ML estimation with given samples

Question

Let $X_i,...,X_n$ be a random independent sample from a distribution with pdf $$ f(x;\theta)= (\theta + 1)x^{-(\theta+2)},$$ where $x>0$, and $\theta > 0$. What is the ML estimate for the parameter $\theta$ given the samples $0.5,1,2,3,$?

SOLUTION

I calculated the log likelihood function:

$$ l(\theta) = n\log(\theta+1)- (\theta +2)\sum \log(x_i)$$

then took the first derivative of $l(\theta)$ and solved for $\theta$, which yields

$$ \hat{\theta}_{MLE}=\frac{n}{\sum \log(X_i)}-1$$

However, I have not used the information that the samples $0.5,1,2,3,$ were given. Am I supposed to use that info, or is it there just for "confusion"?

Answer 1

The $MLE$ estimate is a function of the data. In particular, we can write $\hat{\theta}=\hat{\theta}(X_1,X_2\dotsc,X_n)$. The question asks for $$ \hat{\theta}(0.5,1,2,3)=\frac{4}{\log0.5+\log 1+\log 2+\log 3}-1. $$ which is the MLE for the given sample.

Answer 2

Yes so you have estimated the MLE as a function of your data $$\hat{\theta}_{MLE}=\frac{n}{\sum \log(X_i)}-1$$ Now for your $n = 4$ datapoints, $X_1 = 0.5$, $X_2 = 1$, $X_3 = 2$ and $X_4 = 3$ you will get $$\hat{\theta}_{MLE}=\frac{4}{\log X_1 + \log X_2 + \log X_3 + \log X_4}-1$$ you get $$\hat{\theta}_{MLE}=\frac{4}{\log 0.5 + \log 1 + \log 2 + \log 3}-1$$