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A sample of size two is taken from the distribution $ f(x;\alpha)=\frac{2}{\alpha^2}(\alpha-x)I_{(0,\alpha)}(x)$. We need to find the maximum likelihood estimator for $\alpha$.

The likelihood function $$L(x_1,x_2;\alpha)=\frac{2}{\alpha^2}(\alpha-x_1)\frac{2}{\alpha^2}(\alpha-x_2)I_{(0,\alpha)}(x_1,x_1)$$

i.e. $$L(x_1,x_2;\alpha)=\frac{4}{\alpha^4}(\alpha-x_1)(\alpha-x_2)I_{(0,y_2)}(y_1)I_{(y_1,\alpha)}(y_2)$$

where $y_1=\min(x_1,x_2)$ and $y_2=\max(x_1,x_2)$.

Since we have a higher power of $\alpha$ in the denominator, first I thought of taking $\hat\alpha=y_1$ but that would make $L=0$, similarly for $\hat \alpha=y_2$.

Considering another method - Maximizing the numerator and minimizing the denominator. Maximizing numerator gives $\hat \alpha=\bar{x}$ and minimizing denominator gives $\hat \alpha=y_1$.

What should we do here?

EDIT: I tried another way, first we'll be able to maximize the likelihood function if the indicator functions are $1$, taking their values as $1$ and maximizing the function gives $$\hat \alpha=\frac{3\sum x_i}{4} \pm \frac{\sqrt{9\sum x_i^2-14x_1x_2}}{4}$$

I also plotted $L(x_1,x_2;\alpha)=\frac{4}{\alpha^4}(\alpha-x_1)(\alpha-x_2)I_{(0,y_2)}(y_1)I_{(y_1,\alpha)}(y_2)$ for fixed values $(x_1,x_2)$ and observed that at $\hat \alpha=\frac{3\sum x_i}{4} + \frac{\sqrt{9\sum x_i^2-14x_1x_2}}{4}$ the function indeed attains a maximum, but what about the other root?

Also the question asks whether this statistic is sufficient or not. How do I tackle that one?

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  • $\begingroup$ Because $\alpha$ governs the support, the location, and the scale, I would expect the MLE to be some function of the mean, variance, and minimum. For some $x$'s it will be the min, and for other $x$'s some function of the moments. You will need to split to different regions of $x$. $\endgroup$ – JohnRos May 14 '18 at 10:35
  • $\begingroup$ Same question with a particular sample: stats.stackexchange.com/q/317874/119261. $\endgroup$ – StubbornAtom May 7 at 20:01
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The MLE does exist. You have calculated the likelihood directly, it is positive for $\alpha > \max(x_1, x_2)$. Then pass to the loglikelihood, calculate its derivative and set it to zero. I then find the same root as you. The other root is probably a false solution. You could plot the loglikelihood as function of alpha, and then also plot the two roots to see.

As for sufficiency, can you factor the likelihood function, and then use the factorization theorem?

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  • $\begingroup$ The factorization yields that $(\sum x_i,x_1x_2,y_1,y_n)$ are jointly sufficient, and this should imply that the $\hat \alpha$ obtained above can't be sufficient ? $\endgroup$ – User9523 May 14 '18 at 11:21
  • $\begingroup$ That sounds correct, yes. $\endgroup$ – kjetil b halvorsen May 14 '18 at 11:36

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