Suppose that $a(\cdot)$ and $b(\cdot)$ are two non-negative functions such that $$f(x;\theta)=a(\theta)b(x)$$ is a probability density function for each $\theta > 0$. Find the maximum likelihood estimator of $\theta$.
My try: Our likelihood function is given by $$L(\theta) = \prod_{i=1}^n a(\theta)b(x_i) = a(\theta)^nb(x)^n$$ The log likelihood function is given by $$\ln L(\theta) = n\ln a(\theta) + n \ln b(x)$$ Equating it to zero we get $$\ln a (\theta) = - \ln b(x)$$ which obviously leads to nowhere.
Moreover, the question itself seems weird to me. I am used to the form of "Given a random sample $X_1,...,X_n$ of size $n$ (...)", since this is missing now, does this imply that I cannot use the usual method I demonstrated above?
Lastly, if you want to, could you check the exercise below for errors?
Let $X_1,...,X_n$ denote a random sample from $$f(x;\theta) = f_\theta (x) = \theta f_1(x) + (1-\theta)f_0 (x)$$ where $0 \leq \theta \leq 1$ and $f_0(\cdot)$ and $f_1(\cdot)$ are known densities, estimate $\theta$ by the method of moments.
Answer: First, we need to write $E[X]$ in a better form: \begin{align*} E[x] &= \int_{-\infty}^\infty x\cdot f(x;\theta)dx = \int_{-\infty}^\infty x(\theta f_1(x) + (1-\theta)f_0 (x))dx \\ &= \theta \int_{-\infty}^\infty x\cdot f_1 (x)dx + (1-\theta)\int_{-\infty}^\infty x\cdot f_0 (x)dx \\ &= \theta \int x(f_1-f_0)dx + \int x f_0 dx \\ &=\theta \left(E_1\left[x\right] - E_0\left[x\right]\right) + E_0\left[x\right] \end{align*} Equating this to the first sample moment ($m_1'$) we get: \begin{align*} m_1'= \theta \left(E_1\left[x\right] - E_0\left[x\right]\right) + E_0\left[x\right] \end{align*} which is equivalent to \begin{align*} \theta = \dfrac{m_1' - E_0[x]}{E_1[x] - E_0[x]} \end{align*} Hence, our method of moments estimator for $\theta$ is given by: $$\hat{\theta} = \dfrac{m_1' - E_0[x]}{E_1[x] - E_0[x]}$$
Both are questions from "Introduction to the theory of statistics" by Mood, Graybill and Boes.
Thanks in advance!
Another example where including the (crucial) indicator functions in the densities simplifies everything... Here the PDF is $$ f(x;\theta)=a(\theta)b(x)\mathbf 1_{[0,\theta]}(x), $$ where $$ \frac1{a(\theta)}=\int_0^\theta b(x)\,\mathrm dx, $$ hence the likelihood of a sample $\mathbf x=(x_k)$ is $$ L(\mathbf x,\theta)=\prod_kf(x_k;\theta)=a(\theta)^n\,\mathbf 1_{\theta\geqslant m(\mathbf x)}\,\prod_kb(x_k), $$ where $$ m(\mathbf x)=\max_kx_k. $$ The last product does not depend on $\theta$ hence one can forget it. The indicator function shows that $L(\mathbf x,\theta)$ can be nonzero only when $\theta\geqslant m(\mathbf x)$. And $\theta\mapsto a(\theta)$ is nonincreasing hence one looks for $\theta$ as small as possible. Finally, $L(\mathbf x,\theta)$ is maximal when $\theta=\hat\theta(\mathbf x)$ with $$ \hat\theta(\mathbf x)=m(\mathbf x). $$