We have independent uniform random variables $U_1, U_2, U_3, U_4$ where $U_k ~ [-k\theta,k\theta]$ so these random variables have different supports. Using the observed values, $u_1 = -3.8, u_2 = 8.2, u_3 = -12.6, u_4 = 15.6$, give the maximum likelihood estimate of $\theta$.
I started by determining the likelihood functions for each of the U's. So for U1, we have the support is $[-\theta,\theta]$, and the likelihood function is $$\frac{1}{2\theta^n}$$ Using this same process, the likelihood for U4 is $$\frac{1}{8\theta^n}$$ From these, I can understand that the likelihood function is decreasing, and I have seen in notes that for a uniform random variable with support$(0,\theta)$,$\theta_{MME} = max(X_1,..,X_n)$, but I am having trouble determining how to apply that to this problem.
Here, for every $k$, the PDF $f_k$ of $U_k$ is $$f_k(u)=\frac1{2k\theta}\mathbf 1_{|u|\leqslant k\theta}$$ hence the likelihood of a sample $(U_k)_{1\leqslant k\leqslant n}=(u_k)_{1\leqslant k\leqslant n}$ is $$L(\theta\mid u_1,\ldots,u_n)=\prod_{k=1}^nf_k(u_k)=\prod_{k=1}^n\left(\frac1{2k\theta}\mathbf 1_{|u_k|\leqslant k\theta}\right)=\frac1{2^nn!\theta^n}\mathbf 1_{\theta\geqslant m(u_1,\ldots,u_n)}$$ with $$m(u_1,\ldots,u_n)=\max_{1\leqslant k\leqslant n}\frac{|u_k|}k$$ Thus, $L(\theta\mid u_1,\ldots,u_n)$ is obviously maximal at