# Trouble with a Maximum Likelihood Estimator question

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!

• You forgot the condition that each $X_i$ must be between $0$ and $\theta +1$ in the likelihood. Commented Jan 29, 2020 at 17:07

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 $$\hat\theta = X_{(n)}-1$$.

The density is not $$\frac{2x}{(\theta+1)^2}$$, but is $$\frac{2x}{(\theta+1)^2} \mathbf{1}\{ 0 \le x \le \theta + 1\}$$ - i.e., there's a part in there which limits the support of $$X$$ according to $$\theta$$, and this needs to be included.

So, the log likelihood, given data $$x_1, \dots, x_n$$ is $$\log \mathcal{L}(\theta; x_1,\dots, x_n) = \sum \log(2x_i) - 2n \log(\theta + 1) + \sum \log \mathbf{1}\{0 \le x \le \theta + 1\}.$$

Now, if $$\theta < \max_i(x_i) - 1,$$ then the log likelihood is $$-\infty$$, since one of the indicators is $$0$$.

If $$\theta \ge \max_i(x_i) - 1,$$ then the log likelihood is $$\sum \log(2x_i) - 2n \log(\theta + 1).$$ (Note that its $$-\log(\theta + 1),$$ not $$+\log(\theta + 1)$$. This is clearly decreasing with $$\theta$$ in this domain.

This means that the maximiser of the log-likelihood is precisely $$\max(x_i) - 1$$ - before this we had $$-\infty,$$ and after this we have a smaller value.