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I'm preparing for my exam and I came across this question:

Let $X_1, X_2, ...,X_n$ be iid with PDF $f(x)=\frac{2x}{\theta^2}$ for $0 \leq x \leq \theta.$ Find the MLE of $\theta$

So this is what I did:

$L(\theta)= \prod\frac{2x_i}{\theta^2}I(0 \leq x \leq \theta)$

$= (2n \bar{x})(\frac{1}{\theta^{2n}})I(0 \leq min (x_i)\leq max(x_i) \leq \theta) $

$=(2n \bar{x})(\frac{1}{\theta^{2n}})I(0 \leq X_{(1)}\leq X_{(n)} \leq \theta)$

In order to find MLE, I need to maximize $L(\theta)$, by choosing a small $\theta$. However, $\theta$ can't be smaller than $ X_n$. Thus I conclude that:

$\hat{\theta}= max(x_i)$

Is my calculation and conclusion correct?

dembrownies
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    If you add the reasoning: the likelihood is higher for small $\theta$ but it can't be smaller than ... Then jt is correct – Stan Tendijck Feb 23 '19 at 21:41
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    https://math.stackexchange.com/questions/3113045/maximum-likelihood-estimator-and-cumulative-distribution-for-an-i-i-d-distributi/3113109#3113109 – Robert W. Feb 23 '19 at 23:18

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