Let $X_1, . . . , X_n$ be independent and identically distributed random variables each with probability density function,
$$f(x)=\frac{2x}{(\theta +1)^2} $$
where $\theta$ is an unknown parameter and where $0\leq x \leq \theta+1$.
We have observations of $x_1, . . . , x_n$ how can I show that the maximum likelihood estimator of $\theta$ is given by $\theta=max(x_1, . . . , x_n)-1$
First I logged the PDF to get:
$$\ln(f(x))=\ln(2x)+2(\ln(\theta+1))$$
Then differentiated w.r.t $\theta$:
$$\ln(f(x))'=\frac{2}{\theta+1}$$
But then if I set equal to zero I cannot solve for $\theta$ or even get a max in there at all? Can anyone give me some pointers? Thank you!
Best Answer
I see no need to use log likelihood here. The function itself is decreasing in $\theta$. $$ \mathcal{L}(\theta; X_1,\dots, X_n) = f(X_1)\cdot\ldots\cdot f(X_n) $$ $$= \begin{cases}\frac{2X_1}{(\theta+1)^2}, & 0\leq X_1\leq \theta+1 \cr 0, & \text{else}\end{cases}\;\times\;\ldots\;\times\; \begin{cases}\frac{2X_n}{(\theta+1)^2}, & 0\leq X_n\leq \theta+1 \cr 0, & \text{else}\end{cases} $$ $$ =\begin{cases}\frac{2^n\prod\limits_{i=1}^n X_i}{(\theta+1)^{2n}}, & 0\leq X_1,\ldots,X_n\leq \theta+1 \cr 0, & \text{else}\end{cases} $$ (here $X_{(n)}=\max(X_1,\ldots,X_n)$) $$ =\begin{cases}\frac{2^n\prod\limits_{i=1}^n X_i}{(\theta+1)^{2n}}, & X_{(n)}\leq \theta+1 \cr 0, & \text{else}\end{cases} $$ $$ =\begin{cases}\frac{2^n\prod\limits_{i=1}^n X_i}{(\theta+1)^{2n}}, & \theta\geq X_{(n)}-1 \cr 0, & \theta< X_{(n)}-1 \end{cases} $$
Since $\frac{2^n\prod\limits_{i=1}^n X_i}{(\theta+1)^{2n}}$ decrease as $\theta$ increase, the highest value of $\mathcal{L}(\theta; X_1,\dots, X_n)$ is attained at the smallest value of $\theta$ satisfying the inequality $\theta\geq X_{(n)}-1$. So, MLE is $\hat\theta = X_{(n)}-1$.