# Maximum likelihood estimator of $\theta$ of the following probability distribution

Let $x_{1}=-2,x_{2}=1,x_{3}=3,x_{4}=-4$ be observed values from the following density function:

$f(x|\theta)=\frac{e^{-x}}{e^{\theta}+e^{-\theta}}$ where the support is $-\theta \leq x\leq\theta$.

The likelihood function will be maximum when the denominator of the density is minimum which happens when the $\theta$ is minimum. Range of all the values are $-\theta\leq x_{1} \leq,....,\leq x_{n}\leq\theta$. One thing is sure that the $\theta$ will take values greater than the maximum value and minus of $\theta$ is always less than the least value. From here, I am not able to understand what will be the maximum likelihood estimator of $\theta$ since combining two conditions are giving me $\theta\geq3$ and $\theta\geq-(-4)$. Then, what will be the maximum likelihood estimator of $\theta$.

## 1 Answer

Clearly θ cannot be 3 because then -4 will not be a possible value, so this leaves θ=4 as the estimate. The intersection of the two conditions θ≥3 and θ≥4 is θ≥4 , which makes 4 the estimate.

• So, I have done it correctly? Is there anything you suggest to modify my approach? Jan 24, 2018 at 4:14
• Your approach is correct, but you did not complete it. You are supposed to find the intersection of the two conditions, which is θ≥−(−4) or θ≥4 , resulting in the estimate θ=4. Jan 24, 2018 at 20:31