Maximum likelihood estimator of a uniform distribution $U(\theta,k\theta)$?

Question

So pretty much what's the maximum likelihood estimator of a uniform distribution $U(\theta,k\theta)$ with $k>1$? I have tried a lot of methods but I can't get the answer right. With known k.

Answer 1

Assuming the parameter space for $\theta$ is $(0,\infty)$, let $L(\theta; x_1,\dots, x_n)$ be the likelihood function of a sample of $n$ $\mathcal{U}(\theta, k \theta)$ IIDRVs. Then, $$L(\theta; x_1,\dotsc, x_n)=f(x_1,\dotsc, x_n; \theta)=f(x_1; \theta)\cdot \dotso \cdot f(x_n; \theta)$$ $$=\left(\frac{1}{\theta(k-1)}\right)^n \mathbb{1}_{\theta \leq x_1 \leq k \theta}(x_1) \cdot \dotso \cdot \mathbb{1}_{\theta \leq x_n \leq k \theta}(x_n)$$ $$=\left(\frac{1}{\theta(k-1)}\right)^n \mathbb{1}_{x_{(1)} \geq \theta} \mathbb{1}_{x_{(n)} \leq k\theta},$$ so $L(\theta;\cdot)$ is $\neq 0$ only on $[x_{(n)}/k, x_{(1)}]$, where $x_{(1)}$ is the minimum of $x_1,\dotsc, x_n$ and $x_{(n)}$ is the maximum.

We just need to show that $L$ is decreasing in $\theta$ on $[x_{(n)}/k, x_{(1)}],$ and then it follows $\hat{\theta}=x_{(n)}/k$ is the MLE.