critical region for some hypothesis testing.