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 $\theta$.


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.

| cite | improve this answer | |
  • $\begingroup$ So, I have done it correctly? Is there anything you suggest to modify my approach? $\endgroup$ – user8125394 Jan 24 '18 at 4:14
  • 1
    $\begingroup$ 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. $\endgroup$ – Zahava Kor Jan 24 '18 at 20:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.