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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!

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    $\begingroup$ $\prod_{i=1}^n b(x_i)=b(x)^n$ doesn't make sense. $\endgroup$ – Stefan Hansen Jun 17 '14 at 14:35
  • $\begingroup$ @StefanHansen When looking back, you're obviously right. What would you suggest how I should continue? $\endgroup$ – Nigel Overmars Jun 17 '14 at 14:39
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    $\begingroup$ Continue by keeping $\prod_{i=1}^n b(x_i)$ as it is. Moreover, you want to solve $l'(\theta)=0$, not $l(\theta)=0$, where $l$ is the log-likelihood. $\endgroup$ – Stefan Hansen Jun 17 '14 at 14:49
  • $\begingroup$ @StefanHansen Yes, I also spotted that mistake. It has been some time since I had to do this. Continuing on your advice: $\ln L(\theta) = n \ln a (\theta) + \sum_i \ln b(x_i)$. Differentiate wrt to $\theta$ and set equal to zero to get: $\dfrac{n\cdot a'(\theta)}{a(\theta)} = 0$ since $\sum_i \ln b(x_i)$ is just a constant. Which gives $a'(\theta)=0$. Do you think this is the answer? $\endgroup$ – Nigel Overmars Jun 17 '14 at 14:54
  • $\begingroup$ This assumes that $a$ is differentiable. Are you sure that you haven't incorrectly stated the problem? Exercise 39 in the book you mention has $f(x;\theta)=a(\theta)b(x)I_{(0,\theta)}(x)$. $\endgroup$ – Stefan Hansen Jun 17 '14 at 15:13
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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 $$ \hat\theta(\mathbf x)=m(\mathbf x). $$

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  • $\begingroup$ Could you explain how you got the $1/a(\theta) = \int_0^\theta b(x) dx$ equality? Or (I think equivalently), why including the indicator function is necessary $\endgroup$ – Nigel Overmars Jun 17 '14 at 17:38
  • $\begingroup$ Because for every $\theta$, the function $f(\ ;\theta)$ must integrate to $1$ and $$\int_\mathbb Rf(x;\theta)\mathrm dx=a(\theta)\int_0^\theta b(x)\mathrm dx.$$ $\endgroup$ – Did Jun 17 '14 at 18:16

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