Can MLE be indepedent of the observations?

Question

Let the random variable $X$ follow the distribution:

$$ f(x;\theta) = \theta^2(x+1)(1-\theta)^x$$

where $x$ takes values in $[0, \infty)$ and $\theta$ in $[0, 1]$. The likelihood is defined as:

$$\mathcal{L}(x \mid \theta) = p_\theta(x)$$

and for $N$ i.i.d observations:

$$\mathcal{L}(\theta) = \prod_{i=1}^N p_\theta(x_i)$$

Due to the monoticity of the $\ln$ function, we can instead maximize the log-likelihood:

$$\ln(\mathcal{L}(\theta)) = \ln \left( \prod_{i=1}^N p_\theta(x_i) \right) = \sum_{i=1}^{N} \ln p_\theta(x_i) = \sum_{i=1}^{N} x_i\ln \left[\theta^2 (x+1)(1-\theta)\right]$$

The derivative of the likelihood with respect to $\theta$ is:

$$\frac{d\ln(\mathcal{L}(\theta))}{d\theta} = \sum_{i=1}^{N} x_i \frac{d}{d\theta} \ln\left[\theta^2 (x+1)(1-\theta)\right] = \sum_{i=1}^{N} x_i \frac{d}{d\theta} \left[ \ln(\theta^2) + \ln(x_i + 1) + \ln(1-\theta)\right]$$

Taking the derivatives of the terms inside the brackets we obtain:

$$\sum_{i=1}^{N} x_i \left(\frac{2}{\theta} - \frac{1}{1-\theta}\right)$$

and setting this to zero leads to:

$$\sum_{i=1}^{N} x_i \frac{2}{\theta} = \sum_{i=1}^{N} x_i \frac{1}{1-\theta} \Longleftrightarrow \frac{2}{\theta}\sum_{i=1}^n x_i = \frac{1}{1-\theta} \sum_{i=1}^n x_i \Longleftrightarrow \frac{2}{\theta} = \frac{1}{1-\theta}$$

Solving the last equations leads to $\hat{\theta}=\frac{2}{3}$.

Is it possible for the MLE to be independent of the observations ($x_i$ are not involved in calculation of $\hat{\theta}$)? If yes, is there something in the form of $f(x;\theta)$ that indicates that?

Answer 1

First, it should be noted that your computation is incorrect: you cannot distribute the $x_i$ across all of the terms. The log-likelihood function should instead look like the following: $$\sum_{i=1}^n 2 \log \theta + \log(x_i+1) + x_i\log(1-\theta)$$ so that the derivative of the log-likelihood function is $$\sum_{i=1}^n \frac2{\theta} - \frac{x_i}{1-\theta} = 0$$ or solving, $$\hat\theta = f(\bar x)$$ where $f$ is the function defined such that $f^{-1}(\theta) = \frac{2(1-\theta)}{\theta}$ and $\bar x$ is the sample mean. This does indeed depend on the data.

More broadly speaking, let's take a step back and ask conceptually what it would mean for the MLE to not depend on the data. Intuitively, this amounts to claiming that no matter what the data says, our best guess of the true parameter is the same. Does this make sense? If this was true, what would even be the point of looking at the data?