(a) Given a random variable $X$ with $\ f_{X}(x) = \frac{1+x\theta}{2} I_{[-1,\ 1]}(x)$, $\theta\in [-1, 1]$. Find the formula for maximum likelihood estimate of $\theta$.
(b) Given $U_1, U_2, U_3$ are i.i.d rvs with $f(u) = \frac{1}{2\delta}I_{[a-b, a+b]}(u)$. Use method of moments to estimate $a$ and $b$ if $u_1 = 1.73, u_2 = -0.77$ and $u_3 = 0.59$.
My attempt: (a) We consider $L(\theta|x) = \frac{x}{2}$. Thus, there are $3$ cases: $x = 0$, $x\in [-1,0)$ and $x\in (0,1]$. For the first case, any $\theta\in [-1,1]$ would be a global max of $L(\theta|x)$ (since $L(\theta|x)=\frac{1}{2}$ in this case), so the mle of $\theta = \theta$.
For the second case, $L(\theta|x)$ has maximum at $\theta = -1$, so mle of $\theta = -I_{[-1,0)}(x)$. For the third case, $L(\theta|x)$ has maximum at $\theta = 1$, the mle of $\theta = I_{(0,1]}(x)$.
(b) My current thought is establishing $L(a, b|u_1, u_2, u_3) = \frac{1}{8\delta^{3}}$ for $u_1, u_2, u_3\in [a-b, a+b]$. But how could we proceed further though?
My question: Could someone please help confirm if my answer to part (a) is correct? Any help on part (b) would really be appreciated.
Suppose we observe a single realization $X = x$ from the aforementioned density in part (a). Then a likelihood function of $\theta$ given this observation is $$\mathcal L(\theta \mid x) \propto f_X(x \mid \theta) = \frac{1 + x \theta}{2} \mathbb 1 (-1 \le x \le 1) \mathbb 1 (-1 \le \theta \le 1).$$ For a fixed $x$, what is $\hat \theta$ such that $\mathcal L(\hat \theta \mid x)$ attains a maximum? This is simple: if $x > 0$, then $\mathcal L$ is monotonically increasing with respect to $\theta$, hence $\hat \theta = 1$. On the other hand, if $x < 0$, then $\mathcal L$ is monotonically decreasing w.r.t $\theta$, hence $\hat \theta = -1$. If $x = 0$, $\mathcal L$ is constant, thus every value in the interval $[-1,1]$ has equal likelihood and the MLE is not uniquely defined. We can summarize this as $$\hat \theta = \begin{cases} \frac{x}{|x|}, & 0 < |x| \le 1, \\ [-1,1], & x = 0. \end{cases}$$
For part (b) of your question, you are asked to use the method of moments, not the likelihood, to estimate the parameters. I can only assume that the relationship between $\delta$ and $a, b$ is $\delta = b$; otherwise $f$ is not a density.
In the method of moments, the raw moments of the distribution are equated with the sample moments until one obtains a system of equations sufficient to determine a unique solution. In this case, we have $$\operatorname{E}[U] = a, \quad \operatorname{E}[U^2] = a^2 + \frac{b^2}{3}.$$ Therefore, $$\tilde a = \frac{1}{n} \sum_{i=1}^n U_i = \frac{U_1 + U_2 + U_3}{3},$$ and we substitute this into the second moment to get $$\tilde a^2 + \frac{\tilde b^2}{3} = \frac{1}{n} \sum_{i=1}^n U_i^2,$$ or $$\tilde b = \left( \frac{3}{n} \sum_{i=1}^n U_i^2 - \frac{3}{n^2}\left(\sum_{i=1}^n U_i\right)^2\right)^{1/2}.$$