# Can MLE be indepedent of the observations?

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?

• Your calculation of the likelihood equation is incorrect. It is easiest to work with a single observation when calculating the MLE. Oct 17, 2022 at 18:24
• I think the calculation with logarithm has a mistake: $\ln(p_\theta(x)) = 2\ln \theta + \ln(x+1) + x \ln(1-\theta)$ Oct 17, 2022 at 18:29

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.