I am trying to find the MLE for an Unknown distribution. I have most of the steps but I am confused on how to simplify the $\prod$ of the denominator.
The given function is $f(x;\beta) = \frac{\beta}{(1+x)^{\beta+1}}$.
To find the MLE we first write this as $$f(x_1,x_2,x_3,\ldots,x_n;\beta) = \prod_{i=1}^n\Big(\frac{\beta}{(1+x)^{\beta+1}}\Big)$$ which simplifies to
$$f(x_1,x_2,x_3,\ldots,x_n;\beta) = \Big(\frac{\beta^n}{(1+x_1)^{\beta+1}\cdot(1+x_2)^{\beta+1}\cdot(1+x_3)^{\beta+1}\cdots(1+x_n)^{\beta+1}}\Big)$$
Here I am having trouble simplifying the denominator into easier to a more approachable form.
I tried to continue by taking the log which equals
$$L(\beta) = n\log(\beta)-(\beta+1)(\log((1+x_1)\cdot(1+x_2)\cdot(1+x_3)\cdots(1+x_n)))$$
However I am stuck unable to simplify here too since I can't just sum the values inside the log. How do I simplify this problem so I can solve it?
EDIT: I just realized that every value in the second log will be a constant, does this mean I can proceed with the derivative since the log will go to 0 regardless.
EDIT #2: Since the derivative of a sum is the sum of each i's derivative does that mean $\frac{L}{L(\beta)} = \frac{n}{\beta} – \frac{1}{\sum_{i=1}^n(x_i+1)}$
Best Answer
I copy JimB's solution while turning it into a complete answer. Your likelihood function is given by $$ L(\beta; x_1, \dots, x_n) = \prod_{i=1}^n \left[ \frac{\beta}{(1+x_i)^{\beta+1}}\right] = \beta^n \prod_{i=1}^n \left[\frac{1}{(1+x_i)^{\beta+1}}\right]. $$ The loglikelihood is given as the natural logarith of the likelihood $l(\beta; x_1, \dots, x_n) = \ln L(\beta, x_1, \dots, x_n)$. Remember now that $\ln (a b) = \ln a + \ln b$, $\ln (a^n) = n \ln a$ and that $\ln [1/a] = -\ln a$. $$ \begin{align} l(\beta; x_1, \dots, x_n) &= n \ln \beta + \sum_{i=1}^n \ln \left[\frac{1}{(1+x_i)^{\beta+1}}\right] \\ & = n\ln \beta + \sum_{i=1}^n - \ln [(1+x_i)^{\beta + 1}] \\ & = n\ln \beta + \sum_{i=1}^n - (\beta + 1)\ln [1+x_i] \end{align} $$ When finding the MLE we want to maximize $L$ and because the $\ln$ is a monotone transform the maximum of $l$ will be the maximum of $L$. We thus derivate wrt to $\beta$ $$ \frac{\partial l}{\partial \beta}(\beta; x_1, \dots, x_n) = \frac{n}{\beta} - \sum_{i=1}^n \ln [1+x_i] $$ and find the critical points for which $\frac{\partial l}{\partial \beta} = 0$, $$ \frac{n}{\beta} - \sum_{i=1}^n \ln [1+x_i] = 0 \, \Longrightarrow \, \hat{\beta}_\text{MLE} = \frac{n}{\sum_{i=1}^n \ln [1+x_i]} $$ Now here we can check that $\hat{\beta}_\text{MLE}$ gives the max and not the min of the loglikelihood as $$ \frac{\partial^2 l}{\partial \beta^2} = -\frac{n}{\beta^2} < 0 \quad \forall \beta $$ the double derivative is negative.