# Maximum likelihood estimation for $\alpha$ with population pdf $f(x;\alpha)=\frac{2}{\alpha^2}(\alpha-x)I_{(0,\alpha)}(x)$

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?

• 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 $x$. – JohnRos May 14 '18 at 10:35
• Same question with a particular sample: stats.stackexchange.com/q/317874/119261. – StubbornAtom May 7 '20 at 20:01

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.
• 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 ? – User9523 May 14 '18 at 11:21