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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!

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

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

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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.

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